NCERT Solutions Class 7 Mathematics Ganita Prakash 1 Chapter 06 Number Play

 

6.1 Numbers Tell Us Things

 

Question. What do the numbers in the figure below tell us?
Answer: The numbers each child says match a particular rule in both arrangements. Each child's number represents how many taller children are standing ahead of them in the line.
In simple words: Each number shows how many taller kids are in front of you.

Exam Tip: Understand the rule before solving - the number depends on the position and heights of people ahead, not total height comparisons.

 

Question. Write down the number each child should say based on this rule for the arrangement shown below.
Answer: Based on the height arrangement shown, assign each child a number equal to how many taller children stand before them. Work from left to right, counting only those taller than the current child.
In simple words: Look at each child and count how many taller kids are in front of them. That's their number.

Exam Tip: Always count only the taller people ahead - never count people behind you or people who are shorter.

 

Figure it Out (Page 128)

 

Question 1. Arrange the stick figure cutouts given at the end, or draw a height arrangement such that the sequence reads: (a) 0, 1, 1, 2, 4, 1, 5 (b) 0, 0, 0, 0, 0, 0, 0 (c) 0, 1, 2, 3, 4, 5, 6 (d) 0, 1, 0, 1, 0, 1, 0 (e) 0, 1, 1, 1, 1, 1, 1 (f) 0, 0, 0, 3, 3, 3, 3
Answer:
(a) The required arrangement is FCBGADE.
(b) The required arrangement is AECGBDF.
(c) The required arrangement is FDBGCEA.
(d) The required arrangement is EAGCDBF.
(e) The required arrangement is FAECGBD.
(f) The required arrangement is BDFAECG.
In simple words: Arrange the figures from shortest to tallest (or tallest to shortest) based on what the sequence needs. Each number tells you how many taller figures stand before that position.

Exam Tip: Work backwards - if you see 0, that person must be the tallest so far. If you see a larger number, that person has many tall people ahead.

 

Question 2. For each of the statements given below, think and identify if it is Always True, Only Sometimes True, or Never True. Share your reasoning. (a) If a person says '0', then they are the tallest in the group. (b) If a person is the tallest, then their number is '0'. (c) The first person's number is '0'. (d) If a person is not first or last in line (i.e., if they are standing somewhere in between), then they cannot say '0'. (e) The person who calls out the largest number is the shortest. (f) What is the largest number possible in a group of 8 people?
Answer:
(a) Only Sometimes True - A person says '0' when they see no one taller than themselves. The tallest person will always say '0', but a shorter person can also say '0' if they are at the front or in a position where no one taller is ahead of them. Thus, this statement is only sometimes true.
(b) Always True - If a person is the tallest, then no one is taller than them, so they will always say '0'. This statement is always true.
(c) Always True - Each person is assigned a number that represents how many taller people are ahead of them. Since there is no one ahead of the first person, their number will always be '0'. Hence, this statement is always true.
(d) Only Sometimes True - A person standing in between can still be assigned '0' if there are no taller people ahead of them.
(e) Only Sometimes True - A person who calls out the largest number has many taller people in front but may not be the shortest overall. For example, if the shortest person is standing at the front, they will call out '0'. Meanwhile, the second shortest person could be at the back and might call out the largest number.
(f) If there are 8 people, then the shortest person will see 7 taller people. So the maximum number someone can say is 7.
In simple words: Saying '0' means no taller people are ahead of you. The tallest person always says '0', but shorter people can also say '0' if they're first in line. The first person always says '0' because nobody is ahead of them.

Exam Tip: Remember - the number is about position and people ahead, not about total height. Shorter people at the front can say '0', and the person with the largest number isn't necessarily the shortest.

 

6.2 Picking Parity

 

Question. Can you figure out which 5 cards add to 30? Is it possible?
Answer: No, it is not possible. The sum of 5 odd numbers is always odd, and 30 is an even number. Since all the cards shown are odd numbers, no combination of 5 cards can add up to 30.
In simple words: Odd plus odd plus odd plus odd plus odd always makes odd. But 30 is even. So this is impossible.

Exam Tip: Use the parity rule for sums - an odd number of odd numbers always gives an odd result.

 

Question. Explore what happens to the sum of (a) 4 odd numbers, (b) 5 odd numbers, and (c) 6 odd numbers.
Answer: Based on the given examples using number cards 1, 3, 5, 7, 9, 11, 13:
(a) Sum of 4 odd numbers = 1 + 3 + 5 + 7 = 16 (even) - can be arranged in pairs.
(b) Sum of 5 odd numbers = 1 + 3 + 5 + 7 + 9 = 25 (odd) - cannot be arranged in pairs.
(c) Sum of 6 odd numbers = 1 + 3 + 5 + 7 + 9 + 11 = 36 (even) - can be arranged in pairs.
In simple words: An even count of odd numbers makes an even sum. An odd count of odd numbers makes an odd sum.

Exam Tip: Remember the pattern: even number of odd numbers = even sum; odd number of odd numbers = odd sum.

 

Figure it Out (Page 131)

 

Question 1. Using your understanding of the pictorial representation of odd and even numbers, find out the parity of the following sums: (a) Sum of 2 even numbers and 2 odd numbers (e.g., even + even + odd + odd) (b) Sum of 2 odd numbers and 3 even numbers (c) Sum of 5 even numbers (d) Sum of 8 odd numbers
Answer:
(a) Even + Even = Even and Odd + Odd = Even. Adding the two results, we get Even + Even = Even. The parity of the result is even. Example: 2 + 4 + 3 + 5 = 6 + 8 = 14 (Even)
(b) Odd + Odd = Even and Even + Even + Even = Even. Adding the two results, we get Even + Even = Even. The parity of the result is even. Example: 3 + 5 + 2 + 4 + 6 = 8 + 12 = 20 (Even)
(c) Adding any 5 even numbers always gives an even number. The parity of the result is even. Example: 2 + 4 + 6 + 8 + 10 = 30 (Even)
(d) Odd + Odd = Even (4 such pairs). Adding the 4 such results, we get Even + Even + Even + Even = Even. The parity of the result is even. Example: 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 = 64 (Even)
In simple words: When you add even numbers, you always get even. When you add an even count of odd numbers, you get even. When you add an odd count of odd numbers, you get odd.

Exam Tip: Focus on how many odd numbers you're adding. Count them first - if the count is even, the sum is even; if the count is odd, the sum is odd.

 

Question 2. Lakpa has an odd number of Rs. 1 coins, an odd number of Rs. 5 coins, and an even number of Rs. 10 coins in his piggy bank. He calculated the total and got Rs. 205. Did he make a mistake? If he did, explain why. If he didn't, how many coins of each type could he have?
Answer: Lakpa has an odd number of Rs. 1 coins, so their total value is odd. He also has an odd number of Rs. 5 coins, so their total is also odd. The Rs. 10 coins are even in number, so their total is even. Now, adding two odd sums, we get Odd (from Rs. 1) + Odd (from Rs. 5) = Even. Adding the Rs. 10 coins' total (even sum) to this even sum, Even + Even = Even. Since 205 is an odd number, the total of Rs. 205 is not possible with an odd number of Rs. 1 and Rs. 5 coins and an even number of Rs. 10 coins. Therefore, Lakpa made a mistake.
In simple words: Odd coins times 1 makes odd. Odd coins times 5 makes odd. Odd plus odd makes even. Even coins times 10 makes even. Even plus even makes even. But 205 is odd. So this is impossible.

Exam Tip: Use parity logic with money problems - multiply the count by the coin value to find the parity of the total, then add to check if the final answer makes sense.

 

Question 3. We know that: (a) even + even = even (b) odd + odd = odd (c) even + odd = odd. Similarly, find out the parity for the following scenarios: (d) even - even = ___ (e) odd - odd = ___ (f) even - odd = ___ (g) odd - even = ___
Answer:
(d) even - even = even
Example: 6 - 2 = 4 → even; 8 - 4 = 4 → even
(e) odd - odd = even
Example: 7 - 3 = 4 → even; 9 - 5 = 4 → even
(f) even - odd = odd
Example: 8 - 3 = 5; 12 - 5 = 7
(g) odd - even = odd
Example: 7 - 2 = 5; 9 - 6 = 3
In simple words: Subtracting two even numbers gives even. Subtracting two odd numbers gives even. Subtracting odd from even gives odd. Subtracting even from odd gives odd.

Exam Tip: Notice that subtraction has different parity rules than addition - when you subtract the same parity from each other (even from even or odd from odd), you get even.

 

Question. In a 3 × 3 grid, there are 9 small squares, which is an odd number. Meanwhile, in a 3 × 4 grid, there are 12 small squares, which is an even number. Given the dimensions of a grid, can you tell the parity of the number of small squares without calculating the product?
Answer: Yes, we can determine the parity of the number of small squares in a grid without directly calculating the full product, simply by observing the parity of the dimensions.

Rule: The product of two numbers is even if at least one of the numbers is even. The product of two numbers is odd if both numbers are odd.
In simple words: If either width or height is even, the total squares are even. Only if both are odd does the total become odd.

Exam Tip: Use the parity rule for multiplication - you don't need to multiply the exact numbers, just check if both dimensions are odd.

 

Question. Find the parity of the number of small squares in these grids: (a) 27 × 13 (b) 42 × 78 (c) 135 × 654
Answer:
(a) Both 27 and 13 are odd numbers, and Odd × Odd = Odd. So, the parity of the number of small squares is odd.
(b) Both 42 and 78 are even numbers, and Even × Even = Even. So, the parity of the number of small squares is even.
(c) 135 is odd, 654 is even, and Odd × Even = Even. So, the parity of the number of small squares is even.
In simple words: Check both numbers. Odd times odd makes odd. Any multiplication with an even number makes even.

Exam Tip: Quickly identify each dimension's parity (odd or even), then apply the multiplication rule without doing the actual calculation.

 

Parity of Expressions

 

Question. Come up with an expression that always has even parity. Some examples are: 100p and 48w - 2. Try to find more.
Answer: Expressions that have even parity are 2p + 10, 8n, 6m - 2, and others. Any expression of the form 2k (where k is any integer) or 2a + (even number) will always be even, since multiplying by 2 makes the result even, and adding an even number keeps it even.
In simple words: Any expression that is 2 times something is always even. Also, even + even = even.

Exam Tip: Look for expressions with a factor of 2, or those that add/subtract even numbers only.

 

Question. Come up with expressions that always have odd parity.
Answer: Expressions that always have odd parity are 2m + 1, 8a + 3, 6b + 5, and others. Any expression of the form 2k + (odd number) will always be odd, since 2 times any number is even, and adding an odd number to an even number makes it odd.
In simple words: To get odd, take something even and add an odd number. For example, 2n + 1 is always odd, no matter what n is.

Exam Tip: Odd expressions use the pattern 2(something) + odd number. The even part ensures the base is even, and the odd part flips it to odd.

 

Question. Come up with other expressions, like 3n + 4, which could have either odd or even parity.
Answer: Expressions like 3n + 4 which could have either odd or even parity are: 5k - 2, n + 5, 7m + 2, and others. These expressions change between odd and even depending on what value the variable takes, because the variable is multiplied by an odd number.
In simple words: When a variable with an odd number in front (like 3n, 5k, 7m) is added or subtracted to another number, the parity swaps based on whether the variable is odd or even.

Exam Tip: If the expression has an odd coefficient times the variable, the parity depends on the variable itself - the parity can be both odd and even.

 

Question. Are there expressions that we can use to list all the even numbers? Hint: All even numbers have a factor of 2.
Answer: All even numbers are multiples of 2, so to generate every even number, use 2n (where n = 1, 2, 3,…). This expression can list all the even numbers. By substituting different values of n, we get 2, 4, 6, 8, 10, and so on - every possible even number.
In simple words: The expression 2n gives you all even numbers. Put in 1 and get 2, put in 2 and get 4, put in 3 and get 6. You can make any even number this way.

Exam Tip: The formula 2n is the general form for all even numbers - it covers every single even number that exists.

 

Question. Are there expressions that we can use to list all odd numbers? We saw earlier how to express the nth term of the sequence of multiples of 4, where n is the letter-number that denotes a position in the sequence (e.g., first, twenty third, hundred and seventeenth, etc.).
Answer: To list all odd numbers, we use 2n - 1 (where n = 1, 2, 3, ….). This generates 1, 3, 5, 7, 9, …, capturing all odd numbers. By substituting different values of n starting from 1, we get every odd number in sequence.
In simple words: The expression 2n - 1 gives you all odd numbers. Put in 1 and get 1, put in 2 and get 3, put in 3 and get 5. Every odd number comes from this formula.

Exam Tip: The formula 2n - 1 is the general form for all odd numbers - start from n = 1 and each value of n gives you the next odd number in order.

 

Question. What would be the nth term for multiples of 2? Or, what is the nth even number?
Answer: The nth term for multiples of 2 is 2n. The nth even number is also 2n. For example, when n = 1, the first even number is 2(1) = 2; when n = 2, the second even number is 2(2) = 4; when n = 5, the fifth even number is 2(5) = 10.
In simple words: The nth even number is just 2n. The formula works for any position you want.

Exam Tip: Remember that 2n gives both all even numbers and the nth even number - it's the same formula for both.

 

6.3 Some Explorations in Grids

 

Question. Observe this 3 × 3 grid. It is filled following a simple rule - use numbers from 1 - 9 without repeating any of them. There are circled numbers outside the grid. What do the circled numbers represent?
Answer: The numbers in the yellow circles are the sums of the corresponding rows and columns. Each circled number on the right side shows the sum of the numbers in that row, and each circled number below shows the sum of the numbers in that column.
In simple words: The circles show what each row and column add up to. Row sums go on the right, column sums go on the bottom.

Exam Tip: Use the circled sums to work backwards and fill the grid - if you know the total and some numbers, you can find the missing ones.

 

Question. Fill the grids below based on the rule mentioned above.
Answer: To fill the grids, use the row and column sums provided. Start by identifying which numbers are already given. Then, use the sum clues to determine the missing values. Work systematically, finding cells where only one number is possible based on the available clues.
In simple words: Look at each row and column sum. If you know most of the numbers in a row or column, the sum tells you what the missing number must be.

Exam Tip: Start filling cells where the row or column is almost complete - these are easiest to solve. Work your way to harder cells once you have more information.

 

Question. Can you find the other possible positions for 1 and 9?
Answer: By analyzing the constraints of the grid sums, 1 and 9 (the smallest and largest numbers) have limited positions where they can fit. Different valid arrangements exist depending on how you balance the row and column sums, but not all positions work - only certain combinations satisfy all the sum requirements.
In simple words: 1 and 9 can go in different places, but not anywhere you want. You need to check if the row and column sums still work out.

Exam Tip: Think about which cells would make the most balanced sums - extreme numbers like 1 and 9 work best away from each other to keep sums reasonable.

 

Question. Now, we have one full row or column of the magic square! Try completing it! [Hint: First fill the row or columns containing 1 and 9]
Answer: Once one complete row or column is filled, use the sum constraints to fill the rest of the grid step by step. Start with the row or column containing 1 and 9, since these extreme values greatly limit what other numbers can go with them. This constrains the remaining cells, making it easier to deduce the rest.
In simple words: With one row or column fully known, you can use subtraction to find missing numbers. If a row should sum to 15 and you have 4 + 6 + 5, you know the fourth spot must hold what's missing.

Exam Tip: Work systematically - fill complete rows or columns first, then use those to deduce values in intersecting rows and columns.

 

Figure it Out (Page 136)

 

Question 1. How many different magic squares can be made using numbers 1-9?
Answer: Using the numbers 1-9, there is exactly one unique magic square (excluding rotations and reflections). If transformations like rotations and reflections are allowed, then there are 8 variations of this magic square (4 rotations and 4 reflections).
In simple words: There's one basic magic square using 1-9. But if you spin or flip it, you get 8 versions that look different but have the same pattern.

Exam Tip: When counting magic squares, be clear about whether rotations and reflections count as different or the same - this changes your answer significantly.

 

Question 2. Create a magic square using numbers 2-10. What strategy would you use for this? Compare it with the magic squares made using 1-9.
Answer: The numbers 2-10 are 9 consecutive numbers, just like 1-9, but increased by 1. Strategy: Start with the classic 1-9 magic square, and add 1 to each number. The magic square for numbers 2-10 would have a different magic sum (18) compared to numbers 1-9 (15). The structure remains similar, but the values are shifted up by 1. Every number is exactly 1 more than in the original square.
In simple words: Take a 1-9 magic square and add 1 to every spot. You get a 2-10 magic square. The sums change by 3 (one more in each of 3 positions).

Exam Tip: This transformation method works for any consecutive number set - always use an existing magic square as your starting point and apply the same operation to all cells.

 

Question 3. Take a magic square, and (a) Increase each number by 1 (b) Double each number. In each case, is the resulting grid also a magic square? How do the magic sums change in each case?
Answer: (a) After increasing each number by 1: This is still a magic square. New magic sum = 18. When you add a constant to every number, each row, column, and diagonal sum increases by 3 times that constant (because each line has 3 cells).
(b) After doubling each number: Still a magic square. New magic sum = 30. When you multiply all numbers by a constant, the magic sum is also multiplied by that constant. All rows, columns, and diagonals scale equally, so the magic square property is preserved.
In simple words: Adding the same number to everything keeps it a magic square - the new sum is the old sum plus (3 times what you added). Multiplying everything by the same number keeps it a magic square - the new sum is the old sum times that number.

Exam Tip: Operations on all cells uniformly preserve the magic square property. Addition changes the sum by 3k (where k is what you add); multiplication changes the sum by k times (where k is the multiplier).

 

Question 4. What other operations can be performed on a magic square to yield another magic square?
Answer: Students should explore and discover this themselves. Possible operations include subtraction of a constant from all cells (similar to addition), division by a constant that divides all cells evenly, rotation of the square (which preserves the magic property), reflection across axes or diagonals, and various rearrangements that maintain the sum property.
In simple words: Try many operations and see what keeps the magic square working. Some will work, others won't - discovering which ones teaches you about the magic square structure.

Exam Tip: Any operation that maintains equal sums across all rows, columns, and diagonals preserves the magic property - test operations systematically.

 

Question 5. Discuss ways of creating a magic square using any set of 9 consecutive numbers (like 2-10, 3-11, 9-17, etc.).
Answer: To create a magic square using any set of 9 consecutive numbers, start with the original 1-9 magic square, then add the appropriate constant to shift all numbers to the desired range. For numbers 2-10, add 1 to each cell. For numbers 3-11, add 2 to each cell. For numbers 9-17, add 8 to each cell. The general rule is: add (starting number - 1) to the original magic square. The resulting magic sum will be 15 + 3k, where k is the constant added to each cell. Different consecutive sets produce different magic sums, but the underlying structure remains identical.
In simple words: Pick any 9 consecutive numbers. Figure out how much they're shifted from 1-9. Add that shift to every cell in the original magic square. Done!

Exam Tip: The pattern is universal - all 9-consecutive-number magic squares have the same structure, just shifted. The magic sum always follows the formula 15 + 3k.

 

Generalising a 3 × 3 Magic Square

 

Question. We can describe how the numbers within the magic square are related to each other, i.e., the structure of the magic square. Choose any magic square that you have made so far using consecutive numbers. If m is the letter-number of the number in the centre, express how other numbers are related to m, how much more or less than m.
Answer: Considering the magic square with centre number 5, we can express it using the letter-number m as follows:

m + 3, m - 4, m + 1
m - 2, m, m + 2
m - 1, m + 4, m - 3

Every other number in the square relates to the centre number m by a fixed offset. For example, the corner numbers are m + 3, m + 4, m - 1, m - 3 (or similar patterns depending on rotation). This pattern works for any magic square using consecutive numbers - the offsets remain identical, only the value of m changes.
In simple words: In a magic square, every number can be written as the centre number plus or minus a fixed amount. Once you know the centre, you can describe all other cells as differences from it.

Exam Tip: Learning the general offsets from the centre helps you quickly build new magic squares - just decide what m should be and fill in the rest using the established pattern.

 

Question. Once the generalised form is obtained, share your observations with the class.
Answer: Students should share and discuss their observations from deriving the generalised form. Key observations might include: the symmetry of the magic square (opposite cells sum to twice the centre), the pattern of offsets, how the magic sum relates to the centre number (magic sum = 3m), and how this unified structure allows creating magic squares for any set of consecutive numbers by simply choosing different values of m.
In simple words: The magic square has beautiful symmetry - opposite corners are equidistant from the centre. This makes all rows, columns, and diagonals sum to exactly 3 times the middle number.

Exam Tip: Understanding the generalised form reveals the deep structure of magic squares - this knowledge helps solve and create them quickly without trial and error.

 

Figure it Out (Page 137)

 

Question 1. Using this generalised form, find a magic square if the centre number is 25.
Answer: Using the generalised form with centre m = 25:

28, 21, 26
23, 25, 27
24, 29, 22

By substituting m = 25 into each offset position (m + 3, m - 4, m + 1, etc.), we obtain this magic square. Each row, column, and diagonal sums to 75 (which equals 3 × 25).
In simple words: Put 25 in the middle. Add the fixed offsets to get 28, 23, 27, etc. This creates a magic square where everything sums to 75.

Exam Tip: Use the memorised offset pattern - it's much faster than trying to construct the square from scratch.

 

Question 2. What is the expression obtained by adding the 3 terms of any row, column, or diagonal?
Answer: Row sum (1st row) = 28 + 21 + 26 = 75. Column sum (1st column) = 28 + 23 + 24 = 75. Diagonal sum (1st diagonal) = 28 + 25 + 22 = 75. The expression obtained = 3 × m, where m is the letter-number representing the number in the centre. No matter which row, column, or diagonal you add, the sum is always 3m.
In simple words: Any line in the magic square - across, down, or diagonal - always adds to 3 times the middle number.

Exam Tip: This formula (magic sum = 3m) is a powerful shortcut - if you're told the magic sum, you can instantly find the centre number by dividing by 3.

 

Question 3. Write the result obtained by (a) Adding 1 to every term in the generalised form. (b) Doubling every term in the generalised form.
Answer: (a) After adding 1 to every term in the generalised form, the new magic square becomes:
m + 4, m - 3, m + 2
m - 1, m + 1, m + 3
m, m + 5, m - 2

(b) After doubling every term in the generalised form, the new magic square becomes:
2m + 6, 2m - 8, 2m + 2
2m - 4, 2m, 2m + 4
2m - 2, 2m + 8, 2m - 6

In both cases, the magic square property is preserved. In (a), the new magic sum is 3m + 3 (or 3(m + 1)). In (b), the new magic sum is 6m (or 3(2m)).
In simple words: Apply the same operation to every cell. Adding 1 gives a new magic square with a bigger magic sum. Doubling gives a magic square with a much bigger magic sum.

Exam Tip: This shows the flexibility of generalised forms - algebraic expressions let you transform and verify properties without calculating individual numbers.

 

Question 4. Create a magic square whose magic sum is 60.
Answer: A 3 × 3 magic square's sum is 3 × the middle element. So, for a sum of 60, the middle element should be 60 ÷ 3 = 20. To get a magic sum of 60, multiply the original magic square by 4:

32, 4, 24
12, 20, 28
16, 36, 8

By multiplying each cell of the standard 1-9 magic square by 4, we obtain numbers that form a magic square with sum 60.
In simple words: Divide the target sum (60) by 3 to get the centre (20). Then build the square using the offset pattern with m = 20.

Exam Tip: For any target magic sum, use the formula m = (target sum) ÷ 3 to find the centre, then apply the generalised pattern.

 

Question 5. Is it possible to get a magic square by filling nine non-consecutive numbers?
Answer: Yes, it is possible. Justification: Magic squares can be constructed using non-consecutive numbers as long as the numbers follow a mathematical pattern that preserves the magic property. For example, two magic squares with different magic sums can be combined or transformed to create new magic squares using non-consecutive numbers. The key requirement is that the numbers maintain symmetric relationships about the centre, as expressed in the generalised form.
In simple words: You don't need consecutive numbers. As long as the numbers fit the magic square pattern (opposite cells sum to 2 times the centre, etc.), it works.

Exam Tip: The magic square property relies on mathematical relationships, not the numbers being consecutive - this opens up creative possibilities for construction.

 

The First-ever 4 × 4 Magic Square

 

Question. The first ever recorded 4 × 4 magic square is found in a 10th-century inscription at the Parshvanath Jain temple in Khajuraho, India, and is known as the Chautisa Yantra. Chautis means 34. Why do you think they called it the Chautisa Yantra?
Answer: They called it the Chautisa Yantra because every row, column, and diagonal in this magic square adds up to 34. The word "Chautis" means 34 in the local language, so the name directly refers to the magic sum of this famous square. This naming convention highlights the most distinctive feature of magic squares - the constant sum across all lines.
In simple words: "Chautisa" means 34. Every line in the square sums to 34. So they named it after that magic sum.

Exam Tip: The name of a magic square often refers to its magic sum - this is a useful memory aid for recalling properties of famous historic squares.

 

Question. Can you find other patterns of four numbers in the square that add up to 34?
Answer: Yes, we can find different combinations of 4 numbers that add up to 34 in the given square:

- Sum of 4 corner numbers: 7 + 14 + 4 + 9 = 34
- Sum of 4 central numbers: 13 + 8 + 10 + 3 = 34
- Sum of 4 numbers in any 2 × 2 square: For example, top-left square: 7 + 12 + 13 + 2 = 34

Beyond the obvious rows, columns, and diagonals, many other number combinations in this magic square also yield the magic sum of 34. This hidden symmetry and multiple paths to the same sum demonstrate the deep mathematical harmony in magic squares.
In simple words: Not just rows and columns - corners add to 34, the middle four add to 34, even small 2×2 squares inside add to 34. There are many hidden patterns.

Exam Tip: Magic squares have many more summation patterns than just rows, columns, and diagonals - look for symmetric groupings and smaller subsquares.

 

6.4 Nature's Favourite Sequence: The Virahanka-Fibonacci Numbers!

 

Question. Use the systematic method to write down all 6-beat rhythms, i.e., write 6 as the sum of 1's and 2's in all possible ways. Did you get 13 ways?
Answer: Yes, we get a total of 13 ways. Write a '1+' in front of all rhythms having 5 beats and then a '2+' in front of all rhythms having 4 beats. This gives us all the rhythms having 6 beats. The systematic approach ensures that every possible combination is captured without repetition. The fact that we get 13 ways (which is the 6th term in the Virahanka-Fibonacci sequence beginning 1, 2, 3, 5, 8, 13) reveals the deep connection between rhythm patterns and this famous number sequence.
In simple words: To make rhythms that total 6 beats, take all rhythms totalling 5 beats (add a 1) and all rhythms totalling 4 beats (add a 2). You get 13 different patterns.

Exam Tip: The systematic counting method (adding 1 or 2 to shorter sequences) efficiently generates all patterns - this recursive approach is the key to understanding how the Fibonacci sequence naturally emerges.

 

Question. Write the next 3 numbers in the sequence: 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, ____, ____, ____,…
Answer: The next 3 terms in the sequence are:
55 + 89 = 144
89 + 144 = 233
144 + 233 = 377

So the sequence continues: 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377,…
In simple words: Add the last two numbers to get the next one. Always. That's how the Fibonacci sequence grows.

Exam Tip: The recursive rule (each term = sum of previous two terms) is the entire definition of the Virahanka-Fibonacci sequence.

 

Question. If you have to write one more number in the sequence above, can you tell whether it will be an odd number or an even number (without adding the two previous numbers)?
Answer: To determine if the next number after 377 is odd or even without adding the previous terms, examine the parity of the sequence. Parity pattern: 1 (odd), 2 (even), 3 (odd), 5 (odd), 8 (even), 13 (odd), 21 (odd), 34 (even), 55 (odd), 89 (odd), 144 (even), 233 (odd), 377 (odd). The parity follows a repeating pattern: Odd, Even, Odd, Odd, Odd, Even, and the cycle restarts. Since 377 is odd and the pattern shows Odd - Odd - Even repeating, the next number after 377 will be even.
In simple words: Look at whether each number is odd or even. The pattern repeats every 3 terms: Odd, Odd, Even. Since 377 is odd, the next number will be even.

Exam Tip: The parity pattern is independent of the actual values - use this to predict odd/even quickly without calculating.

 

Question. What is the parity of each term in the sequence? Do you notice any pattern in the sequence of parities?
Answer: The parity of each term: 1 (odd), 2 (even), 3 (odd), 5 (odd), 8 (even), 13 (odd), 21 (odd), 34 (even), 55 (odd), 89 (odd), 144 (even), 233 (odd), 377 (odd). The parity pattern repeats every 3 terms as Odd, Odd, Even. Two odd numbers are followed by one even number, then the pattern restarts. This beautiful symmetry arises from the parity rules of addition: Odd + Odd = Even, Odd + Even = Odd.
In simple words: The pattern is always Odd, Odd, Even, Odd, Odd, Even, Odd, Odd, Even... It repeats forever.

Exam Tip: This repeating parity pattern demonstrates that mathematical sequences have hidden regularities beyond their face values.

 

6.5 Digits in Disguise

 

Question. Let us look at one more example shown on the right. Here, K2 means that the number is a 2-digit number having the digit '2' in the units place and 'K' in the tens place. K2 is added to itself to give a 3-digit sum HMM. What digit should the letter M correspond to?
Answer: When K2 is added to itself (K2 + K2), the result is a 3-digit number HMM where both the tens place and the units place of the sum have the same digit. To find M, we need 2 + 2 = 4 in the units place. However, since both the tens and units place show the same digit M, we need 2K + (any carry) to produce MM. Testing values: If K = 2, then 22 + 22 = 44, giving HMM = 044, so M = 4 and H = 0 (not 3-digit). If K = 5, then 52 + 52 = 104, which doesn't have matching tens/units. If K = 6, then 62 + 62 = 124, no match. Continue testing until K = 9: 92 + 92 = 184. None match the pattern HMM perfectly. The most reasonable interpretation is that the sum's structure requires specific digit relationships, and M must be determined by the constraint that the tens and units digits are equal.
In simple words: Add K2 to itself. The answer should have the same digit in the tens and units places. Figure out which digit K must be for this to work.

Exam Tip: In these puzzles, test each possible digit systematically and check which ones satisfy all the given conditions.

 

Question. What about H? Can it be 2? Can it be 3?
Answer: The digit H (in the hundreds place of the sum HMM) depends on what value K takes. Since K2 + K2 produces a 3-digit number HMM, we know that 2K (the tens part) plus any carry from adding 2 + 2 in the units place must produce a result in the hundreds place. When 2 + 2 = 4 (no carry to the tens), 2K must be at least 100 to reach the hundreds place, which is impossible since K is a single digit (maximum 2K = 18). This suggests either a carry from adding the two K digits, or the problem interpretation needs adjustment. If the sum is HMM with matching M's in both tens and units, then H = 1 is most likely (from a carry situation), making H = 2 or H = 3 unlikely without K being very large.
In simple words: H is the hundreds digit of the answer. Since we're adding two 2-digit numbers, H must be 1 (from the carry). H can't be 2 or 3 with normal addition.

Exam Tip: In addition problems, the hundreds digit comes from a carry from the tens place - so H = 1 unless multiple carries stack up.

 

Question 1. A light bulb is ON. Dorjee toggles its switch 77 times. Will the bulb be on or off? Why?
Answer: Each time the switch is toggled, the bulb changes state (ON becomes OFF, OFF becomes ON). Since 77 is an odd number, after 77 toggles starting from ON, the bulb will be OFF.
In simple words: Odd number of switches means the bulb ends up in the opposite state from where it started. It will be OFF.

Exam Tip: Remember that odd toggles flip the state, even toggles return it to the original state - this is a key pattern for bulb/switch problems.

 

Question 2. Liswini has a large old encyclopaedia. When she opened it, several loose pages fell out of it. She counted 50 sheets in total, each printed on both sides. Can the sum of the page numbers of the loose sheets be 6000? Why or why not?
Answer: Each sheet has two page numbers - an odd number on the front (form 2n - 1) and an even number on the back (form 2n). The sum of both page numbers on one sheet is (2n - 1) + (2n) = 4n - 1. For 50 sheets with numbers n₁, n₂, ..., n₅₀, the total sum becomes 4(n₁ + n₂ + ... + n₅₀) - 50. Since 4(n₁ + n₂ + ... + n₅₀) is always divisible by 4, but 50 is not divisible by 4, the final result cannot be divisible by 4. Since 6000 is divisible by 4 (6000 ÷ 4 = 1500), the sum of page numbers can never equal 6000.
In simple words: Every sheet's two page numbers add up to a number of the form 4n - 1, which leaves a remainder when divided by 4. So 50 sheets together can never add up to 6000, which divides evenly by 4.

Exam Tip: Look for divisibility patterns when the problem asks "can this ever equal..." - if the sum always has a specific remainder, it can never equal a number without that remainder.

 

Question 3. Here is a 2 × 3 grid. For each row and column, the parity of the sum is written in the circle; 'e' for even and 'o' for odd. Fill the 6 boxes with 3 odd numbers ('o') and 3 even numbers ('e') to satisfy the parity of the row and column sums.
Answer: We track only parities, not actual numbers. Label the cells as:

Row 1 (A, B, C): sum is odd. Row 2 (D, E, F): sum is even. Column 1 (A, D): sum is even. Column 2 (B, E): sum is even. Column 3 (C, F): sum is odd.
Setting A = o (odd), B = e (even), C = e (even): then o + e + e = odd ✓. For column 1 to be even with A = o, we need D = o. For column 2 to be even with B = e, we need E = e. For column 3 to be odd with C = e, we need F = o.
Final grid:Or using actual numbers: place 1, 3, 4, 3, 6, 9 in the cells instead of o and e labels.
In simple words: Odd plus even gives odd. Even plus even gives even. Use these rules to place odd and even numbers so each row and column matches its target parity.

Exam Tip: Work backwards from the column constraints - once you set one row's parities, the columns often force the second row's parities automatically.

 

Question 4. Make a 3 × 3 magic square with 0 as the magic sum. All numbers cannot be zero. Use negative numbers, as needed.
Answer: We need a 3 × 3 square where every row, column, and diagonal sums to 0, with at least one non-zero entry. Using the numbers -4 to 4, we can construct:

Each row, column, and diagonal sums to 0: for example, row 1: -3 + 2 + 1 = 0; column 1: -3 + 4 - 1 = 0; main diagonal: -3 + 0 + 3 = 0.
In simple words: A magic square with sum 0 can be built by placing negative numbers opposite to positive numbers so they cancel out.

Exam Tip: Centre cell with 0 often helps - place larger numbers symmetrically around it so opposite pairs cancel to zero.

 

Question 5. Fill in the following blanks with 'odd' or 'even': (a) Sum of an odd number of even numbers is ______ (b) Sum of an even number of odd numbers is ______ (c) Sum of an even number of even numbers is ______ (d) Sum of an odd number of odd numbers is ______
Answer:
(a) Even
(b) Even
(c) Even
(d) Odd
In simple words: Even numbers always stay even when you add them. Odd numbers paired in even quantities become even; odd quantities of odd numbers stay odd.

Exam Tip: Memorise: any count of even numbers → even sum; even count of odds → even sum; odd count of odds → odd sum.

 

Question 6. What is the parity of the sum of numbers from 1 to 100?
Answer: Using the formula for the sum of first n natural numbers: \( 1 + 2 + 3 + \ldots + 100 = \frac{100 \times 101}{2} = 5050 \). Since 5050 is an even number, the parity is even.
In simple words: Add up all numbers from 1 to 100 using the formula. You get 5050, which is even.

Exam Tip: Remember the sum formula \( \frac{n(n+1)}{2} \) - it saves time and avoids arithmetic errors for large ranges.

 

Question 7. Two consecutive numbers in the Virahanka sequence are 987 and 1597. What are the next 2 numbers in the sequence? What are the previous 2 numbers in the sequence?
Answer: In the Virahanka sequence, each number is the sum of the two preceding numbers.
Next two numbers: 987 + 1597 = 2584, then 1597 + 2584 = 4181
Previous two numbers: 1597 - 987 = 610, then 987 - 610 = 377
Complete sequence: ..., 377, 610, 987, 1597, 2584, 4181, ...
In simple words: To go forward, add the last two numbers. To go backward, subtract the smaller from the larger.

Exam Tip: The Virahanka (Fibonacci) sequence follows one rule: each term equals the sum of the two before it - use this to extend in either direction.

 

Question 8. Angaan wants to climb an 8-step staircase. His playful rule is that he can take either 1 step or 2 steps at a time. For example, one of his paths is 1, 2, 2, 1, 2. In how many different ways can he reach the top?
Answer: We count the number of distinct step combinations that sum to 8, where each combination uses 1s and 2s:
- Eight 1s (1+1+1+1+1+1+1+1): 1 way
- Six 1s and one 2 (sum = 8, 6 ones and 1 two): 7 ways to arrange
- Four 1s and two 2s (sum = 8, 4 ones and 2 twos): 15 ways to arrange
- Two 1s and three 2s (sum = 8, 2 ones and 3 twos): 10 ways to arrange
- Four 2s (2+2+2+2): 1 way
Total: 1 + 7 + 15 + 10 + 1 = 34 ways
In simple words: Find all the different combinations of 1s and 2s that add to 8, then count how many different orders each combination can be arranged in.

Exam Tip: Break the problem into cases by the number of 2s used, then apply combinations formula for each case.

 

Question 9. What is the parity of the 20th term of the Virahanka sequence?
Answer: First, we list the Virahanka sequence and observe the parity pattern: 1 (odd), 2 (even), 3 (odd), 5 (odd), 8 (even), 13 (odd), 21 (odd), 34 (even), 55 (odd), 89 (odd), 144 (even), 233 (odd), 377 (odd), 610 (even), 987 (odd), ...
The parity pattern repeats as: odd, even, odd, repeating every 3 terms. Since 20 = 3 × 6 + 2, the 20th term has the same parity as the 2nd term in the cycle, which is even.
In simple words: The parity pattern (odd or even) in the Virahanka sequence repeats every 3 terms. Since 20 divided by 3 leaves remainder 2, the 20th term matches position 2 in the cycle - which is even.

Exam Tip: Always look for repeating patterns in parity - once you find the cycle length, use modular arithmetic to find the answer for large term numbers.

 

Question 10. Identify the true statements. (a) The expression 4m - 1 always gives odd numbers. (b) All even numbers can be expressed as 6j - 4. (c) Both expressions 2p + 1 and 2q - 1 describe all odd numbers. (d) The expression 2f + 3 gives both even and odd numbers.
Answer:
(a) TRUE. When m = 1: 4(1) - 1 = 3 (odd). When m = 2: 4(2) - 1 = 7 (odd). Since 4m is always even, subtracting 1 always leaves an odd result.

(b) FALSE. The expression 6j - 4 produces: j = 1 gives 2, j = 2 gives 8, j = 3 gives 14. This generates only certain even numbers (2, 8, 14, 20, ...) and misses 4, 6, 10, 12, etc.

(c) FALSE. The expression 2p + 1 for p = 1, 2, 3, ... gives 3, 5, 7, 9, .... The expression 2q - 1 for q = 1, 2, 3, ... gives 1, 3, 5, 7, .... The first expression never produces 1, so together they do not describe all odd numbers.

(d) FALSE. The expression 2f + 3 always gives odd results. When f = 1: 2(1) + 3 = 5 (odd). When f = 2: 2(2) + 3 = 7 (odd). Since 2f is even, adding 3 (odd) always produces an odd number.
In simple words: Test each expression by substituting a few values. Check if it covers all numbers of that type or if it leaves some out.

Exam Tip: To prove a statement false, find one counterexample. To prove it true, show why it works for all values mathematically - not just a few examples.

 

Question 11. Solve this cryptarithm:
\[ \begin{array}{r} UT \\ + \; TA \\ \hline TAT \end{array} \]
Answer: Since the sum UT + TA produces a three-digit number TAT (starting with T), there must be a carry from the tens place to create the hundreds digit. This means T = 1.
Looking at the hundreds place, the carry equals T, so T = 1 (confirmed).
From the ones place: T + A = T or T + A = 10 + T. The second case gives A = 0.
From the tens place: U + T = 10 + A or U + T = A. Since we need a carry and A = 0, we must have U + T = 10, so U + 1 = 10, giving U = 9.
Verification: 91 + 10 = 101 ✓
Therefore: U = 9, T = 1, A = 0
In simple words: The result is 101. This means U is 9, T is 1, and A is 0. Check: 91 + 10 = 101.

Exam Tip: In cryptarithmetic, use the column-by-column analysis: the leftmost digit of the sum tells you about carries, and single-digit constraints (like T being both a result digit and a starting digit) force specific values.