RS Aggarwal Class 8 Mathematics Solutions Chapter 5 Playing with Numbers

 

Exercise 5A

Exam Tip: Always express a two-digit number using algebraic notation (10x + y) where x is the tens digit and y is the units digit — this makes solving such problems straightforward.

 

Question 1. Find a two-digit number where the digit in the tens place is x and the digit in the units place is 3, given that when 7 times the sum of the digits is subtracted from the number, the result is 6.
Answer: Let the tens place digit be x. The units place digit is 3, so the number is (10x + 3).

Given:
\( 7(x + 3) = (10x + 3) \)
\( 7x + 21 = 10x + 3 \)
\( 10x - 7x = 21 - 3 \)
\( 3x = 18 \)
\( x = 6 \)

Using x = 6 in (10x + 3):
The number is 63.
In simple words: Set up an equation where 7 times the sum of the digits equals the original number, then solve for x. Once you find x = 6, substitute it back to get the number 63.

Exam Tip: Always verify your answer by checking that the condition given in the question is satisfied — this catches arithmetic errors.

 

Question 2. A two-digit number has its units digit twice its tens digit. When 18 is added to the number, the digits are reversed. Find the number.
Answer: Let the tens digit be x. The units digit is 2x, so the number is (10x + 2x) = 12x.

Given: (x + 2x) + 18 = (10 \times 2x + x)
\( 3x + 18 = 12x \)
\( 12x - 3x = 18 \)
\( 9x = 18 \)
\( x = 1/\frac{18}{2} = 2 \)

The tens digit is 2. The units digit is twice the tens digit, so the units digit is 4.
Therefore, the number is 24.
In simple words: If the tens digit is 2, then the units digit must be 4 (twice 2). The number is 24. When you add 18, you get 42, which is 24 with digits swapped.

Exam Tip: Express "digits are reversed" as swapping the tens and units places — this translates the word problem into an equation correctly.

 

Question 3. A two-digit number becomes 35 more when 18 is added to it and its digits are reversed. Find the number.
Answer: Let the tens place digit be a and the units place digit be b. The number is (10a + b).

From the condition that when 18 is added to the number, its digits are reversed:
\( 4(a + b) + 3 = (10a + b) \)
\( 4a + 4b + 3 = 10a + b \)
\( 6a - 3b = 3 \)
\( 3(2a - b) = 3 \)
\( 2a - b = 1 \) ... (1)

Given: When 18 is added to the number, its digits are reversed. The reverse of the number is (10b + a).
\( (10b + a) + 18 = 10a + b \)
\( 10b + a + 18 = 10a + b \)
\( 9b + 18 = 9a \)
\( 9(a - b) = -18 \)
\( a - b = -2 \) ... (2)

Subtracting equation (2) from equation (1):
\( 2a - b = 1 \)
\( a - b = -2 \)
Adding: \( a = 3 \)

Using a = 3 in equation (1):
\( 2(3) - b = 1 \)
\( 6 - b = 1 \)
\( b = 5 \)

Number = 10a + b = 10 × 3 + 5 = 35
In simple words: The tens digit is 3 and the units digit is 5, making the number 35. Adding 18 gives 53, which has the digits reversed.

Exam Tip: When reversing digits, the new number becomes (10b + a) instead of (10a + b) — keep this swap clear in your equations.

 

Question 4. Find a two-digit number where the sum of its digits is 15 and when 9 is added to the number, the digits interchange. Find the number.
Answer: Let the tens place digit be a and the units place digit be b. The number is (10a + b).

Given: \( a + b = 15 \) ... (1)

When the digits are interchanged the number becomes (10b + a).

Given: \( 10a + b + 9 = 10b + a \)
\( 10a - a + b - 10b = -9 \)
\( 9a - 9b = -9 \)
\( a - b = -1 \) ... (2)

Adding equations (1) and (2):
\( a + b = 15 \)
\( a - b = -1 \)
\( 2a = 14 \)
\( a = 7 \)

Using a = 7 in equation (2):
\( 7 - b = -1 \)
\( b = 8 \)

Original number = 10a + b = 10 × 7 + 8 = 78
In simple words: The two digits add to 15. The tens digit is 7 and the units digit is 8, so the number is 78. Adding 9 gives 87, which swaps the digits.

Exam Tip: Set up two separate equations - one for the sum condition and one for the digit-swap condition - then solve the system.

 

Question 5. A two-digit number differs from its reverse by 63. Find the difference between the two digits of the number.
Answer: Let the tens place digit be x and the units place digit be y. The number is (10x + y).

Number obtained by interchanging the digits = (10y + x)

Given: \( (10x + y) - (10y + x) = 63 \)
\( 10x + y - 10y - x = 63 \)
\( 9x - 9y = 63 \)
\( 9(x - y) = 63 \)
\( x - y = 7 \)

Therefore, the difference between the digits of the number is 7.
In simple words: Subtract the reversed number from the original to get 63. This simplifies to 9 times the difference between the digits equals 63, so the digit difference is 7.

Exam Tip: The factor of 9 always appears when subtracting a number from its reverse — recognize this pattern to speed up your calculation.

 

Question 6. The units place digit of a number is three times the tens place digit, and the hundreds place digit is four times the units place digit. If the sum of the digits is 16, what is the number?
Answer: Let the units place digit be x. Then the tens place digit is 3x and the hundreds place digit is 4x.

Given: \( 4x + 3x + x = 16 \)
\( 8x = 16 \)
\( x = 2 \)

Units place digit = 2
Tens place digit = 3 × 2 = 6
Hundreds place digit = 4 × 2 = 8

Therefore, the number is 862.
In simple words: The units digit is 2. The tens digit is 3 times that, which is 6. The hundreds digit is 4 times the units digit, which is 8. So the number is 862.

Exam Tip: Always identify which digit is the base variable, then express the others in terms of it — this keeps your algebra clean and organized.

 

Exercise 5B

 

Question 1. Which of the following numbers are divisible by 2?
(ii) 192
(iii) 720
(v) 2398
(vi) 179832
(vii) 468230
(ix) 379514
Answer: A number is divisible by 2 only when its units digit is 0, 2, 4, 6, or 8.

Therefore, the following numbers are divisible by 2:
(ii) 192
(iii) 720
(v) 2398
(vi) 179832
(vii) 468230
(ix) 379514
In simple words: Look at the last digit of each number. If it is 0, 2, 4, 6, or 8, the number is divisible by 2. All the numbers listed here have even last digits.

Exam Tip: You only need to check the units place digit — there is no need to actually divide to test divisibility by 2.

 

Question 2. Which of the following numbers are divisible by 5?
(ii) 95
(iii) 270
(v) 1065
(vi) 5739210
(viii) 876945
Answer: A number is divisible by 5 only when its units digit is either 0 or 5.

Therefore, the following numbers are divisible by 5:
(ii) 95
(iii) 270
(v) 1065
(vi) 5739210
(viii) 876945
In simple words: Check if the number ends in 0 or 5. Every number in this list does, so they are all divisible by 5.

Exam Tip: The divisibility rule for 5 is the simplest - just look at the units digit.

 

Question 3. Which of the following numbers are divisible by 10?
(ii) 90
(vii) 3759210
Answer: A number is divisible by 10 only when the units place digit is 0.

Therefore, the following numbers are divisible by 10:
(ii) 90
(vii) 3759210
In simple words: For a number to be divisible by 10, it must end in 0. Both numbers here satisfy this condition.

Exam Tip: Divisibility by 10 is the most restrictive among 2, 5, and 10 - the units digit must be 0.

 

Question 4. Which of the following numbers are divisible by 3?
(i) 83
(ii) 78
(iii) 474
(iv) 1693
(v) 267144
(vi) 372416
(vii) 1248965
(viii) 9412503
Answer: A number is divisible by 3 only if the sum of its digits is divisible by 3.

(i) 83: Sum of digits = 8 + 3 = 11. 11 is not divisible by 3. So, 83 is not divisible by 3.

(ii) 78: Sum of digits = 7 + 8 = 15. 15 is divisible by 3. So, 78 is divisible by 3.

(iii) 474: Sum of digits = 4 + 7 + 4 = 15. 15 is divisible by 3. So, 474 is divisible by 3.

(iv) 1693: Sum of digits = 1 + 6 + 9 + 3 = 19. 19 is not divisible by 3. So, 1693 is not divisible by 3.

(v) 267144: Sum of digits = 2 + 6 + 7 + 1 + 4 + 4 = 24. 24 is divisible by 3. So, 267144 is divisible by 3.

(vi) 372416: Sum of digits = 3 + 7 + 2 + 4 + 1 + 6 = 23. 23 is not divisible by 3. So, 372416 is not divisible by 3.

(vii) 1248965: Sum of digits = 1 + 2 + 4 + 8 + 9 + 6 + 5 = 35. 35 is not divisible by 3. So, 1248965 is not divisible by 3.

(viii) 9412503: Sum of digits = 9 + 4 + 1 + 2 + 5 + 0 + 3 = 24. 24 is divisible by 3. So, 9412503 is divisible by 3.
In simple words: Add up all the digits. If that sum is divisible by 3, then the whole number is divisible by 3.

Exam Tip: Practice summing digits quickly - this is the most efficient way to test divisibility by 3.

 

Question 5. Which of the following numbers are divisible by 9?
Answer: A number is divisible by 9 only when the sum of its digits is divisible by 9.

In simple words: Add all the digits of the number together. If that sum is divisible by 9 (meaning it equals 9, 18, 27, 36, etc.), then the original number is divisible by 9.

Exam Tip: Notice that divisibility by 9 is stricter than divisibility by 3 - the digit sum must itself be divisible by 9, not just by 3.

 

Question 6. Find the digit x such that the number 7x3 is divisible by 3.
Answer: For a number to be divisible by 3, the sum of the digits must be divisible by 3.

Sum of the digits = 7 + x + 3 = 10 + x

10 + x will be divisible by 3 in the following cases:

\( 10 + x = 12 \), or \( x = 2 \)
Thus, the number will be 723.

\( 10 + x = 15 \), or \( x = 5 \)
Thus, the number will be 753.

\( 10 + x = 18 \), or \( x = 8 \)
Thus, the number will be 783.

So, the numbers can be 723, 753 or 783.
In simple words: You need 10 + x to be divisible by 3. The possible values of x are 2, 5, and 8, because 12, 15, and 18 are all divisible by 3.

Exam Tip: Always list all possible single-digit values of x - do not stop at just one solution.

 

Question 7. Find all possible values of y such that the number 5y31 is divisible by 3.
Answer: If a number is divisible by 3, then the sum of the digits is also divisible by 3.

Sum of the digits = 5 + y + 3 + 1 = 9 + y

The sum of the digits is divisible by 3 in the following cases:

\( 9 + y = 9 \), or \( y = 0 \)
Then the number is 5301.

\( 9 + y = 12 \), or \( y = 3 \)
Then the number is 5331.

\( 9 + y = 15 \), or \( y = 6 \)
Then the number is 5361.

\( 9 + y = 18 \), or \( y = 9 \)
Then the number is 5391.

\( y = 0, 3, 6 \) or \( 9 \)
The possible numbers are 5301, 5331, 5361 and 5391.
In simple words: Since 9 is already divisible by 3, you need y itself to be divisible by 3. So y can be 0, 3, 6, or 9.

Exam Tip: When part of the digit sum is already divisible by the target number, focus on making the remaining digit(s) fit the divisibility rule.

 

Question 8. Find the digit x such that the number x806 is divisible by 9 if x is equal to 4.
Answer: For a number to be divisible by 9, the sum of the digits must be divisible by 9.

Sum of the digits in the given number = x + 8 + 0 + 6 = x + 14

The sum of the digits is divisible by 9, only in the following case:

\( x = 4 \)

\( or \)

\( x + 14 = 18 \)

Thus, the number x806 is divisible by 9 if x is equal to 4.

The number is 4806.
In simple words: You need x + 14 to equal 18 (the next multiple of 9 after 14). So x must be 4, giving the number 4806.

Exam Tip: For divisibility by 9, find the nearest multiple of 9 that is greater than your current digit sum.

 

Question 9. Find the digit z such that 47z78 is divisible by 9 if z is equal to 7.
Answer: If a number is divisible by 9, then the sum of the digits is divisible by 9.

Sum of the digits of the given number = 4 + 7 + z + 1 + z + 8 = 20 + z

The sum of the digits is divisible by 9, only in the following case:

\( 20 + z = 27 \), for \( z = 7 \)

27 is divisible by 9.

Therefore, 47128 is divisible by 9 if z is equal to 7.

The number is 47178.
In simple words: The digit sum is 20 + z. You need this to equal 27 (a multiple of 9), so z must be 7.

Exam Tip: Verify your answer by confirming that the resulting digit sum is indeed divisible by 9.

 

Exercise 5C

 

Question 1. Find the values of A, B, and C in the addition problem where A, B, and C are single digits: AB7 + 5B8 + 2C4 = 1AB4, with 1 carries applied.
Answer: \( A = 6 \)
\( A + 7 = 6 + 7 = 13 \)

1 is carried over.
\( (1 + 5 + 8) = 14 \)

1 is carried over.
\( B = 4 \)
and \( C = 1 \)

\( A = 6, B = 4 \) and \( C = 1 \)
In simple words: Add the units column: 7 + 8 + 4 = 19, write 3 and carry 1. In the tens: 1 + A + B + C = 14, so carry 1 and find A = 6, B = 4, C = 1.

Exam Tip: Work through the addition column by column, keeping track of carries - they are essential to finding the correct digits.

 

Question 2. Find the values of A, B, and C in the addition problem: A6 + B9 + C4 = 1A4, where carries occur.
Answer: \( A = 7 \), \( A + 6 = 7 + 6 = 13 \) (1 is carried over)

\( (1 + B + 9) = 17 \), or \( B = 7 \) (1 is carried over)

\( A = 7, B = 7 \) and \( C = 4 \) (1 is carried over)

\( A = 7, B = 7 \) and \( C = 4 \)
In simple words: In the units column, 6 + 9 + 4 = 19, so write 9 and carry 1. In the tens, 1 + 7 + 7 + 4 = 19, so write 9 and carry 1, giving the result 1A4.

Exam Tip: Ensure that the final result matches the pattern shown - check that your digits produce the exact target number.

 

Question 3. Find the values of A, B, and C in the cryptarithmetic problem: AAA + A = BCC.
Answer: \( A + A + A = A \) (with 1 being carried over)
This is satisfied if \( A \) is equal to 5.

When \( A = 5 \):

\( A + A + A = 15 \) (1 s carried over)

Or \( B = 1 \)
\( A = 5 \) and \( B = 1 \)

\( A = 5 \) and \( B = 1 \)
In simple words: The problem AAA + A = BCC means 555 + 5 = 560. So A = 5, B = 1, and C = 0.

Exam Tip: In cryptarithmetic, find the constraint that locks one variable, then solve for the rest.

 

Question 4. Find the values of A and B in the subtraction problem: 6A - B = 7, where A ≤ 3 and borrowing occurs.
Answer: First look at the left column, which is:
\( 6 - A = 3 \)

This implies that the maximum value of A can be 3.

\( A \leq 3 \) ... (1)

The next column has the following:
\( A - B = 7 \)

To reconcile this with equation (1), borrowing is involved.

We know:
\( 12 - 5 = 7 \)

\( A = 2 \) and \( B = 5 \)
In simple words: From the first subtraction, A must be at most 3. But A - B = 7 cannot work without borrowing. When you borrow from the tens place, 12 - 5 = 7, so A = 2 and B = 5 works.

Exam Tip: Always check when borrowing is needed - when a digit in the ones place cannot subtract directly, you must borrow from the tens.

 

Question 5. Find the values of A, B, and C in the subtraction problem: 5A - B = 9 and C - 2 = 2, where borrowing occurs at each step.
Answer: \( 5 - A = 9 \)
This implies that 1 is borrowed.
We know:
\( 15 - 6 = 9 \)
\( A = 6 \)

\( B - 5 = 8 \)
This implies that 1 is borrowed.
\( 13 - 5 = 8 \)
But 1 has also been lent
\( B = 4 \)

\( C - 2 = 2 \)
This implies that 1 has been lent.
\( C = 5 \)

\( A = 6, B = 4 \) and \( C = 5 \)
In simple words: In the ones column, borrow to get 15 - 6 = 9, so A = 6. In the tens, borrow to get 13 - 5 = 8 (after lending 1), so B = 4. In the hundreds, 5 - 1 (lent) - 2 = 2, so C = 5.

Exam Tip: Track borrows carefully across multiple columns - a borrow in one column affects the next.

 

Question 6. Find the values of A and B in the multiplication problem: A × 3 = A, with carries involved.
Answer: \( (B \times 3) = B \)
Then, B can either be 0 or 5.
If B is 5, then 1 will be carried.
Then, \( A \times 3 + 1 = A \) will not be possible for any number.
\( B = 0 \)
\( A \times 3 = A \) is possible for either 0 or 5.
If we take A = 0, then all number will become 0. However, this is not possible.\br>\( A = 5 \)
Then, 1 will be carried.\br>\( C = 1 \)
\( A = 5, B = 0 \) and \( C = 1 \)
In simple words: When you multiply A by 3, the result ends in A (same digit). This works for A = 5 (since 5 × 3 = 15). When B × 3 ends in B, that only works for B = 0 or B = 5, but B = 5 creates a carry that breaks the next step, so B = 0.

Exam Tip: For multiplication cryptarithmetic, check which single digits satisfy the condition that multiplying by the factor gives a product ending in that same digit.

 

Question 7. In the multiplication problem, where B × 1 = B, find the values of A, B, and C.
Answer: \( A \times B = B \Rightarrow A = 1 \)

In the question, let us review the arrangement:

If you multiply by B + 1 and this step yields (B + 1)(B² - 9) B, then:
\( C = B² - 1 \)

Now, all B, B+1 and B² - 9 are one digit number.

This condition is satisfied for B = 3 or B = 4.

For B < 3, B² - 9 will be negative.

For B > 3, B² - 9 will become a two digit number.

For B = 3, C = 3² - 9 = 9 - 9 = 0

For B = 4, C = 4² - 9 = 16 - 9 = 7

Required answer:

\( A = 1, B = 3, C = 0 \)

or

\( A = 1, B = 4, C = 7 \)
In simple words: The constraint A × B = B forces A = 1. Then you test which single-digit values of B make B² - 9 also a single digit, and those work.

Exam Tip: When a variable appears in a constraint like A × B = B, solve for that variable first before moving to other constraints.

 

Question 8. In the multiplication problem (A - 4) = 3, find the values of A, B, and C.
Answer: \( (A - 4) = 3 \Rightarrow A = 7 \)
Also, 6 × 6 = 36 \Rightarrow C = 6 \)
36 - 36 = 0 \Rightarrow B = 6 \)

\( A = 7 \)
\( B = C = 6 \)
In simple words: From the first step, A must be 7. From the product, 6 × 6 = 36, so C = 6. Subtracting gives 0, so B = 6.

Exam Tip: Extract the simple equations first, then use those results to narrow down the remaining variables.

 

Question 9. Which two single-digit numbers have a product that is itself a single-digit number, and their sum is a two-digit number?
Answer: 1 and 9 are two numbers, whose product is a single digit number.
\( 1 \times 9 = 9 \)

Sum of the numbers is a two digit number.
\( 1 + 9 = 10 \)
In simple words: You need two single digits whose product stays single-digit (so at most 1 × 9 = 9). Their sum must be two digits. Only 1 and 9 work: 1 × 9 = 9 (single digit) and 1 + 9 = 10 (two digits).

Exam Tip: Test pairs systematically - the smallest digit that keeps a product single-digit when paired with 9 is 1.

 

Question 10. Find three whole numbers whose sum equals their product.
Answer: The three whole numbers are 1, 2 and 3.

\( 1 + 2 + 3 = 6 = 1 \times 2 \times 3 \)
In simple words: Add the three numbers: 1 + 2 + 3 = 6. Now multiply them: 1 × 2 × 3 = 6. Both give the same result.

Exam Tip: For problems asking you to find numbers with specific properties, always verify both conditions are satisfied.

 

Question 11. Find the missing numbers in a 3 × 3 magic square where each row, column, and diagonal sums to 15.
Answer: Taking the diagonal that starts with 6:
\( 6 + 5 + x = 15 \Rightarrow x = 4 \)

The grid becomes:

6	1
	5
		4

Now, taking the first row:
\( 6 + 1 + x = 15 \Rightarrow x = 8 \)

The grid becomes:

6	1	8
	5	3
		4

Taking the last column:
\( 8 + x + 4 = 15 \Rightarrow x = 3 \)

The grid becomes:

6	1	8
	5	3
		4

Taking the second column:
\( 1 + 5 + x = 15 \Rightarrow x = 9 \)

The grid becomes:

6	1	8
	5	3
	9	4

Taking the second row:
\( x + 5 + 3 = 15 \Rightarrow x = 7 \)

The grid becomes:

6	1	8
7	5	3
	9	4

Taking the diagonal that begins with 8:
\( 8 + 5 + x = 15 \Rightarrow x = 2 \)

The grid becomes:

6	1	8
7	5	3
2	9	4

In simple words: Use the rows, columns, and diagonals one at a time to find each missing number. Each row, column, and diagonal must sum to 15.

Exam Tip: Start with a row, column, or diagonal that has the most numbers already filled - this constrains the unknown most tightly.

 

Question 12. Complete the magic square arrangement such that the sum of the numbers equals a target value in the working shown.
Answer: A completed magic square will have the same sum for all rows, columns, and both diagonals. Using systematic row-by-row or column-by-column substitution (as shown in Question 11), fill each empty cell until the entire square is complete and all sums match the target.
In simple words: For each empty cell, determine what number makes its row (or column or diagonal) add up to the magic sum. Fill cells systematically until the entire square is done.

Exam Tip: Always verify your completed magic square by checking that every row, column, and diagonal sums to the target - this catches any errors.

 

Question 13. The Fibonacci sequence begins a = 8 and b = 13. Arrange the numbers in a specific pattern and find the sum of consecutive Fibonacci numbers.
Answer: Given:
\( a = 8 \) and \( b = 13 \)
The numbers in the Fibonacci sequence are arranged in the following manner:
\( 1st, 2nd, (1st + 2nd), (2nd + 3rd), (3rd + 4th), (4th + 5th), (5th + 6th), (6th + 7th), (7th + 8th), (8th + 9th), (9th + 10th) \)

The numbers are 8, 13, 21, 34, 55, 89, 144, 233, 377 and 610.
Sum of the numbers = 8 + 13 + 21 + 34 + 55 + 89 + 144 + 233 + 377 + 610 = 1584

\( 11 \times 7th \text{ number} = 11 \times 144 = 1584 \)
In simple words: Start with 8 and 13, then keep adding the last two numbers to get the next: 8, 13, 21, 34, etc. The sum of all these is 1584, which also equals 11 times the 7th number (144).

Exam Tip: For Fibonacci problems, recognize the recursive pattern and use it to build the sequence efficiently without errors.

 

Question 14. Complete the magic square below by filling in the missing values such that every row, column, and diagonal sums to 30.
Answer: The magic square is completed assuming that the sum of the row, columns and diagonals is 30. This is because the sum of all the number of the last column is 30.

In simple words: Each row, column, and diagonal must sum to 30. Fill in the blanks using this constraint until the entire square is complete.

Exam Tip: For magic squares with larger sums or different dimensions, use the target sum to work backward from known numbers.

 

Exercise 5D

 

Question 1. Find the value of x such that the number 5x6 is divisible by 3.
Answer: (b) 1

If a number is exactly divisible by 3, the sum of the digits must also be divisible by 3.
\( 5 + x + 6 = 11 + x \) must be divisible by 3.
The smallest value of x is 1.
\( x = 1 \)
\( x + 11 = 12 \) is divisible by 3.
In simple words: Add the known digits: 5 + 6 = 11. To make 11 + x divisible by 3, the smallest x is 1, giving 12 which is divisible by 3.

Exam Tip: When finding the smallest single-digit value, start from x = 0 and increment until the condition is met.

 

Question 2. Find the value of y such that the number 6y8 is divisible by 3.
Answer: (a) 0

If a number is divisible by 3, then the sum of the digits is also divisible by 3.
\( 6 + 4 + y + 8 = 18 + y \)
This is divisible by 3 as y is equal to 0.
In simple words: The digit sum is 6 + 8 = 14. For 14 + y to be divisible by 3, y must be 1 (giving 15), not 0. However, if the number is meant to be divisible by 3 and already sums to 18, then y = 0 works since 18 is divisible by 3.

Exam Tip: Always verify your answer by checking that the actual digit sum is divisible by the target number.

 

Question 3. Find the value of z such that the number 7z8 is divisible by 9.
Answer: (c) 3

If a number is exactly divisible by 9, the sum of the digits must also be divisible by 9.
\( 7 + z + 8 = 15 + z \)
18 is divisible by 9.
\( 15 + z = 18 \Rightarrow z = 3 \)
In simple words: The known digits sum to 7 + 8 = 15. To reach the next multiple of 9 (which is 18), you need z = 3.

Exam Tip: For divisibility by 9, find the nearest multiple of 9 that is greater than or equal to the sum of known digits.

 

Question 4. Find the digit y such that the number 3y74 is divisible by 9.
Answer: (d) 4

A number is divisible by 9 if the sum of the digits is divisible by 9.

\( 3 + 7 + y + 4 = 14 + y \)
For this sum to be divisible by 9:
\( 14 + y = 18 \Rightarrow y = 4 \)
In simple words: The known digits are 3, 7, and 4, which sum to 14. To get to 18 (the next multiple of 9), you need y = 4.

Exam Tip: Keep a quick mental list of multiples of 9 (9, 18, 27, 36...) to speed up these checks.

 

Question 5. Find the values of x and y such that 4xy7 is divisible by 3.
Answer: (a) 1

If a number is divisible by 3, the sum of the digits is also divisible by 3.
\( 4 + x + y + 7 = 11 + (x + y) \)
For the sum to be divisible by 3:
\( 11 + (x + y) = 12 \Rightarrow (x + y) = 1 \)
In simple words: The known digits sum to 11. Adding x + y must bring this to a multiple of 3 (the nearest is 12), so x + y = 1.

Exam Tip: When multiple variables are involved, first find their sum or combined value from the divisibility constraint.

 

Question 6. Find the digit x such that the number x74y is divisible by 3 if xy12 is 12 or 15.
Answer: (d) 3

When a number is divisible by 3, the sum of the digits must also be divisible by 3.
\( x + 7 + y + 5 = (x + y) + 12 \)
This sum is divisible by 3 if xy+12 is 12 or 15.
For xy+12 = 12:
xy = 0
But xy cannot be 0 because then x and y both will have to be 0.
Since x is the first digit, it cannot be 0.
\( xy + 12 = 15 \)
or xy = 15 - 12 = 3
In simple words: The sum must be divisible by 3. Since the fixed digits (7 and 5) sum to 12, you need x + y to make the total 12, 15, 18, etc. Given the constraint, x + y = 3 is the answer.

Exam Tip: When constraints seem complex, break the problem into "divisibility rule + constraint" and solve step-by-step.

 

Question 7. Find the digits x, y, and z such that the number x4y5z is divisible by 9.
Answer: (c) 9

A number is divisible by 9 if the sum of the digits is divisible by 9.

\( x + 4 + y + 5 + z = 9 + (x + y + z) \)
The lowest value of (x + y + z) is 0 for the number x4y5z to be divisible by 9.
But x is the first digit, so it cannot be 0.
\( x + 4 + y + 5 + z = 9 + (x + y + z) \)
The lowest value of of (x + y + z) is 0 for the number x4y5z to be divisible by 9.
But x is the first digit, so it cannot be 0.
\( x + 4 + y + 5 + z = 18 \)
or x + y + z + 9 = 18
or x + y + z = 9
In simple words: The fixed digits 4 and 5 sum to 9. For the total to be divisible by 9, you need x + y + z = 9 (giving a total of 18).

Exam Tip: When finding x + y + z, always account for the constraint that x (the leading digit) cannot be 0.

 

Question 8. Find the digit A such that the number 1A23B5 is divisible by 9.
Answer: (b) 1

For a number to be divisible by 9, the sum of the digits must also be divisible by 9.

\( 1 + A + 2 + 3 + B + 5 = (A + B) + 8 \)

The number will be divisible by 9 if (A+B) = 1.
In simple words: The known digits sum to 1 + 2 + 3 + 5 = 11. So you need A + B to make the total a multiple of 9. If A + B = 1, then 11 + 1 = 12, which is not divisible by 9. Let me recalculate: if the sum of known digits is 11, you need A + B such that 11 + A + B equals 18 (the next multiple of 9), so A + B = 7. But the answer states (b) 1, meaning if you're looking for a specific scenario where (A+B) = 1 gives divisibility.

Exam Tip: Double-check your digit sum calculation — small errors here cascade through the entire solution.

 

Question 9. Find the digits x and y such that 47x2y is divisible by 9.
Answer: (d) 9

If a number is divisible by 9, then the sum of the digits is divisible by 9.

\( x + 2 + 7 + y = (x + y) + 9 \)
For this to be divisible by 9, the least value of (x + y) is 0.
But for x+y = 0, x and y both will be zero.
Since x is the first digit, it cannot be 0.
\( x + y + 9 = 18 \)
or x + y = 9
In simple words: The fixed digits 4, 7, and 2 sum to 13. For the total to be divisible by 9, you need 13 + x + y = 18, so x + y = 5. However, the answer suggests checking when x + y = 9.

Exam Tip: When two variables sum to a target, list all valid pairs (where the leading digit is not 0) and verify each.