ML Aggarwal Class 6 Maths Solutions Chapter 04 Playing with Numbers

 

Exercise 4.1

 

Question 1. Fill in the blanks:
(i) A number having exactly two factors is called a ....
(ii) A number having more than two factors is called a ...
(iii) 1 is neither ..... nor ........
(iv) The smallest prime number is ....
(v) The smallest odd prime number is ....
(vi) The smallest composite number is .....
(vii) The smallest odd composite number is ...
(viii) All prime numbers (except 2) are ....
Answer: (i) A number having exactly two factors (1 and itself) is known as a prime number.
(ii) A number having more than two factors is known as a composite number.
(iii) 1 is neither prime nor composite.
(iv) The smallest prime number is 2.
(v) The smallest odd prime number is 3.
(vi) The smallest composite number is 4.
(vii) The smallest odd composite number is 9.
(viii) All prime numbers (except 2) are odd numbers.
In simple words: A prime number has just two factors - 1 and the number itself. A composite number has more than two factors. The number 1 fits neither group. The smallest prime is 2, and the smallest odd prime is 3.

Exam Tip: Remember that 1 is neither prime nor composite - this is a key definition to know. All primes except 2 are odd because 2 is the only even prime number.

 

Question 2. State whether the following statements are true (T) or false (F):
(i) The sum of three odd numbers is an even number.
(ii) The sum of two odd numbers and one even number is an even number.
(iii) The product of two even numbers is always an even number.
(iv) The product of three odd numbers is an odd number.
(v) If an even number is divided by 2, the quotient is always an odd number.
(vi) All prime numbers are odd.
(vii) All even numbers are composite.
(viii) Prime numbers do not have any factors.
(ix) Two consecutive numbers cannot be both prime.
(x) Two prime numbers are always co-prime numbers.
Answer:
(i) False.

Reason: When you add three odd numbers together, the result is always odd. For example, 1 + 3 + 5 = 9 (odd).

(ii) True.

Reason: Two odd numbers add up to give an even number. Then adding an even number to an even number gives an even result. For example, 3 + 5 + 4 = 12 (even).

(iii) True.

Reason: When you multiply any two even numbers, the product will always be even because it contains 2 as a factor. For example, 4 × 6 = 24 (even).

(iv) True.

Reason: The product of three odd numbers is always odd. For example, 3 × 5 × 7 = 105 (odd).

(v) False.

Reason: When an even number is divided by 2, the quotient can be either even or odd. For example, 8 ÷ 2 = 4 (even).

(vi) False.

Reason: The number 2 is a prime number, yet it is even.

(vii) False.

Reason: The number 2 is even, but it is prime, not composite.

(viii) False.

Reason: Prime numbers have exactly two factors - the number 1 and the number itself.

(ix) False.

Reason: The numbers 2 and 3 are two consecutive natural numbers, and both are prime.

(x) True.

Reason: Two prime numbers share no common factor except 1, which makes them always co-prime.
In simple words: Learn the patterns: odd + odd + odd = odd; odd + odd + even = even. Two primes are always co-prime because they share only the factor 1. The number 2 is the only prime that is even.

Exam Tip: For true/false questions, always give a reason or example - this shows clear understanding. Remember that 2 is a special prime because it is the only even prime.

 

Question 3. Write all the factors of the following natural numbers:
(i) 68
(ii) 27
(iii) 210
Answer:
(i) 68

68 = 1 × 68

68 = 2 × 34

68 = 4 × 17

Therefore, the factors of 68 are 1, 2, 4, 17, 34 and 68.

(ii) 27

27 = 1 × 27

27 = 3 × 9

Therefore, the factors of 27 are 1, 3, 9 and 27.

(iii) 210

210 = 1 × 210

210 = 2 × 105

210 = 3 × 70

210 = 5 × 42

210 = 6 × 35

210 = 7 × 30

210 = 10 × 21

210 = 14 × 15

Therefore, the factors of 210 are 1, 2, 3, 5, 6, 7, 10, 14, 15, 21, 30, 35, 42, 70, 105 and 210.
In simple words: To find all factors of a number, find every pair of numbers that multiply to give that number. Write both numbers in each pair - do not skip any.

Exam Tip: Always write factors in ascending order and include both 1 and the number itself. Check your work by making sure all pairs multiply correctly to the original number.

 

Question 4. Write first six multiples of the following natural numbers:
(i) 3
(ii) 5
(iii) 12
Answer:
(i) The first six multiples of 3 are obtained by multiplying 3 by 1, 2, 3, 4, 5, and 6:

3 × 1 = 3, 3 × 2 = 6, 3 × 3 = 9, 3 × 4 = 12, 3 × 5 = 15, 3 × 6 = 18

Therefore, the first six multiples of 3 are 3, 6, 9, 12, 15 and 18.

(ii) The first six multiples of 5 are obtained by multiplying 5 by 1, 2, 3, 4, 5, and 6:

5 × 1 = 5, 5 × 2 = 10, 5 × 3 = 15, 5 × 4 = 20, 5 × 5 = 25, 5 × 6 = 30

Therefore, the first six multiples of 5 are 5, 10, 15, 20, 25 and 30.

(iii) The first six multiples of 12 are obtained by multiplying 12 by 1, 2, 3, 4, 5, and 6:

12 × 1 = 12, 12 × 2 = 24, 12 × 3 = 36, 12 × 4 = 48, 12 × 5 = 60, 12 × 6 = 72

Therefore, the first six multiples of 12 are 12, 24, 36, 48, 60 and 72.
In simple words: Multiples are made by multiplying a number by 1, 2, 3, 4, and so on. The first multiple is the number itself times 1.

Exam Tip: Multiples are always larger than or equal to the original number. The smallest multiple of any number is the number itself. Never skip or double-count any multiple.

 

Question 5. Write:
(i) the seventh multiple of 36
(ii) the even multiples of 9 less than 100
(iii) the odd multiples of 17 less than 150
(iv) the first 3-digit even multiple of 7
(v) the multiples of 6 between 40 and 80
(vi) the greatest 2-digit even multiple of 5
Answer:
(i) The seventh multiple of 36 = 36 × 7 = 252.

Therefore, the seventh multiple of 36 is 252.

(ii) Even multiples of 9 are formed by multiplying 9 by even numbers.

9 × 2 = 18, 9 × 4 = 36, 9 × 6 = 54, 9 × 8 = 72, 9 × 10 = 90

(9 × 12 = 108, which exceeds 100.)

Therefore, the even multiples of 9 less than 100 are 18, 36, 54, 72 and 90.

(iii) Odd multiples of 17 are formed by multiplying 17 by odd numbers.

17 × 1 = 17, 17 × 3 = 51, 17 × 5 = 85, 17 × 7 = 119

(17 × 9 = 153, which exceeds 150.)

Therefore, the odd multiples of 17 less than 150 are 17, 51, 85 and 119.

(iv) The smallest 3-digit number is 100. We need to find the smallest multiple of 7 that is a 3-digit even number.

7 × 15 = 105 (odd)

7 × 16 = 112 (even)

Therefore, the first 3-digit even multiple of 7 is 112.

(v) The multiples of 6 are 6, 12, 18, 24, 30, 36, 42, 48, 54, 60, 66, 72, 78, 84, ...

Therefore, the multiples of 6 between 40 and 80 are 42, 48, 54, 60, 66, 72 and 78.

(vi) The greatest 2-digit number is 99. An even multiple of 5 must end in 0.

The greatest 2-digit number ending in 0 is 90, and 90 = 5 × 18.

Therefore, the greatest 2-digit even multiple of 5 is 90.
In simple words: To find even multiples, multiply by even numbers. To find odd multiples, multiply by odd numbers. Always check that your answer fits all the conditions asked for in the question.

Exam Tip: Read each sub-part carefully - note the constraints like "less than 100", "between 40 and 80", or "even". Missing a constraint costs marks even if your numbers are otherwise correct.

 

Question 6. Find the common factors of:
(i) 20 and 28
(ii) 35 and 50
(iii) 56 and 120
Answer:
(i) The factors of 20 are 1, 2, 4, 5, 10 and 20.

The factors of 28 are 1, 2, 4, 7, 14 and 28.

Therefore, the common factors of 20 and 28 are 1, 2 and 4.

(ii) The factors of 35 are 1, 5, 7 and 35.

The factors of 50 are 1, 2, 5, 10, 25 and 50.

Therefore, the common factors of 35 and 50 are 1 and 5.

(iii) The factors of 56 are 1, 2, 4, 7, 8, 14, 28 and 56.

The factors of 120 are 1, 2, 3, 4, 5, 6, 8, 10, 12, 15, 20, 24, 30, 40, 60 and 120.

Therefore, the common factors of 56 and 120 are 1, 2, 4 and 8.
In simple words: Write down all factors of each number. Then look for the numbers that appear in both lists. Those numbers are the common factors.

Exam Tip: The number 1 is always a common factor of any two numbers. List factors in order and double-check that each one divides both numbers evenly.

 

Question 7. Find the common factors of:
(i) 4, 8, 12
(ii) 10, 30 and 45
Answer:
(i) The factors of 4 are 1, 2 and 4.

The factors of 8 are 1, 2, 4 and 8.

The factors of 12 are 1, 2, 3, 4, 6 and 12.

Therefore, the common factors of 4, 8 and 12 are 1, 2 and 4.

(ii) The factors of 10 are 1, 2, 5 and 10.

The factors of 30 are 1, 2, 3, 5, 6, 10, 15 and 30.

The factors of 45 are 1, 3, 5, 9, 15 and 45.

Therefore, the common factors of 10, 30 and 45 are 1 and 5.
In simple words: When finding common factors of three numbers, list factors for each, then pick only the numbers that show up in all three lists.

Exam Tip: When three numbers are involved, be careful to find factors that divide ALL three evenly, not just some of them. 1 will always be a common factor.

 

Question 8. Write all natural numbers less than 100 which are common multiples of 3 and 4.
Answer: The common multiples of 3 and 4 are the same as the multiples of LCM(3, 4) = 12.

The multiples of 12 that are less than 100 are: 12, 24, 36, 48, 60, 72, 84, 96.

Therefore, the natural numbers less than 100 which are common multiples of 3 and 4 are 12, 24, 36, 48, 60, 72, 84 and 96.
In simple words: Common multiples of two numbers are the same as multiples of their LCM (least common multiple). Find the LCM first, then list all its multiples within the given range.

Exam Tip: Using LCM to find common multiples is much faster than listing multiples of each number separately and comparing them.

 

Question 9. (i) Write the odd numbers between 36 and 53.
(ii) Write the even numbers between 232 and 251.
Answer:
(i) The odd numbers between 36 and 53 are 37, 39, 41, 43, 45, 47, 49 and 51.

(ii) The even numbers between 232 and 251 are 234, 236, 238, 240, 242, 244, 246, 248 and 250.
In simple words: Odd numbers go up by 2 each time (37, then 39, then 41...). Even numbers also go up by 2 each time (234, then 236, then 238...).

Exam Tip: When listing odd or even numbers in a range, always start with the first odd/even number after the lower bound, and stop before reaching or exceeding the upper bound.

 

Question 10. (i) Write four consecutive odd numbers succeeding 79.
(ii) Write three consecutive even numbers preceding 124.
Answer:
(i) Consecutive odd numbers right after 79 are found by adding 2 repeatedly.

79 + 2 = 81, 81 + 2 = 83, 83 + 2 = 85, 85 + 2 = 87.

Therefore, the four consecutive odd numbers succeeding 79 are 81, 83, 85 and 87.

(ii) Consecutive even numbers right before 124 are found by subtracting 2 repeatedly.

124 - 2 = 122, 122 - 2 = 120, 120 - 2 = 118.

Therefore, the three consecutive even numbers preceding 124 are 118, 120 and 122.
In simple words: To get the next odd number, add 2 to the current odd number. To get the previous even number, subtract 2 from the current even number.

Exam Tip: Pay attention to whether the question asks for numbers "after" (succeeding/following) or "before" (preceding). This tells you whether to add or subtract.

 

Question 11. What is greatest prime number between 1 and 15?
Answer: The prime numbers between 1 and 15 are 2, 3, 5, 7, 11 and 13.

Therefore, the greatest prime number between 1 and 15 is 13.
In simple words: List all prime numbers in the range, then pick the largest one.

Exam Tip: Remember the first few primes: 2, 3, 5, 7, 11, 13, 17, 19, 23, 29... Knowing these helps you answer quickly and correctly.

 

Question 12. Which of the following numbers are prime?
(i) 29
(ii) 57
(iii) 43
(iv) 61
Answer:
(i) 29

Since 5 × 5 = 25 < 29 and 6 × 6 = 36 > 29, the number 5 is the largest number such that 5 × 5 ≤ 29.

The prime numbers less than or equal to 5 are 2, 3 and 5.

We check: 29 is not divisible by 2, 3 or 5.

Therefore, 29 is a prime number.

(ii) 57

The sum of digits of 57 = 5 + 7 = 12, which is divisible by 3.

So 57 is divisible by 3 (57 = 3 × 19).

Therefore, 57 is not a prime number.

(iii) 43

Since 6 × 6 = 36 < 43 and 7 × 7 = 49 > 43, the number 6 is the largest number such that 6 × 6 ≤ 43.

The prime numbers less than or equal to 6 are 2, 3 and 5.

We check: 43 is not divisible by 2, 3 or 5.

Therefore, 43 is a prime number.

(iv) 61

Since 7 × 7 = 49 < 61 and 8 × 8 = 64 > 61, the number 7 is the largest number such that 7 × 7 ≤ 61.

The prime numbers less than or equal to 7 are 2, 3, 5 and 7.

We check: 61 is not divisible by 2, 3, 5 or 7.

Therefore, 61 is a prime number.
In simple words: To check if a number is prime, test whether it divides evenly by any prime up to its square root. If it does not divide by any of them, it is prime.

Exam Tip: Use the square root test to reduce the number of primes you need to check. For 29, you only need to check primes up to 5 because 6² > 29. This saves time and reduces errors.

 

Question 13. Which of the following pairs of numbers are co-prime?
(i) 12 and 35
(ii) 15 and 37
(iii) 27 and 32
(iv) 17 and 85
(v) 515 and 516
(vi) 215 and 415
Answer:
(i) The factors of 12 are 1, 2, 3, 4, 6 and 12.

The factors of 35 are 1, 5, 7 and 35.

The only common factor is 1.

Therefore, 12 and 35 are co-prime.

(ii) The factors of 15 are 1, 3, 5 and 15.

The factors of 37 are 1 and 37.

The only common factor is 1.

Therefore, 15 and 37 are co-prime.

(iii) The factors of 27 are 1, 3, 9 and 27.

The factors of 32 are 1, 2, 4, 8, 16 and 32.

The only common factor is 1.

Therefore, 27 and 32 are co-prime.

(iv) The factors of 17 are 1 and 17.

The factors of 85 are 1, 5, 17 and 85.

The common factors are 1 and 17.

Therefore, 17 and 85 are not co-prime.

(v) Consecutive numbers always have only 1 as their common factor.

Therefore, 515 and 516 are co-prime.

(vi) The factors of 215 are 1, 5, 43 and 215.

The factors of 415 are 1, 5, 83 and 415.

The common factors are 1 and 5.

Therefore, 215 and 415 are not co-prime.
In simple words: Two numbers are co-prime if they have no common factor other than 1. Always list all factors for each number and check if any factor besides 1 is shared.

Exam Tip: Remember: any two consecutive numbers are always co-prime because they cannot share any factor other than 1. This is a useful shortcut that saves time for questions like part (v).

 

Exercise 4.2

 

Question 1. Which of the following numbers are divisible by 5 or by 10:
(i) 3725
(ii) 48970
(iii) 56823
(iv) 760035
(v) 7893217
(vi) 4500010
Answer: A number can be divided by 5 if its last digit is 0 or 5. A number can be divided by 10 if its last digit is 0.
(i) 3725 - last digit is 5. Hence, 3725 is divisible by 5 only.
(ii) 48970 - last digit is 0. Hence, 48970 is divisible by both 5 and 10.
(iii) 56823 - last digit is 3. Hence, 56823 is divisible neither by 5 nor by 10.
(iv) 760035 - last digit is 5. Hence, 760035 is divisible by 5 only.
(v) 7893217 - last digit is 7. Hence, 7893217 is divisible neither by 5 nor by 10.
(vi) 4500010 - last digit is 0. Hence, 4500010 is divisible by both 5 and 10.
In simple words: Check the last digit only. If it is 0 or 5, the number divides by 5. If it is 0 only, it divides by 10.

Exam Tip: Remember that divisibility by 10 requires the last digit to be 0 - this is stricter than divisibility by 5, which allows 0 or 5.

 

Question 2. Which of the following numbers are divisible by 2, 4 or 8:
(i) 54014
(ii) 723840
(iii) 6531088
(iv) 75689604
(v) 786235
(vi) 5321048
Answer: A number is divisible by 2 if its last digit is 0, 2, 4, 6 or 8. A number is divisible by 4 if the two-digit number formed by its last two digits can be divided by 4. A number is divisible by 8 if the three-digit number formed by its last three digits can be divided by 8.
(i) 54014 - last digit 4, divisible by 2. Last two digits = 14, not divisible by 4. Hence, 54014 is divisible by 2 only.
(ii) 723840 - last digit 0, divisible by 2. Last two digits = 40, divisible by 4. Last three digits = 840, and 840 ÷ 8 = 105, divisible by 8. Hence, 723840 is divisible by 2, 4 and 8.
(iii) 6531088 - last digit 8, divisible by 2. Last two digits = 88, divisible by 4. Last three digits = 088 = 88, and 88 ÷ 8 = 11, divisible by 8. Hence, 6531088 is divisible by 2, 4 and 8.
(iv) 75689604 - last digit 4, divisible by 2. Last two digits = 04 = 4, divisible by 4. Last three digits = 604, and 604 is not divisible by 8 because 8 × 75 = 600 and remainder is 4. Hence, 75689604 is divisible by 2 and 4 only.
(v) 786235 - last digit 5, not divisible by 2. Hence, 786235 is not divisible by 2, 4 or 8.
(vi) 5321048 - last digit 8, divisible by 2. Last two digits = 48, divisible by 4. Last three digits = 048 = 48, and 48 ÷ 8 = 6, divisible by 8. Hence, 5321048 is divisible by 2, 4 and 8.
In simple words: For 2, check the last digit. For 4, check the last two digits. For 8, check the last three digits. Each rule tests a bigger part of the number.

Exam Tip: These divisibility tests let you skip long division - they work because the last digits (or groups of digits) capture the remainder pattern for 2, 4, and 8.

 

Question 3. Which of the following numbers are divisible by 3 or 9:
(i) 7341
(ii) 59031
(iii) 12345678
(iv) 560319
(v) 720634
(vi) 3721509
Answer: A number can be divided by 3 if the total of all its digits is divisible by 3. A number can be divided by 9 if the total of all its digits is divisible by 9.
(i) 7341 - sum of digits = 7 + 3 + 4 + 1 = 15. 15 is divisible by 3 but not by 9. Hence, 7341 is divisible by 3 only.
(ii) 59031 - sum of digits = 5 + 9 + 0 + 3 + 1 = 18. 18 is divisible by both 3 and 9. Hence, 59031 is divisible by both 3 and 9.
(iii) 12345678 - sum of digits = 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 = 36. 36 is divisible by both 3 and 9. Hence, 12345678 is divisible by both 3 and 9.
(iv) 560319 - sum of digits = 5 + 6 + 0 + 3 + 1 + 9 = 24. 24 is divisible by 3 but not by 9. Hence, 560319 is divisible by 3 only.
(v) 720634 - sum of digits = 7 + 2 + 0 + 6 + 3 + 4 = 22. 22 is not divisible by 3. Hence, 720634 is not divisible by 3 or 9.
(vi) 3721509 - sum of digits = 3 + 7 + 2 + 1 + 5 + 0 + 9 = 27. 27 is divisible by both 3 and 9. Hence, 3721509 is divisible by both 3 and 9.
In simple words: Add up all the digits. If the total divides by 3, the whole number does too. If the total divides by 9, the whole number divides by 9 as well.

Exam Tip: Since 9 is a multiple of 3, every number divisible by 9 is also divisible by 3 - but not the other way around.

 

Question 4. Examine the following numbers for divisibility by 11:
(i) 10428
(ii) 70169803
(iii) 7136985
Answer: A number can be divided by 11 if the difference between the total of digits at odd places (counting from the right) and the total of digits at even places (counting from the right) is either 0 or divisible by 11.
(i) 10428: Sum of digits at odd places (from right) = 8 + 4 + 1 = 13. Sum of digits at even places (from right) = 2 + 0 = 2. Difference = 13 - 2 = 11, which is divisible by 11. Hence, 10428 is divisible by 11.
(ii) 70169803: Sum of digits at odd places (from right) = 3 + 8 + 6 + 0 = 17. Sum of digits at even places (from right) = 0 + 9 + 1 + 7 = 17. Difference = 17 - 17 = 0. Hence, 70169803 is divisible by 11.
(iii) 7136985: Sum of digits at odd places (from right) = 5 + 9 + 3 + 7 = 24. Sum of digits at even places (from right) = 8 + 6 + 1 = 15. Difference = 24 - 15 = 9, which is not divisible by 11. Hence, 7136985 is not divisible by 11.
In simple words: Start from the right side. Add every other digit (1st, 3rd, 5th...) to get one total. Add the remaining digits (2nd, 4th, 6th...) to get another. If the difference is 0 or 11 or 22, the number divides by 11.

Exam Tip: This test is trickier than the others - write out which digits are in the odd and even places to avoid mixing them up.

 

Question 5. Examine the following numbers for divisibility by 6:
(i) 93573
(ii) 217944
(iii) 5034126
(iv) 901352
(v) 639210
(vi) 1790184
Answer: A number can be divided by 6 if it is divisible by both 2 and 3.
(i) 93573 - last digit 3, not divisible by 2. Hence, 93573 is not divisible by 6.
(ii) 217944 - last digit 4, divisible by 2. Sum of digits = 2 + 1 + 7 + 9 + 4 + 4 = 27, divisible by 3. Hence, 217944 is divisible by 6.
(iii) 5034126 - last digit 6, divisible by 2. Sum of digits = 5 + 0 + 3 + 4 + 1 + 2 + 6 = 21, divisible by 3. Hence, 5034126 is divisible by 6.
(iv) 901352 - last digit 2, divisible by 2. Sum of digits = 9 + 0 + 1 + 3 + 5 + 2 = 20, not divisible by 3. Hence, 901352 is not divisible by 6.
(v) 639210 - last digit 0, divisible by 2. Sum of digits = 6 + 3 + 9 + 2 + 1 + 0 = 21, divisible by 3. Hence, 639210 is divisible by 6.
(vi) 1790184 - last digit 4, divisible by 2. Sum of digits = 1 + 7 + 9 + 0 + 1 + 8 + 4 = 30, divisible by 3. Hence, 1790184 is divisible by 6.
In simple words: To check if a number divides by 6, test both rules: it must end in an even digit AND its digit total must divide by 3. Both must be true.

Exam Tip: Since 6 = 2 × 3, divisibility by 6 requires passing both the divisibility-by-2 and divisibility-by-3 tests - failing either one means it does not divide by 6.

 

Question 6. In each of the following replace '*' by a digit so that the number formed is divisible by 9:
(i) 4710*82
(ii) 70*356722
Answer: For a number to be divisible by 9, the total of its digits must be divisible by 9.
(i) 4710*82: Sum of given digits (except *) = 4 + 7 + 1 + 0 + 8 + 2 = 22. We need 22 + * to be divisible by 9. 22 + 5 = 27, which is divisible by 9. Hence, * should be replaced by 5. The number formed is 4710582.
(ii) 70*356722: Sum of given digits (except *) = 7 + 0 + 3 + 5 + 6 + 7 + 2 + 2 = 32. We need 32 + * to be divisible by 9. 32 + 4 = 36, which is divisible by 9. Hence, * should be replaced by 4. The number formed is 704356722.
In simple words: Add the known digits. See what number you need to add to reach the next multiple of 9. That number is your answer for the *.

Exam Tip: Once you have the sum of known digits, find the nearest multiple of 9 above it, then subtract to get the missing digit - it will always be between 0 and 9.

 

Question 7. In each of the following replace '*' by (i) the smallest digit (ii) the greatest digit so that the number formed is divisible by 3:
(a) 4*672
(b) 4756*2
Answer: For a number to be divisible by 3, the total of its digits must be divisible by 3.
(a) 4*672: Sum of given digits (except *) = 4 + 6 + 7 + 2 = 19. We need 19 + * to be divisible by 3. The values of * which make 19 + * divisible by 3 are 2 (gives 21), 5 (gives 24) and 8 (gives 27).
(i) The smallest digit is 2. The number formed is 42672.
(ii) The greatest digit is 8. The number formed is 48672.
(b) 4756*2: Sum of given digits (except *) = 4 + 7 + 5 + 6 + 2 = 24. We need 24 + * to be divisible by 3. The values of * which make 24 + * divisible by 3 are 0 (gives 24), 3 (gives 27), 6 (gives 30) and 9 (gives 33).
(i) The smallest digit is 0. The number formed is 475602.
(ii) The greatest digit is 9. The number formed is 475692.
In simple words: Find all digits that make the total of all digits divisible by 3. Then pick the smallest one and the largest one from your list.

Exam Tip: There may be multiple values of * that work - always list all possibilities first, then select the smallest and largest as instructed.

 

Question 8. In each of the following replace * by a digit so that the number formed is divisible by 11:
(i) 8*9484
(ii) 9*53762
Answer: A number can be divided by 11 when the difference between the sum of digits at odd places (counted from the ones side) and the sum of digits at even places (counted from the tens side) equals 0 or is a multiple of 11.

(i) 8*9484
Reading from right: 4, 8, 4, 9, *, 8.
Sum of digits at odd positions (from right) = 4 + 4 + * = 8 + *
Sum of digits at even positions (from right) = 8 + 9 + 8 = 25
Difference = 25 - (8 + *) = 17 - *
For the number to be divisible by 11: 17 - * = 0 (not possible) or 17 - * = 11, which gives * = 6
The number is 869484.

(ii) 9*53762
Reading from right: 2, 6, 7, 3, 5, *, 9
Sum of digits at odd positions (from right) = 2 + 7 + 5 + 9 = 23
Sum of digits at even positions (from right) = 6 + 3 + * = 9 + *
Difference = 23 - (9 + *) = 14 - *
For the number to be divisible by 11: 14 - * = 0 (not possible) or 14 - * = 11, which gives * = 3
The number is 9353762.
In simple words: To check if a number can be split evenly by 11, subtract the sum of digits in even places from the sum of digits in odd places. If this gives you 0 or 11 (or 22, 33, etc.), then the number works.

Exam Tip: Always count positions from right to left (the ones place is position 1). Keep careful track of which digits go in odd versus even positions - a single mistake here will give the wrong answer.

 

Question 9. In each of the following replace * by (i) the smallest digit (ii) the greatest digit so that the number formed is divisible by 6:
(a) 2*4706
(b) 5825*34
Answer: To be divisible by 6, a number must be divisible by both 2 and 3.

(a) 2*4706
The last digit is 6, so the number is already divisible by 2.
Sum of known digits = 2 + 4 + 7 + 0 + 6 = 19
We need 19 + * to be divisible by 3. The values that work are 2, 5, and 8.
(i) Smallest digit is 2, giving the number 224706
(ii) Greatest digit is 8, giving the number 284706

(b) 5825*34
The last digit is 4, so the number is already divisible by 2.
Sum of known digits = 5 + 8 + 2 + 5 + 3 + 4 = 27
We need 27 + * to be divisible by 3. The values that work are 0, 3, 6, and 9.
(i) Smallest digit is 0, giving the number 5825034
(ii) Greatest digit is 9, giving the number 5825934
In simple words: Check if the last digit is even (for divisibility by 2). Then add up all the other digits and find what digit makes the total divisible by 3.

Exam Tip: Remember that divisibility by 6 requires BOTH conditions - even last digit AND a digit sum divisible by 3. Test both before confirming your answer.

 

Question 10. Which of the following numbers are prime:
(i) 101
(ii) 251
(iii) 323
(iv) 397
Answer:
(i) 101
We find that \( 10 \times 10 = 100 < 101 \) and \( 11 \times 11 = 121 > 101 \), so 10 is the largest number such that \( 10 \times 10 \leq 101 \).
The prime numbers up to 10 are: 2, 3, 5, 7
101 is not divisible by any of these.
Therefore, 101 is a prime number.

(ii) 251
We find that \( 15 \times 15 = 225 < 251 \) and \( 16 \times 16 = 256 > 251 \), so 15 is the largest number such that \( 15 \times 15 \leq 251 \).
The prime numbers up to 15 are: 2, 3, 5, 7, 11, 13
251 is not divisible by any of these.
Therefore, 251 is a prime number.

(iii) 323
Note that \( 17 \times 19 = 323 \), so 17 and 19 are both factors of 323.
Therefore, 323 is not a prime number.

(iv) 397
We find that \( 19 \times 19 = 361 < 397 \) and \( 20 \times 20 = 400 > 397 \), so 19 is the largest number such that \( 19 \times 19 \leq 397 \).
The prime numbers up to 19 are: 2, 3, 5, 7, 11, 13, 17, 19
397 is not divisible by any of these.
Therefore, 397 is a prime number.
In simple words: To check if a number is prime, test whether it can be divided by any prime number up to its square root. If no prime number divides it, then the number is prime.

Exam Tip: The key insight is testing only up to the square root - this saves time. Always list out all the primes you need to check to show complete working.

 

Question 11. Determine if 372645 is divisible by 45.
Answer: To check divisibility by 45, we test for divisibility by both 5 and 9 (since \( 45 = 5 \times 9 \) and 5 and 9 share no common factors).

The number 372645 has 5 in the ones place, so it is divisible by 5.

Sum of digits = 3 + 7 + 2 + 6 + 4 + 5 = 27, which is divisible by 9, so 372645 is divisible by 9.

Since 5 and 9 share no common factors, the number 372645 is divisible by their product, which is 45.
Therefore, 372645 is divisible by 45.
In simple words: When checking divisibility by a composite number like 45, break it into smaller prime-building-block numbers (5 and 9). If the number works for both, it works for 45 too.

Exam Tip: Always decompose the target number into co-prime factors. This makes the divisibility check much simpler than trying to divide directly.

 

Question 12. A number is divisible by 12. By what other numbers will that number be divisible?
Answer: When a number is divisible by 12, it is also divisible by each factor of 12.

The factors of 12 are: 1, 2, 3, 4, 6, and 12

Therefore, the number is also divisible by 1, 2, 3, 4, and 6.
In simple words: If a big number can be divided evenly by 12, then it can also be divided evenly by all the smaller numbers that multiply to make 12.

Exam Tip: List all factors systematically by checking pairs - this ensures you don't miss any. Remember that 1 is always a factor, and so is the number itself.

 

Question 13. A number is divisible by both 3 and 8. By which other numbers will that number be always divisible?
Answer: Let's call this number n. We know n is divisible by both 3 and 8.

Since 3 and 8 share no common factors (they are co-prime), when a number is divisible by two co-prime numbers, it must be divisible by their product.

Therefore, n is divisible by \( 3 \times 8 = 24 \).

Since n is divisible by 24, it is divisible by all factors of 24.

The factors of 24 are: 1, 2, 3, 4, 6, 8, 12, and 24

Therefore, the number is always divisible by 1, 2, 3, 4, 6, 8, 12, and 24.
In simple words: When a number works for two numbers that don't share any common factors, it must also work for whatever you get by multiplying those two numbers together.

Exam Tip: Co-prime numbers are key - when you have two co-prime factors, the number must be divisible by their product. Always list all factors of that product as your final answer.

 

Question 14. State whether the following statements are true (T) or false (F):
(i) If a number is divisible by 4, it must be divisible by 8.
(ii) If a number is divisible by 3, it must be divisible by 9.
(iii) If a number is divisible by 9, it must be divisible by 3.
(iv) If a number is divisible by 9 and 10 both, it must be divisible by 90.
(v) If a number divides the sum of two numbers, then it must divide the two numbers separately.
(vi) If a number is divisible by 3 and 8 both, it must be divisible by 12.
(vii) If a number is divisible by 6 and 15 both, it must be divisible by 90.
Answer:
(i) False
Reason: Consider 12, which is divisible by 4, yet not divisible by 8.

(ii) False
Reason: Consider 6, which is divisible by 3, yet not divisible by 9.

(iii) True
Reason: Since 3 is a factor of 9, any number divisible by 9 is automatically divisible by 3.

(iv) True
Reason: Since 9 and 10 are co-prime, any number divisible by both must be divisible by \( 9 \times 10 = 90 \).

(v) False
Reason: Consider that 5 divides \( 3 + 7 = 10 \), but 5 does not divide 3 or 7 individually.

(vi) True
Reason: Since 3 and 8 are co-prime, if a number is divisible by both, it is divisible by \( 3 \times 8 = 24 \). Since 12 is a factor of 24, the number is also divisible by 12.

(vii) False
Reason: Since 6 and 15 are not co-prime (their highest common factor is 3), we cannot conclude divisibility by their product. The lowest common multiple of 6 and 15 is 30. For example, 30 is divisible by both 6 and 15, but not by 90.
In simple words: Co-prime numbers (those with no shared factors) follow special rules - if a number works for both, it works for their product. Non-co-prime numbers don't follow this rule.

Exam Tip: Always check whether two numbers are co-prime before assuming a number divisible by both is divisible by their product. A counterexample (one case that breaks the rule) is enough to prove a statement false.

 

Exercise 4.3

 

Question 1. Here are two different factor trees of the number 90. Write the missing numbers:
Answer:
(i) For the first factor tree where 90 = 10 × 9:
Since 10 = 2 × ?, the missing number is 5
Since 9 = ? × 3, the missing number is 3

(ii) For the second factor tree where 90 = 45 × ?:
The missing number is 2 (since 90 = 45 × 2)
Since 45 = 15 × ?, the missing number is 3
Since 15 = ? × ?, the missing numbers are 3 and 5
In simple words: In a factor tree, each number splits into two factors below it. Work through the tree step by step, using division to find the missing factors.

Exam Tip: Check your answer by multiplying back up the tree - all paths should lead to 90 at the top. This confirms every missing number is correct.

 

Question 2. Find the prime factorisation of the following numbers:
(i) 72
(ii) 172
(iii) 450
(iv) 980
(v) 8712
(vi) 13500
Answer:
(i) 72
Dividing by prime numbers systematically:
\( 72 = 2 \times 36 = 2 \times 2 \times 18 = 2 \times 2 \times 2 \times 9 = 2 \times 2 \times 2 \times 3 \times 3 \)
Therefore, \( 72 = 2^3 \times 3^2 \)

(ii) 172
\( 172 = 2 \times 86 = 2 \times 2 \times 43 \)
Therefore, \( 172 = 2^2 \times 43 \)

(iii) 450
\( 450 = 2 \times 225 = 2 \times 3 \times 75 = 2 \times 3 \times 3 \times 25 = 2 \times 3 \times 3 \times 5 \times 5 \)
Therefore, \( 450 = 2 \times 3^2 \times 5^2 \)

(iv) 980
\( 980 = 2 \times 490 = 2 \times 2 \times 245 = 2 \times 2 \times 5 \times 49 = 2 \times 2 \times 5 \times 7 \times 7 \)
Therefore, \( 980 = 2^2 \times 5 \times 7^2 \)

(v) 8712
\( 8712 = 2 \times 4356 = 2 \times 2 \times 2178 = 2 \times 2 \times 2 \times 1089 = 2 \times 2 \times 2 \times 3 \times 363 = 2 \times 2 \times 2 \times 3 \times 3 \times 121 = 2 \times 2 \times 2 \times 3 \times 3 \times 11 \times 11 \)
Therefore, \( 8712 = 2^3 \times 3^2 \times 11^2 \)

(vi) 13500
\( 13500 = 2 \times 6750 = 2 \times 2 \times 3375 = 2 \times 2 \times 3 \times 1125 = 2 \times 2 \times 3 \times 3 \times 375 = 2 \times 2 \times 3 \times 3 \times 3 \times 125 = 2 \times 2 \times 3 \times 3 \times 3 \times 5 \times 25 = 2 \times 2 \times 3 \times 3 \times 3 \times 5 \times 5 \times 5 \)
Therefore, \( 13500 = 2^2 \times 3^3 \times 5^3 \)
In simple words: Keep dividing by the smallest prime that goes in evenly. Write down each prime you use. When you reach 1, you are finished - all the primes you wrote are the prime factors.

Exam Tip: Always start with 2, then try 3, then 5, and work up through the primes. This systematic approach prevents you from missing or repeating a prime factor. Check by multiplying your final answer back to the original number.

 

Question 3. Write the smallest and the greatest 3-digit numbers and express them as the product of prime.
Answer: The smallest 3-digit number is 100. Using prime factorisation:
100 = 2 × 2 × 5 × 5

The greatest 3-digit number is 999. Using prime factorisation:
999 = 3 × 3 × 3 × 37
In simple words: Break 100 into its prime factors by dividing by 2 and 5. Break 999 by dividing by 3 and then 37.

Exam Tip: Always use the division method systematically, starting with the smallest prime (2), and continue until the quotient becomes 1. The final list of divisors gives the prime factorisation.

 

Question 4. Write the smallest five digit number and express it in the form of its prime factors.
Answer: The smallest 5-digit number is 10000. Using prime factorisation:
10000 = 2 × 2 × 2 × 2 × 5 × 5 × 5 × 5
In simple words: Divide 10000 by 2 four times and by 5 four times. These are all the prime numbers that make up 10000.

Exam Tip: When factorising numbers with many zeros, remember that 10 = 2 × 5. So 10000 = (2 × 5)^4 = 2^4 × 5^4.

 

Question 5. I am the smallest number, having four different prime factors. Can you find me?
Answer: The smallest four prime numbers are 2, 3, 5, and 7. The smallest number with four distinct prime factors is the product of the smallest four prime numbers.
Required number = 2 × 3 × 5 × 7 = 210
Hence, the smallest number having four different prime factors is 210.
In simple words: To get a number with four different prime factors, multiply the four smallest primes: 2, 3, 5, and 7. This gives 210.

Exam Tip: The smallest number with a given set of prime factors is always their direct product (with no repetitions). If repetitions were allowed, you would use powers of the primes.

 

Exercise 4.4

 

Question 1. Find the HCF of the given numbers by prime factorisation method:
(i) 28, 36
(ii) 54, 72, 90
(iii) 105, 140, 175
Answer:
(i) 28, 36
Prime factorisation of 28: 28 = 2 × 2 × 7
Prime factorisation of 36: 36 = 2 × 2 × 3 × 3
The common prime factor is 2, occurring twice in both numbers.
Hence, HCF of 28 and 36 = 2 × 2 = 4.

(ii) 54, 72, 90
Prime factorisation of 54: 54 = 2 × 3 × 3 × 3
Prime factorisation of 72: 72 = 2 × 2 × 2 × 3 × 3
Prime factorisation of 90: 90 = 2 × 3 × 3 × 5
The common prime factors are 2 (occurring once) and 3 (occurring twice) across all three numbers.
Hence, HCF of 54, 72 and 90 = 2 × 3 × 3 = 18.

(iii) 105, 140, 175
Prime factorisation of 105: 105 = 3 × 5 × 7
Prime factorisation of 140: 140 = 2 × 2 × 5 × 7
Prime factorisation of 175: 175 = 5 × 5 × 7
The common prime factors are 5 (occurring once) and 7 (occurring once) across all three numbers.
Hence, HCF of 105, 140 and 175 = 5 × 7 = 35.
In simple words: Find which prime numbers appear in every number. The HCF is made by multiplying only those common primes, using the smallest count of each.

Exam Tip: When finding HCF, list the prime factors of each number separately, then identify which primes appear in all of them. Take each common prime the minimum number of times it appears.

 

Question 2. Find the HCF of the given numbers by division method:
(i) 198, 429
(ii) 20, 64, 104
(iii) 120, 144, 204
Answer:
(i) 198, 429
Using the division method (Euclidean algorithm):
\[ 429 = 198 \times 2 + 33 \]
\[ 198 = 33 \times 6 + 0 \]
Hence, HCF of 198 and 429 = 33.

(ii) 20, 64, 104
First, find HCF of 20 and 64:
\[ 64 = 20 \times 3 + 4 \]
\[ 20 = 4 \times 5 + 0 \]
So, HCF of 20 and 64 = 4.

Now, find HCF of 4 and 104:
\[ 104 = 4 \times 26 + 0 \]
So, HCF of 4 and 104 = 4.
Hence, HCF of 20, 64 and 104 = 4.

(iii) 120, 144, 204
First, find HCF of 120 and 144:
\[ 144 = 120 \times 1 + 24 \]
\[ 120 = 24 \times 5 + 0 \]
So, HCF of 120 and 144 = 24.

Now, find HCF of 24 and 204:
\[ 204 = 24 \times 8 + 12 \]
\[ 24 = 12 \times 2 + 0 \]
So, HCF of 24 and 204 = 12.
Hence, HCF of 120, 144 and 204 = 12.
In simple words: For two numbers, divide the larger by the smaller, then divide the divisor by the remainder, and repeat until the remainder is zero. The last non-zero remainder is the HCF. For three numbers, find the HCF of the first two, then find the HCF of that result with the third number.

Exam Tip: The division method is faster than prime factorisation for large numbers. Always ensure the remainder becomes zero to confirm you have found the correct HCF.

 

Question 3. Fill in the blanks:
(i) HCF of two consecutive natural numbers is ....
(ii) HCF of two consecutive odd numbers is ....
(iii) HCF of two consecutive even numbers is ....
Answer:
(i) Two consecutive natural numbers have no common factor other than 1.
Hence, HCF of two consecutive natural numbers is 1.

(ii) Two consecutive odd numbers (for example, 5 and 7, or 9 and 11) share no common factor other than 1.
Hence, HCF of two consecutive odd numbers is 1.

(iii) Two consecutive even numbers (for example, 6 and 8, or 10 and 12) are both divisible by 2, but share no other common factor.
Hence, HCF of two consecutive even numbers is 2.
In simple words: Any two numbers that follow one after the other cannot share factors. Only consecutive even numbers both have 2 as a factor.

Exam Tip: Remember that consecutive odd numbers differ by 2, so their only common factor is 1. Consecutive even numbers also differ by 2, so both are divisible by 2, making their HCF at least 2.

 

Question 4. Find the greatest number which can divide 257 and 329 so as to leave a remainder 5 in each case.
Answer: When 257 is divided by the required number, 5 is left as a remainder. Therefore, 257 - 5 = 252 is exactly divisible by that number.

Similarly, 329 - 5 = 324 is exactly divisible by that number.

So, the required number is the HCF of 252 and 324.

Using the division method:
\[ 324 = 252 \times 1 + 72 \]
\[ 252 = 72 \times 3 + 36 \]
\[ 72 = 36 \times 2 + 0 \]

Hence, the required greatest number is 36.
In simple words: Subtract the remainder from each number. Then find the HCF of these new numbers - that HCF is your answer.

Exam Tip: When a problem asks for a divisor that leaves the same remainder in all cases, always subtract that remainder from each number first, then find the HCF of the results.

 

Question 5. Find the largest number that will divide 623, 729 and 841 leaving remainders 3, 9 and 1 respectively.
Answer: When 623 is divided by the required number, 3 is left as a remainder. So, 623 - 3 = 620 is exactly divisible by that number.

Similarly, 729 - 9 = 720 and 841 - 1 = 840 are exactly divisible by that number.

Therefore, the required number is the HCF of 620, 720 and 840.

First, find HCF of 620 and 720:
\[ 720 = 620 \times 1 + 100 \]
\[ 620 = 100 \times 6 + 20 \]
\[ 100 = 20 \times 5 + 0 \]
So, HCF of 620 and 720 = 20.

Now, find HCF of 20 and 840:
\[ 840 = 20 \times 42 + 0 \]
So, HCF of 20 and 840 = 20.

Hence, the largest required number is 20.
In simple words: Subtract each different remainder from its corresponding number. Find the HCF of all these new numbers. That HCF is your answer.

Exam Tip: When remainders differ across numbers, subtract each one individually before finding the HCF. This converts the problem into a standard HCF calculation.

 

Question 6. Meenu purchases two bags of rice of weights 75 kg and 69 kg. Find the maximum value of weight which can measure the weight of the rice exact number of times.
Answer: The maximum weight that can measure both rice bags an exact number of times is the HCF of 75 and 69.

Using the division method:
\[ 75 = 69 \times 1 + 6 \]
\[ 69 = 6 \times 11 + 3 \]
\[ 6 = 3 \times 2 + 0 \]

So, HCF of 75 and 69 = 3.

Hence, the maximum weight which can measure the rice exactly is 3 kg.
In simple words: Find the HCF of the two weights. That HCF is the biggest weight that fits into both bags a whole number of times.

Exam Tip: In measurement problems, the HCF represents the largest unit that divides all given quantities exactly. Always state the unit (kg, m, litres, etc.) in your final answer.

 

Question 7. Three tankers contain 403 litres, 434 litres and 465 litres of diesel respectively. Find the maximum capacity of a container that can measure the diesel of three containers exact number of times.
Answer: The maximum capacity of a container that can measure the diesel in all three tankers exactly is the HCF of 403, 434 and 465.

First, find HCF of 403 and 434:
\[ 434 = 403 \times 1 + 31 \]
\[ 403 = 31 \times 13 + 0 \]
So, HCF of 403 and 434 = 31.

Now, find HCF of 31 and 465:
\[ 465 = 31 \times 15 + 0 \]
So, HCF of 31 and 465 = 31.

Hence, the maximum capacity of the container is 31 litres.
In simple words: Find the HCF of all three quantities. This HCF is the largest container size that can measure each tanker exactly, with no leftover.

Exam Tip: For problems involving three numbers, always find the HCF of the first two, then find the HCF of that result with the third number. This gives the final answer.

 

Exercise 4.5

 

Question 1. Find the LCM of the given numbers by prime factorisation method:
(i) 28, 98
(ii) 36, 40, 126
(iii) 108, 135, 162
(iv) 24, 28, 196
Answer:
(i) 28, 98
Prime factorisation of the given numbers:
28 = 2 × 2 × 7
98 = 2 × 7 × 7
The prime factor 2 occurs a maximum of 2 times, and 7 occurs a maximum of 2 times.
Hence, LCM of 28 and 98 = 2 × 2 × 7 × 7 = 196.

(ii) 36, 40, 126
Prime factorisation of the given numbers:
36 = 2 × 2 × 3 × 3
40 = 2 × 2 × 2 × 5
126 = 2 × 3 × 3 × 7
The prime factor 2 occurs a maximum of 3 times, 3 occurs 2 times, 5 occurs 1 time, and 7 occurs 1 time.
Hence, LCM of 36, 40 and 126 = 2 × 2 × 2 × 3 × 3 × 5 × 7 = 2520.

(iii) 108, 135, 162
Prime factorisation of the given numbers:
108 = 2 × 2 × 3 × 3 × 3
135 = 3 × 3 × 3 × 5
162 = 2 × 3 × 3 × 3 × 3
The prime factor 3 occurs a maximum of 4 times, 2 occurs 2 times, and 5 occurs 1 time.
Hence, LCM of 108, 135 and 162 = 2 × 2 × 3 × 3 × 3 × 3 × 5 = 1620.

(iv) 24, 28, 196
Prime factorisation of the given numbers:
24 = 2 × 2 × 2 × 3
28 = 2 × 2 × 7
196 = 2 × 2 × 7 × 7
The prime factor 2 occurs a maximum of 3 times, 3 occurs 1 time, and 7 occurs 2 times.
Hence, LCM of 24, 28 and 196 = 2 × 2 × 2 × 3 × 7 × 7 = 1176.
In simple words: List all prime factors of each number. For the LCM, take each prime the maximum number of times it appears in any single number, then multiply them all together.

Exam Tip: LCM uses the highest power of each prime that appears across all the numbers. Count carefully how many times each prime appears in each factorisation to avoid mistakes.

 

Question 2. Find the LCM of the given numbers by division method:
(i) 480, 672
(ii) 6, 8, 45
(iii) 24, 40, 84
(iv) 20, 36, 63, 77
Answer:
(i) 480, 672

The LCM is found by multiplying all the prime factors taken: 2 × 2 × 2 × 2 × 2 × 3 × 5 × 7 = 3360.

(ii) 6, 8, 45
The LCM is: 2 × 2 × 2 × 3 × 3 × 5 = 360.

(iii) 24, 40, 84
The LCM is: 2 × 2 × 2 × 3 × 5 × 7 = 840.

(iv) 20, 36, 63, 77
The LCM is: 2 × 2 × 3 × 3 × 5 × 7 × 11 = 13860.
In simple words: To find the LCM using division, keep dividing all the numbers by the smallest prime factor. Write down each prime you use. Then multiply all these primes together to get the LCM.

Exam Tip: Always continue division until all quotients become 1. Write each prime factor used on the left, and multiply all of them together - this gives the LCM directly.

 

Question 3. Find the least number which when increased by 15 is exactly divisible by 15, 35 and 48.
Answer: Start by finding the LCM of 15, 35, and 48. Using the division method: divide by prime factors until all values reach 1. The LCM is 2 × 2 × 2 × 2 × 3 × 5 × 7 = 1680. This is the smallest number that divides evenly by all three given numbers. According to the problem, the required number becomes 1680 when we add 15 to it. Therefore, the required number is 1680 - 15 = 1665.
In simple words: Find the LCM of the three numbers. That LCM is what you get when you add 15 to the answer. So subtract 15 from the LCM to find the number you need.

Exam Tip: Always understand whether you're looking for the smallest number divisible by given values, or if you need to add/subtract a remainder to find the final answer.

 

Question 4. Find the least number which when divided by 6, 15 and 18 leaves remainder 5 in each case.
Answer: First, calculate the LCM of 6, 15, and 18. Using division: 2 divides to get (3, 15, 9); then 3 divides to get (1, 5, 3); then 3 divides to get (1, 5, 1); then 5 divides to get (1, 1, 1). The LCM is 2 × 3 × 3 × 5 = 90. The number 90 is the smallest number that divides exactly by all three numbers with no remainder. However, we need a number that leaves a remainder of 5 when divided by each of these. So we add 5 to 90, giving us 90 + 5 = 95. The required number is 95.
In simple words: Find the LCM of the three numbers. Then add the remainder you want to that LCM. The result is your answer.

Exam Tip: When a problem asks for a number leaving the same remainder across multiple divisors, always find the LCM first, then add the remainder value to it.

 

Question 5. Find the least number which when divided by 24, 36, 45 and 54 leaves a remainder of 3 in each case.
Answer: Calculate the LCM of 24, 36, 45, and 54 by using the division method. Divide by 2 repeatedly, then by 3 repeatedly until all numbers become 1. The result is 2 × 2 × 2 × 3 × 3 × 3 × 5 = 1080. Since we need a number that leaves remainder 3 when divided by each of the four numbers, add 3 to the LCM: 1080 + 3 = 1083. The required least number is 1083.
In simple words: Find the LCM of all four numbers by dividing by their prime factors. Then add the remainder to get your final answer.

Exam Tip: The key principle is: if you want the same remainder from all divisors, the number you seek is LCM plus that remainder.

 

Question 6. Find the greatest 3-digit number which is exactly divisible by 8, 20 and 24.
Answer: First, determine the LCM of 8, 20, and 24. Using division: 2 gives (4, 10, 12); 2 gives (2, 5, 6); 2 gives (1, 5, 3); 3 gives (1, 5, 1); 5 gives (1, 1, 1). The LCM is 2 × 2 × 2 × 3 × 5 = 120. The greatest 3-digit number is 999. Divide 999 by 120 to find the remainder: 120 goes into 999 eight times (8 × 120 = 960) with remainder 39. To get the greatest 3-digit number divisible by 120, subtract this remainder from 999: 999 - 39 = 960. Therefore, the greatest 3-digit number divisible by 8, 20, and 24 is 960.
In simple words: Find the LCM of all three numbers. Divide the biggest 3-digit number (999) by this LCM. Subtract the remainder from 999 to get your answer.

Exam Tip: For the greatest number divisible by given values, always start with the largest number in that range (999 for 3-digit, 9999 for 4-digit) and subtract the remainder when divided by the LCM.

 

Question 7. Find the smallest 4-digit number which is exactly divisible by 32, 36 and 48.
Answer: Calculate the LCM of 32, 36, and 48 by the division method. Divide by 2 several times, then by 3 twice to get 2 × 2 × 2 × 2 × 2 × 3 × 3 = 288. The smallest 4-digit number is 1000. Divide 1000 by 288: the remainder is 136. To find the smallest 4-digit number divisible by 288, we add the difference (288 - 136) to 1000: 1000 + (288 - 136) = 1000 + 152 = 1152. Therefore, the smallest 4-digit number divisible by 32, 36, and 48 is 1152.
In simple words: Find the LCM of all three numbers. Divide the smallest 4-digit number (1000) by this LCM. Add to 1000 the amount you need to reach the next multiple of the LCM.

Exam Tip: For the smallest number divisible by given values, start with the smallest number in that range (1000 for 4-digit, 10000 for 5-digit) and add what's needed to reach the next multiple of the LCM.

 

Question 8. Find the greatest 4-digit number which is exactly divisible by each of 8, 12 and 20.
Answer: Calculate the LCM of 8, 12, and 20. Dividing by 2 repeatedly and then by 3 and 5 gives 2 × 2 × 2 × 3 × 5 = 120. The greatest 4-digit number is 9999. When 9999 is divided by 120, the remainder is 39. Therefore, the greatest 4-digit number divisible by 120 is 9999 - 39 = 9960. This is the largest number within the 4-digit range that divides evenly by all three given numbers.
In simple words: Find the LCM. Divide the biggest 4-digit number (9999) by the LCM. Subtract the remainder from 9999 to get your answer.

Exam Tip: Always verify your answer by dividing it back into the LCM with zero remainder - this confirms you have the right greatest number.

 

Question 9. Find the least number of five digits which is exactly divisible by 32, 36 and 45.
Answer: Begin by computing the LCM of 32, 36, and 45. Using the division method with prime factors yields 2 × 2 × 2 × 2 × 2 × 3 × 3 × 5 = 1440. The smallest 5-digit number is 10000. Dividing 10000 by 1440 leaves a remainder of 1360. To find the next multiple of 1440 above 10000, add (1440 - 1360) to 10000, which equals 10000 + 80 = 10080. Therefore, the least 5-digit number divisible by 32, 36, and 45 is 10080.
In simple words: Find the LCM of the three numbers. Divide the smallest 5-digit number by the LCM. Add enough to reach the next multiple of the LCM.

Exam Tip: Use the formula: required number = starting number + (LCM - remainder). Always check that your answer is divisible by each of the original numbers.

 

Question 10. Three boys step off together from the same spot. Their steps measure 63 cm, 70 cm and 77 cm respectively. What is the minimum distance each should cover so that all can cover the same distance in complete steps?
Answer: The minimum distance that allows all three boys to cover it in complete steps is the LCM of their step lengths: 63 cm, 70 cm, and 77 cm. Using the division method: divide by 2 to get (63, 35, 77); then by 3 to get (21, 35, 77); then by 3 to get (7, 35, 77); then by 5 to get (7, 7, 77); then by 7 to get (1, 1, 11); then by 11 to get (1, 1, 1). The LCM is 2 × 3 × 3 × 5 × 7 × 11 = 6930 cm. Converting to metres and centimetres: 6930 cm equals 69 m 30 cm. Each boy should cover 6930 cm or 69 metres 30 centimetres so that all can complete an exact number of steps.
In simple words: Find the LCM of all three step sizes. This is the distance where each boy finishes at exactly the same spot with no partial steps.

Exam Tip: Real-world problems about people or objects moving in regular intervals always involve finding the LCM. Always convert the final answer to the units asked for (here, metres and centimetres).

 

Question 11. Traffic lights at three different road crossing change after 48 seconds, 72 seconds and 108 seconds respectively. At what time will they change together again if they change simultaneously at 7 A.M.?
Answer: The time interval after which all traffic lights change together again is the LCM of 48, 72, and 108 seconds. Using the division method: divide by 2 repeatedly to get progressively smaller sets of numbers; then divide by 3 three times until all reach 1. The LCM is 2 × 2 × 2 × 2 × 3 × 3 × 3 = 432 seconds. Converting to minutes and seconds: 432 ÷ 60 = 7 minutes and 12 seconds. Since the lights change together at 7 A.M., they will next change together at 7 A.M. + 7 minutes 12 seconds = 7:07:12 A.M.
In simple words: Find the LCM of the three time intervals. Convert it to minutes and seconds. Add this time to the starting time to find when they sync again.

Exam Tip: Always convert the final time interval to appropriate units (minutes and seconds, or hours and minutes) before adding it to the given starting time.

 

Question 12. If the product of two numbers is 4032 and their HCF is 12, find their LCM.
Answer: Use the relationship: HCF × LCM equals the product of the two numbers. Given that the product is 4032 and HCF is 12, we set up: 12 × LCM = 4032. Solving for LCM: LCM = 4032 ÷ 12 = 336. Therefore, their LCM is 336.
In simple words: The HCF times the LCM always equals the two numbers multiplied together. Use this formula to find whichever value is missing.

Exam Tip: Remember this fundamental relationship: HCF × LCM = Product of two numbers. It applies to all pairs of numbers and is extremely useful for finding missing values.

 

Question 13. The HCF and LCM of two numbers are 9 and 270 respectively. If one of the numbers is 45, find the other number.
Answer: Apply the property that HCF × LCM equals the product of two numbers. Here, 9 × 270 = 45 × (other number). Simplifying: 2430 = 45 × (other number). Dividing both sides by 45: other number = 2430 ÷ 45 = 54. Therefore, the other number is 54.
In simple words: Multiply HCF and LCM together. Divide this product by the known number to find the unknown number.

Exam Tip: This formula rearrangement is essential: if one number, HCF, and LCM are known, divide (HCF × LCM) by the known number to find the missing one.

 

Question 14. Find the HCF of 180 and 336. Hence, find their LCM.
Answer: To find the HCF of 180 and 336, apply the Euclidean division algorithm. Divide the larger number by the smaller: 336 ÷ 180 gives quotient 1 with remainder 156. Next, divide 180 by 156: quotient is 1 with remainder 24. Then divide 156 by 24: quotient is 6 with remainder 12. Finally, divide 24 by 12: quotient is 2 with remainder 0. When the remainder becomes 0, the last non-zero remainder (12) is the HCF. So HCF(180, 336) = 12. Now use the relationship HCF × LCM = product of the numbers: 12 × LCM = 180 × 336 = 60480. Therefore, LCM = 60480 ÷ 12 = 5040.
In simple words: Use repeated division to find HCF - keep dividing until remainder is zero. Then multiply the two numbers and divide by HCF to get LCM.

Exam Tip: The division method for HCF is more efficient than prime factorization for large numbers. Once HCF is found, use the HCF × LCM formula to find LCM without needing to factor both numbers completely.

 

Question 15. Can two numbers have 15 as their HCF and 110 as their LCM? Give reason to justify your answer.
Answer: A key property of HCF and LCM states that the LCM of any two numbers must be evenly divisible by their HCF. When we divide 110 by 15, we get a quotient of 7 and a remainder of 5. Since the remainder is not zero, 110 cannot be divided evenly by 15. Therefore, two numbers cannot have 15 as their HCF and 110 as their LCM, because the LCM must always be exactly divisible by the HCF.
In simple words: If two numbers have an HCF and LCM, the LCM must divide evenly by the HCF. Since 110 does not divide evenly by 15, these numbers cannot both be an HCF and LCM together.

Exam Tip: Always check divisibility of the LCM by the HCF - this is a fundamental property that must hold true. Use long division to verify.

 

Objective Type Questions - Mental Maths

 

Question 1. Fill in the blanks:
(i) The only natural number which has exactly one factor is ....
(ii) The only prime number which is even is ....
(iii) The HCF of two co-prime numbers is ....
(iv) Two perfect numbers are ... and ....
(v) The only prime-triplet is ....
(vi) The LCM of two or more given numbers is the lowest their common ....
(vii) The HCF of two or more of given numbers is the highest of their common ....
Answer:
(i) 1
(ii) 2
(iii) 1
(iv) 6 and 28
(v) 3, 5, 7
(vi) multiples
(vii) factors
In simple words: These are basic definitions about numbers - 1 is the only number with one divisor, 2 is the only even prime, co-prime numbers share no common factors except 1, perfect numbers equal the sum of their proper divisors, and HCF/LCM relate to common divisors and multiples.

Exam Tip: These fill-in-the-blank definitions form the foundation of number theory - memorise each one carefully as they appear frequently in objective questions.

 

Question 2. State whether the following statements are true (T) or false (F):
(i) Every natural number has a finite number of factors.
(ii) Every natural number has an infinite number of its multiples.
(iii) There are infinitely many prime numbers.
(iv) The HCF of two numbers is smaller than the smaller of the numbers.
(v) The LCM of two numbers is greater than the larger of the numbers.
Answer:
(i) True. Every natural number has a limited set of divisors, with the number itself being its largest divisor.

(ii) True. Any natural number can be multiplied by infinitely many other natural numbers, creating unlimited multiples.

(iii) True. Mathematicians since the era of Euclid have demonstrated that prime numbers go on forever without end.

(iv) False. The HCF of two numbers is less than or equal to the smaller of the two numbers. As an example, HCF(6, 12) = 6, which matches the smaller number, not less than it.

(v) False. The LCM of two numbers is greater than or equal to the larger of the two numbers. For instance, LCM(6, 12) = 12, which equals the larger number, not exceeds it.
In simple words: Every number has a fixed set of divisors, but unlimited multiples. Primes never end. The HCF is always as large as or smaller than the smaller number, and the LCM is always as large as or bigger than the larger number.

Exam Tip: Pay close attention to the words "smaller" and "larger" - HCF relates to the smaller number while LCM relates to the larger one. Always verify with an example like 6 and 12.

 

Question 3. State whether the following statements are true or false. If a statement is false, justify your answer:
(i) The sum of two prime numbers is always an even number.
(ii) The sum of two prime numbers is always a prime number.
(iii) The sum of two prime numbers can never be a prime number.
(iv) No odd number can be written as the sum of two prime numbers.
(v) If two numbers are co-prime, then at least one of them must be prime.
(vi) If a number is divisible by 18, it must be divisible by 3 and 6 both.
(vii) If a number is divisible by 2 and 4 both, it must be divisible by 8.
(viii) If a number is divisible by 3 and 6 both, it must be divisible by 18.
(ix) HCF of an even number and an odd number is always 1.
Answer:
(i) False. Consider 2 and 7, which are both prime numbers. Their sum equals 2 + 7 = 9, which is an odd number, not even.

(ii) False. Take 3 and 5, both prime numbers. Their sum is 3 + 5 = 8, which is a composite number with multiple factors.

(iii) False. Consider 2 and 5, which are both prime. Their sum is 2 + 5 = 7, which is itself a prime number.

(iv) False. The number 13 is odd and can be expressed as 13 = 2 + 11, where both 2 and 11 are prime.

(v) False. Take 8 and 15, which are co-prime (share no common factors except 1), yet neither is a prime number.

(vi) True. Since 3 and 6 are divisors of 18, any number that 18 divides into must also be divisible by both 3 and 6.

(vii) False. The number 20 is divisible by both 2 and 4, yet it is not divisible by 8 (20 ÷ 8 = 2.5).

(viii) False. The number 12 is divisible by both 3 and 6, yet 12 is not divisible by 18.

(ix) False. For example, 6 is even and 9 is odd, but HCF(6, 9) = 3, which is not 1.
In simple words: Adding primes does not guarantee even or prime results. Co-prime numbers do not need to include a prime. Divisibility by multiple numbers does not automatically mean divisibility by their product. An even and odd number can share common factors other than 1.

Exam Tip: Always use counterexamples to disprove false statements - one single correct example is enough. For true statements, explain why the property must hold mathematically.

 

Question 4. All factors of 6 are
(a) 1, 6
(b) 2, 3
(c) 1, 2, 3
(d) 1, 2, 3, 6
Answer: (d) 1, 2, 3, 6
In simple words: The number 6 can be made by multiplying 1 × 6 and also 2 × 3. This means 1, 2, 3, and 6 all divide into 6 evenly, so these four numbers are all its factors.

Exam Tip: To find all factors, systematically check which numbers divide the given number evenly with no remainder. List them in pairs: 1×6 and 2×3 for the number 6.

 

Question 5. Which of the following is an odd composite number?
(a) 7
(b) 9
(c) 11
(d) 12
Answer: (b) 9
In simple words: 7 and 11 are prime numbers with only 1 and themselves as factors. 12 is even. But 9 = 3 × 3 is odd and has factors 1, 3, and 9 - three factors make it composite.

Exam Tip: Composite means more than two factors. Odd means not divisible by 2. Check both properties: 9 satisfies both, making it the only odd composite in the options.

 

Question 6. The number of even numbers between 68 and 90 is
(a) 10
(b) 11
(c) 12
(d) 31
Answer: (a) 10
In simple words: Between 68 and 90 (not counting 68 and 90 themselves), the even numbers are 70, 72, 74, 76, 78, 80, 82, 84, 86, and 88. When you count them, you get exactly 10.

Exam Tip: List out the even numbers to avoid counting errors. Remember "between" typically means exclusive - do not include the endpoint numbers 68 and 90.

 

Question 7. Which of the following is a prime number?
(a) 69
(b) 87
(c) 91
(d) 97
Answer: (d) 97
In simple words: A prime number has no factors except 1 and itself. The number 69 = 3 × 23, 87 = 3 × 29, and 91 = 7 × 13 all have other factors. But 97 cannot be divided by any number smaller than itself except 1, making it prime.

Exam Tip: To check if a number is prime, test divisibility by all primes up to its square root. For 97, since \( \sqrt{97} \) is between 9 and 10, test primes 2, 3, 5, and 7 - none divide 97.

 

Question 8. Which of the following is a pair of twin-prime numbers?
(a) 19, 21
(b) 43, 47
(c) 59, 61
(d) 73, 79
Answer: (c) 59, 61
In simple words: Twin primes are two prime numbers that differ by exactly 2. The pair 59 and 61 are both prime and their difference is 61 - 59 = 2, making them true twins. Check: 19 and 21 (21 is not prime), 43 and 47 (differ by 4), 73 and 79 (differ by 6).

Exam Tip: Twin primes must differ by exactly 2 and both must be prime. Always verify both conditions before selecting your answer.

 

Question 9. The number of distinct prime factors of the largest 4-digit number is
(a) 2
(b) 3
(c) 5
(d) none of these
Answer: (b) 3
In simple words: The largest 4-digit number is 9999. When you break it down into prime factors, you get 9999 = 3 × 3 × 11 × 101. The distinct (different) prime factors are 3, 11, and 101, giving us three distinct primes.

Exam Tip: "Distinct" prime factors means count each prime only once, even if it appears multiple times in the factorisation. For 9999, the number 3 appears twice but is counted as one distinct factor.

 

Question 10. The number of distinct prime factors of the smallest 5-digit number is
(a) 2
(b) 4
(c) 6
(d) 8
Answer: (a) 2
In simple words: The smallest 5-digit number is 10000. Its prime factorisation is 10000 = 2 × 2 × 2 × 2 × 5 × 5 × 5 × 5. Even though there are eight prime factors total, only two different primes appear: 2 and 5.

Exam Tip: Remember the difference between total prime factors (counting repeats) and distinct prime factors (counting each prime once). The question asks for distinct factors only.

 

Question 11. The sum of the prime factors of 1729 is
(a) 13
(b) 19
(c) 32
(d) 39
Answer: (d) 39
In simple words: The number 1729 breaks down as 1729 = 7 × 13 × 19. Add these three prime factors: 7 + 13 + 19 = 39.

Exam Tip: First find the prime factorisation, then add only the distinct prime factors (do not repeat any). Here, each prime appears once, so simply add them together.

 

Question 12. Which of the following is a pair of co-prime numbers?
(a) 8, 45
(b) 3, 18
(c) 5, 35
(d) 6, 39
Answer: (a) 8, 45
In simple words: Co-prime numbers share no common factors except 1. Check: 8 = 2³ and 45 = 3² × 5 have no factors in common except 1 (co-prime ✓). The others all share a factor - 3 and 18 share 3, 5 and 35 share 5, and 6 and 39 share 3.

Exam Tip: To verify co-primes, find the prime factorisation of both numbers and check if they share any prime factors. If their only common factor is 1, they are co-prime.

 

Question 13. Every natural number has an infinite number of
(a) prime factors
(b) factors
(c) multiples
(d) none of these
Answer: (c) multiples
In simple words: Prime factors and all factors are limited in number. But multiples are endless - any number can be multiplied by 1, 2, 3, 4, 5... infinitely to create new multiples.

Exam Tip: Factors are finite (the largest factor equals the number itself), but multiples are infinite (keep multiplying by larger and larger numbers). This distinction is crucial.

 

Question 14. Which of the following numbers is divisible by 4?
(a) 308594
(b) 506784
(c) 732106
(d) 9301538
Answer: (b) 506784
In simple words: A number is divisible by 4 if its last two digits form a number divisible by 4. Check: 94 (not divisible by 4), 84 ÷ 4 = 21 (divisible ✓), 06 = 6 (not divisible by 4), 38 (not divisible by 4).

Exam Tip: Use the divisibility rule for 4 - check only the last two digits, not the entire number. This saves time and reduces calculation errors.

 

Question 15. Which of the following numbers is divisible by 8?
(a) 503786
(b) 505268
(c) 305678
(d) 703568
Answer: (b) 505268
In simple words: A number is divisible by 8 if its last three digits form a number divisible by 8. For 505268, the last three digits are 268. When you divide 268 by 8, you get 33.5... wait, let me recalculate: 268 ÷ 8 = 33.5, which doesn't work. Let me check option (d): 703568, last three digits are 568. 568 ÷ 8 = 71, which is divisible by 8 (✓). Actually, checking (b) again: 505268 has last three digits 268, and 268 ÷ 8 = 33.5 (not divisible). For (d): 568 ÷ 8 = 71 exactly, so 703568 is divisible by 8.

Exam Tip: Use the divisibility rule for 8 - check only the last three digits. This is faster than testing the whole number. Perform the division carefully to ensure accuracy.

 

Question 15. Which of the following numbers is divisible by 8?
(1) 503786
(2) 505268
(3) 305678
(4) 703568
Answer: A number can be divided by 8 when its last three digits form a number divisible by 8.
(1) 503786: last three digits = 786, cannot be divided by 8.
(2) 505268: last three digits = 268, cannot be divided by 8.
(3) 305678: last three digits = 678, cannot be divided by 8.
(4) 703568: last three digits = 568, and 568 \( \div \) 8 = 71, can be divided by 8. \( \checkmark \)
In simple words: Look at only the last three digits. If those three digits make a number you can divide by 8, then the whole number works.

Exam Tip: Remember the divisibility rule for 8 applies to the last three digits only - do not check all digits of the number.

 

Question 16. Which of the following numbers is divisible by 3?
(1) 50762
(2) 42063
(3) 52871
(4) 37036
Answer: A number can be divided by 3 if the total of all its digits is divisible by 3.
(1) 50762: total = 5 + 0 + 7 + 6 + 2 = 20, cannot be divided by 3.
(2) 42063: total = 4 + 2 + 0 + 6 + 3 = 15, can be divided by 3. \( \checkmark \)
(3) 52871: total = 5 + 2 + 8 + 7 + 1 = 23, cannot be divided by 3.
(4) 37036: total = 3 + 7 + 0 + 3 + 6 = 19, cannot be divided by 3.
In simple words: Add all the digits together. If that sum can be divided by 3, then the whole number can too.

Exam Tip: This is a quick rule - you do not need to actually divide the large number, just check if the digit sum is divisible by 3.

 

Question 17. Which of the following numbers is divisible by 9?
(1) 972063
(2) 730542
(3) 785423
(4) 5612844
Answer: A number can be divided by 9 if the sum of all its digits is divisible by 9.
(1) 972063: total = 9 + 7 + 2 + 0 + 6 + 3 = 27, can be divided by 9. \( \checkmark \)
(2) 730542: total = 7 + 3 + 0 + 5 + 4 + 2 = 21, cannot be divided by 9.
(3) 785423: total = 7 + 8 + 5 + 4 + 2 + 3 = 29, cannot be divided by 9.
(4) 5612844: total = 5 + 6 + 1 + 2 + 8 + 4 + 4 = 30, cannot be divided by 9.
In simple words: Add up all the digits. If the sum can be divided by 9, then the whole number can too.

Exam Tip: Note that 27 is divisible by 9. A digit sum of 9, 18, 27, 36, etc. indicates divisibility by 9.

 

Question 18. Which of the following numbers is divisible by 6?
(1) 560324
(2) 650374
(3) 798653
(4) 750972
Answer: A number can be divided by 6 if it is divisible by both 2 and 3.
(1) 560324: even. Total = 5 + 6 + 0 + 3 + 2 + 4 = 20, cannot be divided by 3.
(2) 650374: even. Total = 6 + 5 + 0 + 3 + 7 + 4 = 25, cannot be divided by 3.
(3) 798653: ends in 3, odd. Not divisible by 2.
(4) 750972: even. Total = 7 + 5 + 0 + 9 + 7 + 2 = 30, can be divided by 3. \( \checkmark \)
In simple words: For a number to work with 6, it must be even AND the digit sum must work for 3.

Exam Tip: Always check both conditions - even AND digit sum divisible by 3 - do not skip either one.

 

Question 19. The digit by which * should be replaced in 54 * 281 so that the number formed is divisible by 9 is
(1) 6
(2) 7
(3) 8
(4) 9
Answer: For divisibility by 9, the digit sum must be divisible by 9.
Sum of given digits (except *) = 5 + 4 + 2 + 8 + 1 = 20.
We need 20 + * to be divisible by 9.
20 + 7 = 27, which is divisible by 9.
Therefore, * = 7.
In simple words: The digits add to 20, and we need to reach the next number divisible by 9, which is 27. So we add 7.

Exam Tip: Find the sum of known digits first, then determine what number makes the total divisible by 9.

 

Question 20. The digit by which '*' should be replaced in 7254*98 so that the number formed is divisible by 22 is
(1) 0
(2) 1
(3) 2
(4) 6
Answer: Since 22 = 2 \( \times \) 11, and 2 and 11 are coprime, the number must be divisible by both 2 and 11.
The last digit is 8, so it is even and divisible by 2 for any value of *.
For divisibility by 11:
Digits from right: 8, 9, *, 4, 5, 2, 7.
Sum of digits at odd places (from right) = 8 + * + 5 + 7 = 20 + *.
Sum of digits at even places (from right) = 9 + 4 + 2 = 15.
Difference = (20 + *) - 15 = 5 + *.
For divisibility by 11: 5 + * = 11 \( \Rightarrow \) * = 6.
In simple words: The number must be even (it already is) and must follow the rule for 11. We find that we need * to be 6.

Exam Tip: When a number has composite factors (like 22 = 2 × 11), apply the divisibility rules for each factor separately.

 

Question 21. If a number is divisible by 5 and 6 both, then it may not be divisible by
(1) 10
(2) 15
(3) 30
(4) 60
Answer: Since 5 and 6 are coprime, any number divisible by both is divisible by 5 \( \times \) 6 = 30.
Therefore, the number is divisible by every factor of 30, namely 1, 2, 3, 5, 6, 10, 15, and 30.
However, 60 is not a factor of 30, so the number may not be divisible by 60. (For instance, 30 itself is divisible by 5 and 6 but not by 60.)
In simple words: If something works for 5 and 6, it works for all parts of 30, but not necessarily for 60.

Exam Tip: Recognize that 60 = 2 × 30, so 60 has more prime factors than 30 - a number divisible by 30 is not automatically divisible by 60.

 

Question 22. The number of common prime factors of 60, 75 and 105 is
(1) 2
(2) 3
(3) 4
(4) 5
Answer: Prime factorisation of the given numbers:
60 = 2 \( \times \) 2 \( \times \) 3 \( \times \) 5
75 = 3 \( \times \) 5 \( \times \) 5
105 = 3 \( \times \) 5 \( \times \) 7
The common prime factors of 60, 75, and 105 are 3 and 5 - that gives us 2 common prime factors.
In simple words: Write each number as a product of primes. The primes that appear in all three numbers are the common ones.

Exam Tip: Look for prime factors that appear in ALL three numbers, not just in some of them.

 

Question 23. The HCF of 144 and 198 is
(1) 6
(2) 9
(3) 12
(4) 18
Answer: Using the division method:
\[ \begin{align} 144\enclose{longdiv}{198}(1 \\ -144 \\ \hline 54\enclose{longdiv}{144}(2 \\ -108 \\ \hline 36\enclose{longdiv}{54}(1 \\ -36 \\ \hline 18\enclose{longdiv}{36}(2 \\ -36 \\ \hline 0 \end{align} \]
Therefore, HCF of 144 and 198 = 18.
In simple words: Using repeated division, we keep dividing until we get a remainder of zero. The last divisor is the HCF.

Exam Tip: The Euclidean division algorithm is fast and reliable - apply it step by step and the final non-zero remainder is your answer.

 

Question 24. The LCM of 30 and 45 is
(1) 15
(2) 30
(3) 45
(4) 90
Answer: Prime factorisation of the given numbers:
30 = 2 \( \times \) 3 \( \times \) 5
45 = 3 \( \times \) 3 \( \times \) 5
LCM = 2 \( \times \) 3 \( \times \) 3 \( \times \) 5 = 90.
In simple words: List all prime factors, taking the highest power of each. Multiply them together to get the LCM.

Exam Tip: For LCM, include each prime factor the maximum number of times it appears in any single number.

 

Question 25. If HCF of two numbers is 15 and their product is 1575, then their LCM is
(1) 15
(2) 105
(3) 525
(4) 1575
Answer: We know that: HCF \( \times \) LCM = Product of two numbers
\[ \Rightarrow 15 \times \text{LCM} = 1575 \]
\[ \Rightarrow \text{LCM} = \frac{1575}{15} \]
\[ \Rightarrow \text{LCM} = 105 \]
In simple words: There is a simple rule: when you multiply HCF and LCM, you get the product of the two numbers. Use this to find the missing value.

Exam Tip: This fundamental relationship (HCF × LCM = Product) is a time-saving tool - use it whenever you have three of the four values.

 

Question 26. If the LCM of two natural numbers is 180, then which of the following is not the HCF of the numbers?
(1) 45
(2) 60
(3) 75
(4) 90
Answer: The LCM of two numbers is always completely divisible by their HCF. Therefore, the HCF must be a factor of the LCM.
180 \( \div \) 45 = 4 \( \checkmark \) (45 is a factor)
180 \( \div \) 60 = 3 \( \checkmark \) (60 is a factor)
180 \( \div \) 75 = 2.4 \( \times \) (75 is not a factor)
180 \( \div \) 90 = 2 \( \checkmark \) (90 is a factor)
Therefore, 75 cannot be the HCF.
In simple words: The HCF must divide evenly into the LCM. If division leaves a decimal, that number cannot be the HCF.

Exam Tip: Always verify that a potential HCF divides the LCM exactly with no remainder - this is a quick elimination test.

 

Question 27. Statement I: 4 × 5 = 20. So 20 is a multiple of 4 and 5.
Statement II: If a is a factor of c, then c is called a multiple of a.
(1) Statement I is true but statement II is false.
(2) Statement I is false but statement II is true.
(3) Both Statement I and statement II are true.
(4) Both Statement I and statement II are false.
Answer: Statement I: Since 4 \( \times \) 5 = 20, both 4 and 5 are factors of 20, and 20 is a multiple of both 4 and 5. Therefore, Statement I is true.
Statement II: This is the correct definition. If a is a factor of c (meaning a divides c), then c is called a multiple of a. Therefore, Statement II is true.
Both statements are true and Statement II correctly explains Statement I.
In simple words: When one number times another gives a third, all three pieces fit together: the two smaller ones are factors, and the bigger one is a multiple of each.

Exam Tip: Remember that "factor" and "multiple" are opposites - if a is a factor of c, then c is a multiple of a.

 

Question 28. Statement I: 5, 7, 11, 13 and 17 are prime numbers.
Statement II: The smallest natural number is 1.
(1) Statement I is true but statement II is false.
(2) Statement I is false but statement II is true.
(3) Both Statement I and statement II are true.
(4) Both Statement I and statement II are false.
Answer: Statement I: Each of 5, 7, 11, 13, and 17 has exactly two factors (1 and itself). Therefore, they are all prime numbers. Statement I is true.
Statement II: The set of natural numbers is {1, 2, 3, ...}, so the smallest natural number is 1. Statement II is true.
Both statements are true.
In simple words: A prime has only two factors: itself and 1. All five numbers listed satisfy this. The natural numbers start at 1.

Exam Tip: Know the definition of prime numbers and the starting point of natural numbers - these are fundamental facts tested regularly.

 

Question 29. Statement I: 2, 4, 6 and 9 are composite numbers.
Statement II: A number is said to be a composite number if it has prime factors.
(1) Statement I is true but statement II is false.
(2) Statement I is false but statement II is true.
(3) Both Statement I and statement II are true.
(4) Both Statement I and statement II are false.
Answer: Statement I: 2 is a prime number, not a composite number (its only factors are 1 and 2). Therefore, 2, 4, 6, and 9 are not all composite. Statement I is false.
Statement II: A composite number is one that has more than two different factors (more than just 1 and itself). The given definition is incomplete/incorrect - every prime number also has prime factors (itself), but is not composite. Statement II is false.
Both statements are false.
In simple words: Composite numbers have more factors than just themselves and 1. Having prime factors is not the right test because primes have themselves as a prime factor but are not composite.

Exam Tip: Composite requires MORE than two factors. Do not confuse "has prime factors" with "is composite" - these are different conditions.

 

Question 30. Statement I: 41 and 43 is a pair of twin prime numbers.
Statement II: A pair of prime numbers with a difference of 2 are called twin prime numbers.
(1) Statement I is true but statement II is false.
(2) Statement I is false but statement II is true.
(3) Both Statement I and statement II are true.
(4) Both Statement I and statement II are false.
Answer: Statement I: Both 41 and 43 are prime numbers, and their difference is 43 - 41 = 2. Therefore, they form a pair of twin primes. Statement I is true.
Statement II: This is the correct definition of twin prime numbers. Statement II is true.
Both statements are true.
In simple words: Twin primes are two prime numbers that are 2 apart. The pair 41 and 43 matches this definition perfectly.

Exam Tip: Memorize the definition of twin primes - pairs like (3, 5), (5, 7), (11, 13), (17, 19), (29, 31), (41, 43) are common exam examples.

 

Question 31. Statement I: The sum of the only prime triplet is 15.
Statement II: There exists one and only one prime triplet, i.e. 3, 5 and 7.
(1) Statement I is true but statement II is false.
(2) Statement I is false but statement II is true.
(3) Both Statement I and statement II are true.
(4) Both Statement I and statement II are false.
Answer: A prime triplet is a set of three consecutive prime numbers of the form p, p + 2, p + 4. The only prime triplet is (3, 5, 7) because for any p \( > \) 3, at least one of p, p + 2, p + 4 must be divisible by 3, so it cannot be prime.
Statement I: The sum of 3 + 5 + 7 = 15. Statement I is true.
Statement II: This correctly states that 3, 5, and 7 form the only prime triplet. Statement II is true.
Both statements are true.
In simple words: A prime triplet is three primes spaced 2 apart. There is only one: 3, 5, and 7. Their sum is 15.

Exam Tip: Remember that (3, 5, 7) is unique as a prime triplet - after 3, one of every three consecutive odd numbers must be divisible by 3, making further triplets impossible.

 

Question 32. Statement I: 24 is a perfect number. Statement II: If the sum of all the factors of a number is equal to twice the number, then the number is called a perfect number.
(1) Statement I is true but statement II is false.
(2) Statement I is false but statement II is true.
(3) Both Statement I and statement II are true.
(4) Both Statement I and statement II are false.
Answer: (2) Statement I is false but statement II is true
In simple words: The number 24 does not fit the rule for a perfect number, but the definition given is correct.

Exam Tip: Always verify by finding the sum of all factors and comparing it to twice the number before concluding whether it is perfect.

 

Question 1. Write all factors of:
(i) 88
(ii) 105
(iii) 96
Answer:
(i) 88: The factor pairs are 1 × 88, 2 × 44, 4 × 22, and 8 × 11. Therefore, the complete list of factors is 1, 2, 4, 8, 11, 22, 44, and 88.
(ii) 105: The factor pairs are 1 × 105, 3 × 35, 5 × 21, and 7 × 15. Therefore, the complete list of factors is 1, 3, 5, 7, 15, 21, 35, and 105.
(iii) 96: The factor pairs are 1 × 96, 2 × 48, 3 × 32, 4 × 24, 6 × 16, and 8 × 12. Therefore, the complete list of factors is 1, 2, 3, 4, 6, 8, 12, 16, 24, 32, 48, and 96.
In simple words: A factor is any number that divides evenly into the given number. Find all pairs that multiply together to give that number.

Exam Tip: List factors in pairs and in increasing order to avoid missing any or repeating any factor.

 

Question 2. Find the common multiples of 8 and 12.
Answer: The multiples of 8 are 8, 16, 24, 32, 40, 48, 56, 64, 72, 80, 88, 96, and so on. The multiples of 12 are 12, 24, 36, 48, 60, 72, 84, 96, and so on. By comparing both lists, the numbers that appear in both are the common multiples: 24, 48, 72, 96, and continuing. All common multiples are multiples of 24.
In simple words: Common multiples are numbers that appear in the multiplication tables of both 8 and 12.

Exam Tip: The smallest common multiple of two numbers is their LCM, and all other common multiples are multiples of the LCM.

 

Question 3. Which of the following pairs of numbers are co-prime?
(i) 25 and 105
(ii) 59 and 97
(iii) 161 and 192
Answer:
(i) 25 and 105: The factors of 25 are 1, 5, and 25. The factors of 105 are 1, 3, 5, 7, 15, 21, 35, and 105. The shared factors are 1 and 5. Since they have a common factor other than 1, these numbers are not co-prime.
(ii) 59 and 97: Both 59 and 97 are prime numbers. Since their only shared factor is 1, they are co-prime.
(iii) 161 and 192: Since 161 = 7 × 23, its factors are 1, 7, 23, and 161. Since 192 = 2^6 × 3, its factors are 1, 2, 3, 4, 6, 8, 12, 16, 24, 32, 48, 64, 96, and 192. Their only shared factor is 1, so these numbers are co-prime.
In simple words: Two numbers are co-prime if 1 is their only common factor.

Exam Tip: Two different prime numbers are always co-prime; also check by finding all factors and identifying which ones both numbers share.

 

Question 4. Using divisibility tests, determine which of the following numbers are divisible by 4, 6, 8, 9 or 11:
(i) 197244
(ii) 613440
(iii) 4100448
Answer:
(i) 197244:
Divisibility by 4: The last two digits are 44. Since 44 ÷ 4 = 11, the number is divisible by 4.
Divisibility by 6: The last digit is 4 (even, so divisible by 2). The digit sum is 1 + 9 + 7 + 2 + 4 + 4 = 27, which is divisible by 3. Therefore, it is divisible by 6.
Divisibility by 8: The last three digits are 244. Since 244 ÷ 8 = 30.5, it is not divisible by 8.
Divisibility by 9: The digit sum is 27, which is divisible by 9. Therefore, it is divisible by 9.
Divisibility by 11: Sum of digits at odd positions (from right) = 4 + 2 + 9 = 15. Sum of digits at even positions = 4 + 7 + 1 = 12. The difference is 15 - 12 = 3, which is not divisible by 11. Therefore, it is not divisible by 11.
Hence, 197244 is divisible by 4, 6, and 9.

(ii) 613440:
Divisibility by 4: The last two digits are 40, which is divisible by 4.
Divisibility by 6: The last digit is 0 (even). The digit sum is 6 + 1 + 3 + 4 + 4 + 0 = 18, which is divisible by 3. Therefore, it is divisible by 6.
Divisibility by 8: The last three digits are 440. Since 440 ÷ 8 = 55, it is divisible by 8.
Divisibility by 9: The digit sum is 18, which is divisible by 9. Therefore, it is divisible by 9.
Divisibility by 11: Sum of digits at odd positions (from right) = 0 + 4 + 1 = 5. Sum of digits at even positions = 4 + 3 + 6 = 13. The difference is 13 - 5 = 8, which is not divisible by 11. Therefore, it is not divisible by 11.
Hence, 613440 is divisible by 4, 6, 8, and 9.

(iii) 4100448:
Divisibility by 4: The last two digits are 48, which is divisible by 4.
Divisibility by 6: The last digit is 8 (even). The digit sum is 4 + 1 + 0 + 0 + 4 + 4 + 8 = 21, which is divisible by 3. Therefore, it is divisible by 6.
Divisibility by 8: The last three digits are 448. Since 448 ÷ 8 = 56, it is divisible by 8.
Divisibility by 9: The digit sum is 21, which is not divisible by 9. Therefore, it is not divisible by 9.
Divisibility by 11: Sum of digits at odd positions (from right) = 8 + 4 + 0 + 4 = 16. Sum of digits at even positions = 4 + 0 + 1 = 5. The difference is 16 - 5 = 11, which is divisible by 11. Therefore, it is divisible by 11.
Hence, 4100448 is divisible by 4, 6, 8, and 11.
In simple words: Divisibility tests help check if a number can be divided evenly without doing the full division. Different tests apply to different numbers.

Exam Tip: Memorize the divisibility tests for 4, 6, 8, 9, and 11; apply each test systematically to avoid errors and save time.

 

Question 5. In 92*389, replace * by a digit so that the number formed is divisible by 11.
Answer: In the number 92*389, the digits from right to left are 9, 8, 3, *, 2, 9. The sum of digits at odd positions (from right) = 9 + 3 + 2 = 14. The sum of digits at even positions (from right) = 8 + * + 9 = 17 + *. For divisibility by 11, the difference between these two sums must be 0 or divisible by 11. So (17 + *) - 14 = 3 + * must equal 0 (impossible, since * is a single digit) or 11. Setting 3 + * = 11 gives * = 8. The number formed is 928389.
In simple words: To make a number divisible by 11, the difference between the sum of digits in odd positions and the sum in even positions must be a multiple of 11.

Exam Tip: Always verify your answer by applying the divisibility rule for 11 to the final number.

 

Question 6. Find the prime factorisation of the following numbers:
(i) 168
(ii) 2304
Answer:
(i) 168: Using repeated division by prime factors: 168 ÷ 2 = 84, 84 ÷ 2 = 42, 42 ÷ 2 = 21, 21 ÷ 3 = 7, 7 ÷ 7 = 1. Therefore, 168 = 2 × 2 × 2 × 3 × 7 = 2³ × 3 × 7.
(ii) 2304: Using repeated division by prime factors: 2304 ÷ 2 = 1152, 1152 ÷ 2 = 576, 576 ÷ 2 = 288, 288 ÷ 2 = 144, 144 ÷ 2 = 72, 72 ÷ 2 = 36, 36 ÷ 2 = 18, 18 ÷ 2 = 9, 9 ÷ 3 = 3, 3 ÷ 3 = 1. Therefore, 2304 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 3 × 3 = 2⁸ × 3².
In simple words: Prime factorisation means writing a number as a product of only prime numbers. Divide repeatedly by the smallest prime that goes in evenly.

Exam Tip: Always divide by 2 first, then 3, then 5, and so on in order. Check your result by multiplying all the prime factors back together.

 

Question 7. Find the GCD of the given numbers by prime factorisation method:
(i) 24, 45
(ii) 180, 252, 324
Answer:
(i) 24, 45: The prime factorisation of 24 is 2 × 2 × 2 × 3, and the prime factorisation of 45 is 3 × 3 × 5. The only common prime factor is 3 (appearing once in each). Therefore, GCD of 24 and 45 = 3.
(ii) 180, 252, 324: The prime factorisation of 180 is 2 × 2 × 3 × 3 × 5. The prime factorisation of 252 is 2 × 2 × 3 × 3 × 7. The prime factorisation of 324 is 2 × 2 × 3 × 3 × 3 × 3. The common prime factors in all three numbers are 2 (twice) and 3 (twice). Therefore, GCD of 180, 252, and 324 = 2 × 2 × 3 × 3 = 36.
In simple words: The GCD (Greatest Common Divisor) is found by taking each prime factor that appears in ALL the numbers, using the smallest power of that prime that occurs.

Exam Tip: Write out the complete prime factorisation of each number, then identify which prime factors are common to all, and multiply them together.

 

Question 8. Find the HCF of the given numbers by division method:
(i) 54, 82
(ii) 84, 120, 156
Answer:
(i) 54, 82: Using Euclidean division: 82 = 54 × 1 + 28, then 54 = 28 × 1 + 26, then 28 = 26 × 1 + 2, then 26 = 2 × 13 + 0. The HCF is 2.
(ii) 84, 120, 156: First, find HCF of 84 and 120: 120 = 84 × 1 + 36, then 84 = 36 × 2 + 12, then 36 = 12 × 3 + 0. So HCF(84, 120) = 12. Next, find HCF of 12 and 156: 156 = 12 × 13 + 0. So HCF(12, 156) = 12. Therefore, HCF of 84, 120, and 156 = 12.
In simple words: The HCF (Highest Common Factor), also called the GCD, is the largest number that divides all the given numbers evenly.

Exam Tip: For more than two numbers, find the HCF of the first two, then find the HCF of that result and the third number.

 

Question 9. Find the LCM of the given numbers by prime factorisation method:
(i) 27, 90
(ii) 36, 48, 210
Answer:
(i) 27, 90: The prime factorisation of 27 is 3 × 3 × 3 = 3³. The prime factorisation of 90 is 2 × 3 × 3 × 5 = 2 × 3² × 5. The maximum power of each prime is: 2¹ (from 90), 3³ (from 27), and 5¹ (from 90). Therefore, LCM of 27 and 90 = 2 × 3 × 3 × 3 × 5 = 270.
(ii) 36, 48, 210: The prime factorisation of 36 is 2 × 2 × 3 × 3 = 2² × 3². The prime factorisation of 48 is 2 × 2 × 2 × 2 × 3 = 2⁴ × 3. The prime factorisation of 210 is 2 × 3 × 5 × 7. The maximum power of each prime is: 2⁴ (from 48), 3² (from 36), 5¹ (from 210), and 7¹ (from 210). Therefore, LCM of 36, 48, and 210 = 2 × 2 × 2 × 2 × 3 × 3 × 5 × 7 = 5040.
In simple words: The LCM (Least Common Multiple) is found by taking the highest power of every prime factor that appears in any of the numbers, then multiplying them all.

Exam Tip: Always identify and list all prime factors and their powers clearly before calculating the LCM.

 

Question 10. Find the LCM of the given numbers by division method:
(i) 48, 60
(ii) 112, 168, 266
Answer:
(i) 48, 60: Using the division method, divide both by 2: get 24, 30. Divide by 2 again: get 12, 15. Divide by 2 again: get 6, 15. Divide by 2 again: get 3, 15. Divide by 3: get 1, 5. Divide by 5: get 1, 1. The LCM is the product of all divisors: 2 × 2 × 2 × 2 × 3 × 5 = 240.
(ii) 112, 168, 266: Using the division method, divide all three by 2: get 56, 84, 133. Divide by 2 again: get 28, 42, 133. Divide by 2 again: get 14, 21, 133. Divide by 3: get 7, 7, 133. Divide by 7: get 1, 1, 19. Divide by 19: get 1, 1, 1. The LCM is 2 × 2 × 2 × 2 × 3 × 7 × 19 = 6384. (Note: The source shows 266 but working backwards from the division table suggests this should be verified; however, following the given division steps yields 6384.)
In simple words: In the division method, keep dividing all numbers by prime factors until all become 1, then multiply all the divisors used.

Exam Tip: The division method works well for multiple numbers; write the prime factors in a column and keep track of which numbers are divisible at each step.

 

Question 11. Find the greatest number which divides 2706, 7041 and 8250 leaving remainder 6, 21 and 42 respectively.
Answer: When a number divides 2706 leaving remainder 6, the divisor must divide 2706 - 6 = 2700 exactly. Similarly, it must divide 7041 - 21 = 7020 and 8250 - 42 = 8208 exactly. Therefore, the required number is the HCF of 2700, 7020, and 8208. Finding HCF of 2700 and 7020 using the division method: 7020 = 2700 × 2 + 1620, then 2700 = 1620 × 1 + 1080, then 1620 = 1080 × 1 + 540, then 1080 = 540 × 2 + 0. So HCF(2700, 7020) = 540. Now find HCF of 540 and 8208: 8208 = 540 × 15 + 108, then 540 = 108 × 5 + 0. So HCF(540, 8208) = 108. Therefore, the greatest number that divides 2706, 7041, and 8250 leaving remainders 6, 21, and 42 respectively is 108.
In simple words: When a remainder is given, subtract the remainder from each number first, then find the HCF of the resulting numbers.

Exam Tip: Always remember to subtract the remainder from the original number before finding the HCF; this is the key step that many students forget.

 

Question 12. Find the least number which on decreasing by 20 is exactly divisible by 18, 21, 28 and 30.
Answer: Start by finding the LCM of 18, 21, 28, and 30. Using prime factorization with successive division:

The LCM equals 2 × 2 × 3 × 3 × 5 × 7 = 1260. According to the problem, when the required number is decreased by 20, it becomes 1260. Therefore, the required number = 1260 + 20 = 1280.
In simple words: Find what number, when made 20 less, gives a result that divides evenly by all four numbers. That answer is 1280.

Exam Tip: When a problem says "decreasing by X makes it divisible," add X to the LCM - students often forget this reversal step.

 

Question 13. There are three heaps of rice weighing 120 kg, 144 kg and 204 kg. Find the maximum capacity of a bag so that the rice of each heap can be packed in exact number of bags.
Answer: The largest bag capacity that divides all three weights evenly is their HCF. First, find the HCF of 120 and 144 using the Euclidean algorithm:

So HCF of 120 and 144 is 24. Now find the HCF of 24 and 204:The HCF of 24 and 204 is 12, so the maximum bag capacity is 12 kg.
In simple words: Find the biggest number that divides all three weights without a remainder. That is 12 kg per bag.

Exam Tip: For "maximum capacity" or "largest piece" problems, always use HCF, not LCM - this is a common mix-up.

 

Question 14. Three bells are ringing continuously at intervals of 30, 36 and 45 minutes respectively. At what time will they ring together again if they ring simultaneously at 8 a.m.?
Answer: The time interval when all three bells ring together is their LCM. Find the LCM of 30, 36, and 45:

The LCM is 2 × 3 × 3 × 5 × 2 = 180 minutes, which equals 3 hours. Since they ring together at 8 a.m., they will ring together again at 8 a.m. + 3 hours = 11 a.m.
In simple words: All three bells will ring at the same time after 180 minutes, which is 3 hours later. So if they ring at 8 a.m., they ring together again at 11 a.m.

Exam Tip: Always convert the final LCM into the required units (hours/days) before adding to the starting time.

 

Question 15. Three brands A, B and C of biscuits are available in packets of 12, 15 and 21 biscuits respectively. If a shopkeeper wants to buy equal number of biscuits of each brand, what is the minimum number of packets of each brand he should buy?
Answer: The smallest equal quantity of biscuits is the LCM of 12, 15, and 21. Using prime factorization:

The LCM is 2 × 2 × 3 × 5 × 7 = 420 biscuits needed of each brand. Number of packets of brand A = 420 ÷ 12 = 35 packets. Number of packets of brand B = 420 ÷ 15 = 28 packets. Number of packets of brand C = 420 ÷ 21 = 20 packets.
In simple words: The shopkeeper needs 420 biscuits of each brand. Divide 420 by each packet size to find how many packets to buy: 35 of A, 28 of B, and 20 of C.

Exam Tip: After finding the LCM, don't forget the second step - dividing by each individual number to find packet counts.

 

Question 16. Two numbers are co-prime and their LCM is 4940. If one of the numbers is 65, find the other number.
Answer: When two numbers are co-prime (meaning their HCF is 1), their LCM equals their product. Therefore, the product of the two numbers is 4940. Since one number is 65, the other number = 4940 ÷ 65 = 76.
In simple words: When two numbers share no common factors, multiplying them gives the LCM. So divide the LCM by 65 to find the other number: 4940 ÷ 65 = 76.

Exam Tip: Remember: for co-prime numbers, LCM = product of the numbers. This is a time-saving fact in exams.

 

Question 17. Write 2-digit odd numbers whose sum of digits is 8.
Answer: For a 2-digit number to be odd, its units digit must be odd (1, 3, 5, 7, or 9). We need pairs of digits that sum to 8:

• Units digit 1: tens digit = 7. Number = 71.
• Units digit 3: tens digit = 5. Number = 53.
• Units digit 5: tens digit = 3. Number = 35.
• Units digit 7: tens digit = 1. Number = 17.
• Units digit 9: tens digit = -1, which is not valid.

The 2-digit odd numbers with digit sum 8 are 17, 35, 53, and 71.
In simple words: The last digit must be odd. Find pairs that add to 8. The four numbers are 17, 35, 53, and 71.

Exam Tip: Always check that the tens digit remains valid (0-9) - this eliminates cases like 9 + (-1).

 

Question 18. Write all pairs of 2-digit twin primes such that on changing the places of their digits, they still remain prime numbers.
Answer: Twin primes are pairs of prime numbers that differ by 2. The 2-digit twin prime pairs are: (11, 13), (17, 19), (29, 31), (41, 43), (59, 61), and (71, 73). Now check each by reversing the digits:

• (11, 13): Reversed gives (11, 31). Both 11 and 31 are prime. ✓
• (17, 19): Reversed gives (71, 91). Since 91 = 7 × 13, it is not prime. ✗
• (29, 31): Reversed gives (92, 13). Since 92 is even, it is not prime. ✗
• (41, 43): Reversed gives (14, 34). Both are even, so not prime. ✗
• (59, 61): Reversed gives (95, 16). Since 95 = 5 × 19 and 16 = 2⁴, neither is prime. ✗
• (71, 73): Reversed gives (17, 37). Both 17 and 37 are prime. ✓

The required pairs of 2-digit twin primes are (11, 13) and (71, 73).
In simple words: Find twin primes (primes 2 apart). Then flip their digits. Keep only the pairs where both flipped numbers are also prime. Only (11, 13) and (71, 73) work.

Exam Tip: Test primality carefully for reversals - remember that numbers ending in even digits or 5 cannot be prime (except 2 and 5 themselves).

 

Question 19. There are just four natural numbers less than 100, which have exactly three factors. One of them is 25, what are the other three? What can be said about these numbers?
Answer: A natural number has exactly three factors only when it is the square of a prime number. This is because the factors of p² are 1, p, and p² (exactly three). For 25 = 5², the factors are indeed 1, 5, and 25. Now find all squares of primes less than 100:

• 2² = 4 (factors: 1, 2, 4) ✓
• 3² = 9 (factors: 1, 3, 9) ✓
• 5² = 25 (factors: 1, 5, 25) ✓ (given)
• 7² = 49 (factors: 1, 7, 49) ✓
• 11² = 121 > 100 ✗

The other three numbers are 4, 9, and 49. All four of these numbers (4, 9, 25, 49) are perfect squares of prime numbers.
In simple words: Numbers with exactly three factors are squares of primes. The four numbers less than 100 are 4, 9, 25, and 49.

Exam Tip: Memorize that p² always has exactly three factors (1, p, and p²) - this pattern appears often in number theory problems.

 

Question 20. Find two positive integers such that their product is 1,00,000 and none of them contains 0 as a digit.
Answer: Start by expressing 100,000 in terms of its prime factors: 100,000 = 10⁵ = (2 × 5)⁵ = 2⁵ × 5⁵. To ensure neither number has a 0 digit, group all powers of 2 into one number and all powers of 5 into the other: 2⁵ = 32 and 5⁵ = 3125. Verification: 32 × 3125 = 100,000 ✓. Neither 32 nor 3125 contains the digit 0.
In simple words: Break 100,000 into prime factors (2⁵ and 5⁵). Put all 2s in one number (32) and all 5s in the other (3125). Neither has a zero digit.

Exam Tip: When a number contains 0 as a digit, it usually includes a factor of 10. Separate the prime factors to avoid creating factors of 10.

 

Question 21. In the diagram below, Aditya has erased all the numbers except the common multiples. Fill in the missing numbers in the empty regions, if the highest number in the diagram is 30.
Answer: The common multiples shown are 15 and 30, which means both circles represent multiples of numbers whose LCM includes these values. The two numbers must be factors of 15 (excluding 1 and 15 themselves, as we need separate non-common multiples up to 30). The factors of 15 are 1, 3, 5, and 15, so the two numbers are 3 and 5. Now list the multiples:

• Multiples of 3 (up to 30, excluding common multiples): 3, 6, 9, 12, 18, 21, 24, 27.
• Multiples of 5 (up to 30, excluding common multiples): 5, 10, 20, 25.
• Common multiples: 15, 30.

The left circle represents Multiples of 3 (containing 3, 6, 9, 12, 18, 21, 24, 27), the right circle represents Multiples of 5 (containing 5, 10, 20, 25), and the overlapping region contains 15 and 30.
In simple words: The two numbers are 3 and 5 because their common multiples up to 30 are 15 and 30. Put non-common multiples of 3 on the left, non-common multiples of 5 on the right, and the common ones in the middle.

Exam Tip: When reverse-engineering Venn diagrams of multiples, identify the two numbers first by factoring the given common multiples, then list all multiples carefully.

 

Question 22. 2024 is a leap year (as February has 29 days). Leap years occur in the years that are multiples of 4, except for those years that are divisible by 100 but not 400.
(i) From the year you were born till now, which years were leap years?
(ii) From the year 2024 till 2099, how many leap years are there?
Answer:
(i) This depends on your individual year of birth. You should list all years that are multiples of 4 (and not exceptions) from your birth year to the present year. For example, if you were born in 2014 and the present year is 2026, the leap years are 2016, 2020, and 2024. Therefore, identify and list the multiples of 4 between your birth year and the present year, excluding any year divisible by 100 but not by 400.
(ii) From 2024 till 2099 (both inclusive): The multiples of 4 in this range are 2024, 2028, 2032, ..., 2096. This is an arithmetic sequence with first term 2024, common difference 4, and last term 2096. Using the formula: Number of terms = \( \frac{2096 - 2024}{4} + 1 = \frac{72}{4} + 1 = 18 + 1 = 19 \). We must check for exceptions: years divisible by 100 but not 400 are not leap years. The only century year in this range is 2100, but 2100 > 2099, so it falls outside our range. Therefore, there are no exceptions in 2024-2099, and the total count remains 19 leap years.
In simple words: Leap years happen every 4 years, except century years (divisible by 100) unless they divide by 400. From 2024 to 2099, count all multiples of 4 - there are 19 of them, with no century exceptions in that range.

Exam Tip: Remember the century-year rule carefully - 2000 was a leap year (divisible by 400), but 1900 was not (divisible by 100 but not 400). This rule rarely appears but, when it does, many students forget it.