Number Play

Write down the number each child should say based on this rule for the arrangement shown below.

(The rule is — each child calls out the number of children in front of them who are taller than them.)

Answer

Each child counts the number of children standing in front of them who are taller than they are.

Starting from the front and moving towards the back:

Child 1: No one is in front ⇒ 0
Child 2: No taller child is in front ⇒ 0
Child 3: One taller child is in front ⇒ 1
Child 4: No taller child is in front ⇒ 0
Child 5: Three taller children are in front ⇒ 3
Child 6: No taller child is in front ⇒ 0
Child 7: Three taller children are in front ⇒ 3

Therefore, the numbers called out by the children are: 0, 0, 1, 0, 3, 0, 3

Hence, the required sequence is 0, 0, 1, 0, 3, 0, 3.

Arrange the stick figure cutouts given at the end of the book 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) 0, 1, 1, 2, 4, 1, 5

Heights (front → back): F, C, B, G, A, D, E

(b) 0, 0, 0, 0, 0, 0, 0

Heights (front → back): A, E, C, G, B, D, F

(c) 0, 1, 2, 3, 4, 5, 6

Heights (front → back): F, D, B, G, C, E, A

(d) 0, 1, 0, 1, 0, 1, 0

Heights (front → back): E, A, G, C, D, B, F

(e) 0, 1, 1, 1, 1, 1, 1

Heights (front → back): F, A, E, C, G, B, D

(f) 0, 0, 0, 3, 3, 3, 3

Heights (front → back): B, D, F, A, E, C, G

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

Recall the rule: each person’s number is the count of taller people standing in front of them.

(a) If a person says ‘0’, then they are the tallest in the group. → Only Sometimes True

Reason — Saying ‘0’ only means no one in front is taller; a taller person could still be standing behind them. For example, the front person always says ‘0’ but need not be the tallest.

(b) If a person is the tallest, then their number is ‘0’. → Always True

Reason — The tallest person has no one taller anywhere in the line, so the number of taller people in front of them is 0.

(c) The first person’s number is ‘0’. → Always True

The first person has no one in front of them, so there are 0 taller people in front, and they must say ‘0’.

(d) If a person is in between, then they cannot say ‘0’. → Only Sometimes True

Reason — A person standing in the middle can say ‘0’ if no one taller than them is standing in front. For example, in the arrangement 3, 7, 5, 4, 6, 2, 1, the second person (height 7) is not first or last, yet says ‘0’ because there is no taller person in front. However, in many arrangements, a person in the middle may have a taller person in front and therefore not say ‘0’.

(e) The person who calls out the largest number is the shortest. → Only Sometimes True

Reason — A large number means that many taller people are standing in front of that person. This may happen when the shortest person stands near the back of the line. However, the shortest person could also stand near the front and say a smaller number. Therefore, the person calling out the largest number is not always the shortest person.

(f) What is the largest number possible in a group of 8 people?

A person can have at most all the people in front of them being taller. The maximum number of people in front of anyone in a line of 8 is 7 (for the last person).

∴ The largest number possible is 7, said by the shortest child when placed at the very back.

Hence, the largest possible number in a group of 8 people is 7.

Kishor has some number cards and is working on a puzzle: There are 5 boxes, and each box should contain exactly 1 number card. The numbers in the boxes should sum to 30. Can you help him find a way to do it?

Can you figure out which 5 cards add to 30? Is it possible?

(The number cards available are 13, 9, 1, 7, 11, 5, and 3, with several copies of each.)

Answer

Look at the cards available: 13, 9, 1, 7, 11, 5 and 3.

Every one of these numbers is odd.

Kishor has to place 5 cards, so he is adding 5 odd numbers.

Now, odd + odd = even (a pair of odd numbers), and adding one more odd number makes the sum odd again. Adding odd numbers, the sum is even after every even count of them and odd after every odd count of them.

Since 5 is an odd count, the sum of any 5 of these odd cards is always odd.

For example:

11 + 9 + 5 + 3 + 1 = 29 (odd)

13 + 9 + 5 + 3 + 1 = 31 (odd)

But 30 is an even number.

⇒ An odd sum can never equal the even number 30.

∴ It is not possible to choose 5 of these cards that add up to 30.

Hence, Kishor cannot fill the 5 boxes to make the sum 30, because the sum of any 5 odd numbers is always odd, while 30 is even.

Explore what happens to the sum of (a) 4 odd numbers, (b) 5 odd numbers, and (c) 6 odd numbers.

Answer

Odd numbers can be paired up with exactly one left over. When odd numbers are added, every two of them combine their leftovers into a complete pair (giving an even total), so the parity of the sum depends only on how many odd numbers are added.

(a) Sum of 4 odd numbers

4 is an even count, so the leftovers pair up completely.

Example: 3 + 5 + 7 + 9 = 24 → even

(b) Sum of 5 odd numbers

5 is an odd count, so one leftover remains unpaired.

Example: 3 + 5 + 7 + 9 + 11 = 35 → odd

(c) Sum of 6 odd numbers

6 is an even count, so all leftovers pair up.

Example: 1 + 3 + 5 + 7 + 9 + 11 = 36 → even

Two siblings, Martin and Maria, were born exactly one year apart. Today they are celebrating their birthday. Maria says that the sum of their ages is 112. Is this possible? Why or why not?

Answer

Since Martin and Maria were born one year apart, their ages are consecutive numbers. Of any two consecutive numbers, one is even and the other is odd.

Therefore:

even + odd = odd

So the sum of their ages must be odd. However, 112 is even.

Hence, the sum of Martin’s and Maria’s ages cannot be 112.

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

We know that:

• even + even = even
• odd + odd = even
• even + odd = odd

(a)

2 even numbers and 2 odd numbers

(even + even) + (odd + odd)

= even + even

= even

(b)

2 odd numbers and 3 even numbers

(odd + odd) + (even + even + even)

= even + even

= even

(c)

5 even numbers

The sum of any number of even numbers is always even.

(d)

8 odd numbers

An even number of odd numbers always adds up to an even number.

Hence, all four sums are even.

Lakpa has an odd number of ₹1 coins, an odd number of ₹5 coins and an even number of ₹10 coins in his piggy bank. He calculated the total and got ₹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

Let the number of ₹1 coins be an odd number, the number of ₹5 coins be an odd number, and the number of ₹10 coins be an even number.

The value of the ₹1 coins is:

odd × 1 = odd

The value of the ₹5 coins is:

odd × 5 = odd

The value of the ₹10 coins is:

even × 10 = even

Therefore, the total amount is:

odd + odd + even

= even + even

= even

⇒ The total amount in the piggy bank must be an even number.

But ₹205 is an odd number.

∴ Lakpa made a mistake. It is not possible to get a total of ₹205 under the given conditions.

We know that:

(a) even + even = even
(b) odd + odd = even
(c) even + odd = odd

Similarly, find out the parity for the scenarios below:

(d) even – even = ...............
(e) odd – odd = ...............
(f) even – odd = ...............
(g) odd – even = ...............

Answer

Subtraction follows the same parity pattern as addition, because subtracting a number changes parity in exactly the same way as adding it.

(d) even – even = even

Example: 8 – 2 = 6 (even).

(e) odd – odd = even

Example: 9 – 3 = 6 (even).

(f) even – odd = odd

Example: 8 – 3 = 5 (odd).

(g) odd – even = odd

Example: 9 – 2 = 7 (odd).

Given the dimensions of a grid, can you tell the parity of the number of small squares without calculating the product?

(i) Find the parity of the number of small squares in these grids:

(a) 27 × 13
(b) 42 × 78
(c) 135 × 654

Answer

The number of small squares in a grid is the product of its two dimensions:

number of small squares = (number of rows) × (number of columns).

For a product:

A product is even if at least one of the two numbers is even (because that even factor supplies a factor of 2).

A product is odd only when both numbers are odd.

⇒ So we can tell the parity just by looking at the parities of the two dimensions:

odd × odd = odd
odd × even = even
even × even = even

∴ The number of small squares is odd only when both dimensions are odd; otherwise it is even.

(i)

We know that:

• odd × odd = odd
• odd × even = even
• even × even = even

(a) 27 × 13

27 and 13 are both odd.

∴ odd × odd = odd.

(b) 42 × 78

42 and 78 are both even.

∴ even × even = even.

(c) 135 × 654

135 is odd and 654 is even.

∴ odd × even = even.

Come up with an expression that always has even parity.

Answer

An expression always has even parity when 2 is a factor of its variable term and its constant term is also even (or zero). Then the whole expression is a multiple of 2 for every integer value of the letter-number.

Some such expressions are:

100p → 2 is a factor → always even

48w – 2 = 2(24w – 1) → 2 is a factor → always even

2m → 2 is a factor → always even

4n + 6 = 2(2n + 3) → 2 is a factor → always even

6k + 2 = 2(3k + 1) → 2 is a factor → always even

For example, taking 4n + 6:

If n = 1:

4n + 6 = 4(1) + 6 = 10 (even)

If n = 2:

4n + 6 = 4(2) + 6 = 14 (even)

Hence, expressions such as 100p, 48w – 2, 2m, 4n + 6 and 6k + 2 always have even parity, because 2 is a factor of each of them.

Come up with expressions that always have odd parity.

Answer

An expression will always have odd parity if it is obtained by adding an odd number to an even expression.

Some such expressions are:

2m + 1 = (even) + 1 → always odd

8a + 3 = (even) + 3 → always odd

4n + 3 = (even) + 3 → always odd

6n + 5 = (even) + 5 → always odd

For example, taking 4n + 3:

If n = 1:

4n + 3 = 4(1) + 3 = 7 (odd)

If n = 2:

4n + 3 = 4(2) + 3 = 11 (odd)

Hence, expressions such as 2m + 1, 8a + 3, 4n + 3 and 6n + 5 always have odd parity, because each is an even term plus an odd constant.

Come up with other expressions, like 3n + 4, which could have either odd or even parity.

Answer

An expression can have either parity when the coefficient of the letter-number is odd. Then the variable term takes the same parity as n, so the value is odd for some values of n and even for others.

Some such expressions are 3n + 4, 5n + 2, 9n + 4, n + 4 and 5n – 2.

Checking 3n + 4:

If n = 1 (odd):

3n + 4 = 3(1) + 4 = 7 (odd)

If n = 2 (even):

3n + 4 = 3(2) + 4 = 10 (even)

Checking 5n + 2:

If n = 1 (odd):

5n + 2 = 5(1) + 2 = 7 (odd)

If n = 2 (even):

5n + 2 = 5(2) + 2 = 12 (even)

Hence, expressions such as 3n + 4, 5n + 2, 9n + 4, n + 4 and 5n – 2 can have either odd or even parity, because the coefficient of n is odd.

Are there expressions using which we can list all the even numbers?

Hint: All even numbers have a factor 2.

Answer

Yes. Every even number has 2 as a factor. Therefore, all even numbers can be written in the form 2n where n = 1, 2, 3, 4,…

For example:

n = 1 ⇒ 2n = 2
n = 2 ⇒ 2n = 4
n = 3 ⇒ 2n = 6
n = 4 ⇒ 2n = 8

Continuing in this way, 2n gives 2, 4, 6, 8, 10, … and no even number is left out.

Hence, the expression 2n lists all the even numbers.

Are there expressions using which we can list all odd numbers?

Answer

Yes. Every odd number is 1 less than the next even number.

Since the even numbers are given by → 2n

The odd numbers are given by → 2n − 1 where n = 1, 2, 3, 4,…

For example:

n = 1 ⇒ 2n − 1 = 1
n = 2 ⇒ 2n − 1 = 3
n = 3 ⇒ 2n − 1 = 5
n = 4 ⇒ 2n − 1 = 7

Continuing in this way, 2n – 1 gives 1, 3, 5, 7, 9, … and no odd number is left out.

Hence, the expression 2n – 1 lists all the odd numbers.

What would be the n^th term for multiples of 2? Or, what is the n^th even number?

Answer

The multiples of 2 are 2, 4, 6, 8, 10, …

Each term is 2 times its position number:

1^st term = 2 × 1 = 2

2^nd term = 2 × 2 = 4

3^rd term = 2 × 3 = 6

So the term at position n is 2 × n.

Hence, the n^th term for multiples of 2, that is the n^th even number, is 2n.

Write a formula to find the n^th odd number.

Answer

Consider the sequences:

Even numbers: 2, 4, 6, 8, 10, …

Odd numbers: 1, 3, 5, 7, 9, …

We know that the n^th even number is 2n

At each position, the odd number is 1 less than the corresponding even number.

Therefore, the n^th odd number is 2n − 1.

For example:

• n = 1 ⇒ 2(1) − 1 = 1
• n = 2 ⇒ 2(2) − 1 = 3
• n = 3 ⇒ 2(3) − 1 = 5

and so on.

Hence, the formula for the n^th odd number is 2n – 1.

Fill the grids below based on the rule mentioned above (the numbers in the yellow circles are the sums of the corresponding rows and columns):

Grid 1:

Grid 2:

Answer

Each grid must be filled with the numbers 1 – 9 without repetition, so that every row adds up to its row-sum circle and every column adds up to its column-sum circle.

Grid 1

Grid 2

Try solving the problem below.

Answer

The numbers 1–9 are to be used without repetition.

Smallest possible sum of three different numbers = 1 + 2 + 3 = 6

Largest possible sum of three different numbers = 7 + 8 + 9 = 24

Therefore, every row sum and column sum must lie between 6 and 24.

But in this grid the first row sum is 5, which is less than 6, and the third column sum is 26, which is greater than 24. Neither of these is possible.

∴ No valid arrangement of the numbers can produce these circled sums.

Hence, the grid cannot be solved.

Using such reasoning, find out which other numbers 1 – 9 cannot occur at the centre of a magic square.

Answer

In a 3 × 3 magic square using the numbers 1–9, the magic sum is 15. Every row, column and diagonal passing through the centre must add up to 15.

We have already seen that 1 and 9 cannot be placed at the centre.

Using the same reasoning:

If 2 is at the centre:

1 + 2 + (third number) = 15

third number = 12

But 12 is not among the numbers 1–9.

∴ 2 cannot be at the centre.

If 3 is at the centre:

1 + 3 + (third number) = 15

third number = 11

But 11 is not among the numbers 1–9.

∴ 3 cannot be at the centre.

If 4 is at the centre:

1 + 4 + (third number) = 15

third number = 10

But 10 is not among the numbers 1–9.

∴ 4 cannot be at the centre.

If 6 is at the centre:

6 + 9 + (third number) = 15

third number = 0

But 0 is not among the numbers 1–9.

∴ 6 cannot be at the centre.

If 7 is at the centre:

7 + 9 + (third number) = 15

third number = −1

But −1 is not among the numbers 1–9.

∴ 7 cannot be at the centre.

If 8 is at the centre:

8 + 9 + (third number) = 15

third number = −2

But −2 is not among the numbers 1–9.

∴ 8 cannot be at the centre.

Thus, the numbers 2, 3, 4, 6, 7, and 8 also cannot occur at the centre.

The only number left is 5.

Hence, apart from 1 and 9, the numbers 2, 3, 4, 6, 7 and 8 also cannot occur at the centre. Therefore, 5 is the only number that can be placed at the centre of a 3 × 3 magic square formed using the numbers 1–9.

Can 1 occur in a corner position of a magic square? If yes, then there should exist three ways of adding 1 with two other numbers to give 15. We have 1 + 5 + 9 = 1 + 6 + 8 = 15. Is any other combination possible?

Answer

A corner cell lies on three lines — one row, one column and one diagonal. So, if 1 were placed in a corner, there would have to be three different pairs of numbers (from 2 – 9) that each add with 1 to give 15, that is, three pairs summing to 15 – 1 = 14.

Listing all pairs of distinct numbers from 2 – 9 that add up to 14:

14 = 5 + 9

14 = 6 + 8

(7 + 7 is not allowed, since a number cannot repeat.)

So there are only two such pairs, giving 1 + 5 + 9 and 1 + 6 + 8. No other combination is possible.

Since a corner needs three such lines but only two pairs exist, 1 cannot be placed in a corner.

Hence, no other combination is possible; only two pairs of numbers add with 1 to give 15, so 1 cannot occur in a corner position.

Similarly, can 9 be placed in a corner position?

Answer

As with 1, a corner cell lies on three lines, so placing 9 in a corner would require three different pairs of numbers (from 1 – 8) that each add with 9 to give 15, that is, three pairs summing to 15 – 9 = 6.

Listing all pairs of distinct numbers from 1 – 8 that add up to 6:

6 = 1 + 5

6 = 2 + 4

(3 + 3 is not allowed, since a number cannot repeat.)

So there are only two such pairs, giving 9 + 1 + 5 and 9 + 2 + 4. As a corner needs three lines but only two pairs exist, 9 cannot be placed in a corner either.

This confirms Observation 3: the numbers 1 and 9 cannot occur in any corner, so they must occur in one of the middle positions.

Hence, only two pairs of numbers add with 9 to give 15, so 9 also cannot occur in a corner position.

Can you find the other possible positions for 1 and 9?

Answer

We know two facts:

• 1 and 9 cannot be at the centre (Observation 2), and
• 1 and 9 cannot be in any corner (Observation 3).

The only cells that remain are the four edge positions — the middle cells of the four sides. So both 1 and 9 must lie in edge positions.

Furthermore, the centre number is 5, and 1 + 5 + 9 = 15. This means 1 and 9 must lie on the same line through the centre — that is, directly opposite each other with 5 in between.

So 1 and 9 occupy the middle cells of two opposite sides: either the two ends of the middle row (with 5 between them), or the two ends of the middle column.

Hence, 1 and 9 must be placed in the middle positions of two opposite sides, directly across from each other through the centre 5, so that 1 + 5 + 9 = 15.

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

Place the line containing 1, 5 and 9 as the middle row, with 5 at the centre:

The remaining numbers to be placed are 2, 3, 4, 6, 7 and 8, in the four corners and the two remaining edge cells.

Using the column through 9 (left column):

9 + top + bottom = 15

⇒ top + bottom = 6

⇒ the pair is {2, 4}

Using the column through 1 (right column):

1 + top + bottom = 15

⇒ top + bottom = 14

⇒ the pair is {6, 8}

The middle column then takes the remaining numbers 3 and 7, since 3 + 5 + 7 = 15.

Adjusting the corners so that the two diagonals also add to 15 gives one valid completion:

Check:

Rows: 2 + 7 + 6 = 9 + 5 + 1 = 4 + 3 + 8 = 15

Columns: 2 + 9 + 4 = 7 + 5 + 3 = 6 + 1 + 8 = 15

Diagonals: 2 + 5 + 8 = 6 + 5 + 4 = 15

Every row, column and diagonal adds up to 15, so this is a valid magic square.

How many different magic squares can be made using the numbers 1 – 9?

Answer

There is only one unique magic square that can be formed using the numbers 1–9 (ignoring rotations and reflections):

All other magic squares are obtained by rotating or reflecting this square.

Hence, there is exactly one unique magic square using the numbers 1–9 (ignoring rotations and reflections).

Create a magic square using the 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 just the numbers 1 – 9 with 1 added to each. So the simplest strategy is to take any magic square made from 1 – 9 and add 1 to every entry.

Starting from a 1 – 9 magic square:

adding 1 to each number gives a magic square for 2 – 10:

Check: every row, column and diagonal adds up to 18 (for example, 3 + 8 + 7 = 18 and 10 + 6 + 2 = 18).

Hence, adding 1 to every entry of a 1 – 9 magic square gives a magic square for 2 – 10 — same structure, centre 6 and magic sum 18.

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

Take the magic square with magic sum 15:

(a) Increasing each number by 1:

Every line has 3 numbers, so each line gains 1 + 1 + 1 = 3. The grid is still a magic square, and the magic sum becomes 15 + 3 = 18.

(b) Doubling each number:

Doubling every number doubles each line total. The grid is still a magic square, and the magic sum becomes 2 × 15 = 30.

Hence, both operations give a magic square.

What other operations can be performed on a magic square to yield another magic square?

Answer

Several operations turn a magic square into another magic square:

• Adding the same number to every entry — if each entry is increased by a number k, every line gains 3k, so the magic sum increases by 3k.
• Subtracting the same number from every entry — the reverse of the above; the magic sum decreases by 3k.
• Multiplying every entry by the same number — if each entry is multiplied by k, every line total is multiplied by k, so the magic sum is also multiplied by k.
• Dividing every entry by the same non-zero number — every line total is divided by that number.
• Rotating the grid by 90°, 180° or 270°.
• Reflecting (flipping) the grid across a row, a column or a diagonal.

More generally, any of these can be combined (for example, double every number and then add 1), and the result is still a magic square.

Hence, adding or subtracting a fixed number to every entry, multiplying or non-zero division on every entry by a fixed number, and rotating or reflecting the grid all produce another magic square.

Discuss ways of creating a magic square using any set of 9 consecutive numbers (like 2 – 10, 3 – 11, 9 – 17, etc.).

Answer

Start with a magic square using the numbers 1–9, for example:

To make a magic square using any other set of 9 consecutive numbers:

• Add the same number to every entry of the square.
• The magic property is preserved because each row, column and diagonal increases by the same amount .

For example:

Using 2–10: Add 1 to every entry.

Similarly:

• For 3–11, add 2 to every entry.
• For 9–17, add 8 to every entry.

Hence, a magic square using any set of 9 consecutive numbers can be obtained by adding the same number to every entry of a magic square using 1–9.

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.

[Hint: Remember, how we described a 2 × 2 grid of a calendar month in the Algebraic Expressions chapter].

(i) Once the generalised form is obtained, share your observations with the class.

Answer

Let us take the magic square made using the consecutive numbers 1 – 9:

The number in the centre is 5, so let m = 5.

Now, express each of the other numbers as "how much more or less than m":

8 = 5 + 3 = m + 3

1 = 5 – 4 = m – 4

6 = 5 + 1 = m + 1

3 = 5 – 2 = m – 2

7 = 5 + 2 = m + 2

4 = 5 – 1 = m – 1

9 = 5 + 4 = m + 4

2 = 5 – 3 = m – 3

Replacing each number by its expression in terms of m gives the generalised form of the magic square:

(i)

Observing the generalised form:

we can share the following observations:

1. The centre entry is m, and the nine entries are the nine consecutive numbers m – 4, m – 3, m – 2, m – 1, m, m + 1, m + 2, m + 3 and m + 4. So m is always the middle (median) of the nine consecutive numbers.

2. Any two entries placed symmetrically opposite each other through the centre add up to 2m (that is, twice the centre number):

(m + 3) + (m – 3) = 2m

(m – 4) + (m + 4) = 2m

(m + 1) + (m – 1) = 2m

(m – 2) + (m + 2) = 2m

3. Adding the three entries of any row, column or diagonal gives 3m, so the magic sum is three times the centre number.

Hence, the whole magic square is controlled by the single centre value m: the entries are nine consecutive numbers centred on m, opposite entries average to m, and the magic sum is 3m.

Using this generalised form, find a magic square if the centre number is 25.

Answer

The generalised form is:

Here the centre number is 25, so m = 25.

Substituting m = 25 in each entry:

m + 3 = 28, m – 4 = 21, m + 1 = 26

m – 2 = 23, m = 25, m + 2 = 27

m – 1 = 24, m + 4 = 29, m – 3 = 22

This gives the magic square:

Each row, column and diagonal adds up to 3m = 3 × 25 = 75.

Hence, the required magic square with centre number 25 is the grid above, whose magic sum is 75.

What is the expression obtained by adding the 3 terms of any row, column or diagonal?

Answer

Consider the generalised form:

Adding the three terms of the first row:

(m + 3) + (m – 4) + (m + 1)

= 3m + (3 – 4 + 1)

= 3m + 0

= 3m

The same is true for every other row, column and diagonal, since the numbers added to and subtracted from m always cancel out.

∴ Sum of any row, column or diagonal = 3m.

Hence, adding the three terms of any row, column or diagonal gives the expression 3m, where m is the centre number.

Write the result obtained by—

(a) adding 1 to every term in the generalised form.
(b) doubling every term in the generalised form.

Answer

The generalised form is:

(a) Adding 1 to every term:

Each entry increases by 1 (for example, m + 3 becomes m + 4, and m becomes m + 1):

(b) Doubling every term:

Each entry is multiplied by 2 (for example, m + 3 becomes 2m + 6, and m – 4 becomes 2m – 8):

Create a magic square whose magic sum is 60.

Answer

The magic sum of the generalised form is 3m.

We need the magic sum to be 60, so:

3m = 60

⇒ m = 20

Substituting m = 20 in the generalised form:

m + 3 = 23, m – 4 = 16, m + 1 = 21

m – 2 = 18, m = 20, m + 2 = 22

m – 1 = 19, m + 4 = 24, m – 3 = 17

This gives the magic square:

Each row, column and diagonal adds up to 60.

Hence, the magic square above, made from the consecutive numbers 16 – 24, has the required magic sum of 60.

Is it possible to get a magic square by filling nine non-consecutive numbers?

Answer

Yes, it is possible.

Starting from the magic square with numbers 1 – 9 and multiplying every number by 3, we get:

The numbers used are 3, 6, 9, 12, 15, 18, 21, 24 and 27, which are non-consecutive (they go up in steps of 3).

Checking the sums:

Rows:

24 + 3 + 18 = 45
9 + 15 + 21 = 45
12 + 27 + 6 = 45

Columns:

24 + 9 + 12 = 45
3 + 15 + 27 = 45
18 + 21 + 6 = 45

Diagonals:

24 + 15 + 6 = 45
18 + 15 + 12 = 45

Every row, column and diagonal adds up to 45, so it is a magic square.

Hence, it is possible to make a magic square using nine non-consecutive numbers, for example the multiples of 3 shown above.

The first-ever recorded 4 × 4 magic square, the Chautīsā Yantra, is shown below. Every row, column and diagonal in this magic square adds up to 34. Can you find other patterns of four numbers in the square that add up to 34?

Answer

Besides the rows, columns and diagonals, several other groups of four numbers also add up to 34:

The four corner numbers:

7 + 14 + 9 + 4 = 34

The four numbers in the centre 2 × 2 block:

13 + 8 + 3 + 10 = 34

The four numbers in each corner 2 × 2 block:

Top-left: 7 + 12 + 2 + 13 = 34

Top-right: 1 + 14 + 8 + 11 = 34

Bottom-left: 16 + 3 + 9 + 6 = 34

Bottom-right: 10 + 5 + 15 + 4 = 34

The two middle numbers of the top row with the two middle numbers of the bottom row:

12 + 1 + 6 + 15 = 34

The two middle numbers of the left column with the two middle numbers of the right column:

2 + 16 + 11 + 5 = 34

Hence, many groups of four numbers — the four corners, the centre four, each 2 × 2 corner block, and the middle numbers of opposite rows or columns — also add up to 34.

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

Using the systematic method: every 6-beat rhythm begins either with a ‘1 +’ (followed by a 5-beat rhythm) or with a ‘2 +’ (followed by a 4-beat rhythm).

Writing a ‘1 +’ in front of all the 8 five-beat rhythms:

1 + 1 + 1 + 1 + 1 + 1

1 + 1 + 1 + 1 + 2

1 + 1 + 1 + 2 + 1

1 + 1 + 2 + 1 + 1

1 + 1 + 2 + 2

1 + 2 + 1 + 1 + 1

1 + 2 + 1 + 2

1 + 2 + 2 + 1

Writing a ‘2 +’ in front of all the 5 four-beat rhythms:

2 + 1 + 1 + 1 + 1

2 + 1 + 1 + 2

2 + 1 + 2 + 1

2 + 2 + 1 + 1

2 + 2 + 2

Counting all the rhythms:

8 (beginning with 1) + 5 (beginning with 2) = 13

Hence, there are exactly 13 ways of writing 6 as a sum of 1’s and 2’s, so there are 13 rhythms having 6 beats.

Write the next 3 numbers in the sequence:

1, 2, 3, 5, 8, 13, 21, 34, 55, 89, .........., .........., ..........

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

In this sequence, each number is the sum of the two numbers before it.

Continuing from 89:

55 + 89 = 144

89 + 144 = 233

144 + 233 = 377

So the next three numbers are 144, 233 and 377.

Predicting the parity of the next number (after 377):

Let us look at the parity (odd/even) of the numbers in the sequence:

1(O), 2(E), 3(O), 5(O), 8(E), 13(O), 21(O), 34(E), 55(O), 89(O), 144(E), 233(O), 377(O), …

The parities repeat in the block odd, even, odd:

O, E, O | O, E, O | O, E, O | ...

Thus, the even terms occur at positions 2, 5, 8, 11, …, while all the other terms are odd.

This pattern follows from the rules odd + even = odd, even + odd = odd and odd + odd = even.

The two numbers just before the required number are 233 (odd) and 377 (odd). So the next number is:

odd + odd = even

∴ Without adding, we can tell that the next number will be even.

Hence, the next three numbers are 144, 233 and 377, and the number after them will be even.

What is the parity of each number in the sequence? Do you notice any pattern in the sequence of parities?

Answer

The Virahāṅka sequence is:

1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, ...

Writing the parity (O for odd, E for even) of each term:

1 → O

2 → E

3 → O

5 → O

8 → E

13 → O

21 → O

34 → E

55 → O

89 → O

144 → E

So the sequence of parities is:

O, E, O, O, E, O, O, E, O, O, E, ...

Reading these in groups of three, the block odd, even, odd keeps repeating.

This happens because each term is the sum of the two before it, and the parities follow the rules O + E = O, E + O = O and O + O = E, which cycle back to the start after every three terms.

Hence, the parities repeat in the cycle odd, even, odd; equivalently, every third term (the 2nd, 5th, 8th, 11th, ...) is even, and all the other terms are odd.

Here are some more questions like this for you to try out. Find out what each letter stands for. (Each letter stands for a particular digit 0 – 9.)

(a) $\begin{array}{r} YY \\ + \space \space Z \\ \hline ZOO \end{array}$

(b) $\begin{array}{r} B5 \\ +\ 3D \\ \hline ED5 \end{array}$

(c) $\begin{array}{r} KP \\ +\ KP \\ \hline PRR \end{array}$

(d) $\begin{array}{r} C1 \\ + \space \space C \\ \hline 1FF \end{array}$

Answer

(a) $\begin{array}{r} YY \\ + \space \space Z \\ \hline ZOO \end{array}$

Adding the two-digit number YY and the single digit Z gives a three-digit number ZOO. The smallest three-digit number is 100, and the largest value of YY + Z is 99 + 9 = 108, so the leading digit of the sum must be Z = 1.

Now YY + Z = 100 + 11 × O, that is 11Y + 1 = 100 + 11O.

⇒ 11Y – 11O = 99

⇒ Y – O = 9

The only digits with a difference of 9 are Y = 9 and O = 0.

Check: 99 + 1 = 100 = ZOO, with Z = 1, O = 0.

Hence, Y = 9, Z = 1 and O = 0.

(b) $\begin{array}{r} B5 \\ +\ 3D \\ \hline ED5 \end{array}$

The largest value of B5 + 3D is 95 + 39 = 134, so the leading digit of the sum is E = 1.

Units column: 5 + D must end in 5, so D = 0 (with no carry).

Tens column: B + 3 = a two-digit result whose tens digit is E = 1 and whose units digit is D = 0, so B + 3 = 10.

⇒ B = 7

Check: 75 + 30 = 105 = ED5, with E = 1, D = 0.

Hence, B = 7, D = 0 and E = 1.

(c) $\begin{array}{r} KP \\ +\ KP \\ \hline PRR \end{array}$

Here KP is added to itself, so PRR = 2 × KP. Since 2 × KP is a three-digit number, KP ≥ 50; and its largest value is 2 × 99 = 198, so the leading digit is P = 1.

So 2 × KP = 100 × P + 11 × R = 100 + 11R, that is 2(10K + 1) = 100 + 11R.

⇒ 20K + 2 = 100 + 11R

⇒ 20K – 98 = 11R

Trying K = 6: 20(6) – 98 = 22 = 11 × 2, so R = 2. No other digit for K works.

Check: 61 + 61 = 122 = PRR, with P = 1, R = 2.

Hence, K = 6, P = 1 and R = 2.

(d) $\begin{array}{r} C1 \\ + \space \space C \\ \hline 1FF \end{array}$

Adding the two-digit number C1 and the single digit C gives a three-digit number 1FF, so the sum is 100 + 11 × F.

C1 + C = 100 + 11F, that is (10C + 1) + C = 100 + 11F.

⇒ 11C + 1 = 100 + 11F

⇒ 11C – 11F = 99

⇒ C – F = 9

The only digits with a difference of 9 are C = 9 and F = 0.

Check: 91 + 9 = 100 = 1FF, with F = 0.

Hence, C = 9 and F = 0.

A light bulb is ON. Dorjee toggles its switch 77 times. Will the bulb be on or off? Why?

Answer

Each toggle of the switch changes the bulb’s state: ON becomes OFF, and OFF becomes ON.

So after an even number of toggles the bulb returns to its starting state, while after an odd number of toggles it is in the opposite state.

The bulb starts ON, and 77 is an odd number.

∴ After 77 toggles the bulb is in the opposite state, that is OFF.

Hence, the bulb will be OFF, because 77 is odd and an odd number of toggles reverses the bulb’s original state.

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

Every sheet of a book is printed on both sides, so it carries two consecutive page numbers — one odd and one even.

For 50 sheets, there are therefore 50 odd page numbers and 50 even page numbers.

The sum of the 50 even page numbers is even.

The sum of the 50 odd page numbers is a sum of an even count of odd numbers, which is also even.

So the total = even + even = even.

Since 6000 is an even number, it does not violate this parity condition.

Hence, the sum of the page numbers of the loose sheets is always even, and because 6000 is even, it is possible for the sum to be 6000.

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 must fill the grid using 3 odd numbers and 3 even numbers so that:

Row 1 sum is odd, Row 2 sum is even, Column 1 sum is even, Column 2 sum is even, Column 3 sum is odd.

Working out which cells must be odd and which must be even from the row and column conditions, and then using the odd numbers 1, 3, 5 and the even numbers 2, 4, 6, one such filling is:

Column checks: 1 + 3 = 4 (even), 2 + 6 = 8 (even), 4 + 5 = 9 (odd). All parities match, and exactly 3 odd numbers (1, 3, 5) and 3 even numbers (2, 4, 6) have been used.

Make a 3 × 3 magic square with 0 as the magic sum. All numbers can not be zero. Use negative numbers, as needed.

Answer

In a 3 × 3 magic square, the magic sum equals 3 times the centre number. For the magic sum to be 0, the centre number must be 0. Using the generalised form of a 3 × 3 magic square with centre m and taking m = 0 gives one such square:

Checking every line:

Rows: 3 – 4 + 1 = 0; –2 + 0 + 2 = 0; –1 + 4 – 3 = 0

Columns: 3 – 2 – 1 = 0; –4 + 0 + 4 = 0; 1 + 2 – 3 = 0

Diagonals: 3 + 0 + (–3) = 0; 1 + 0 + (–1) = 0

Every row, column and diagonal adds up to 0, and not all the numbers are zero.

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

We use the facts that any sum of even numbers is even, and that odd numbers pair up to give an even sum (odd + odd = even).

(a) Sum of an odd number of even numbers = even (a sum of even numbers is always even, whatever the count).

(b) Sum of an even number of odd numbers = even (the odd numbers pair up, and each pair adds to an even number).

(c) Sum of an even number of even numbers = even (a sum of even numbers is always even).

(d) Sum of an odd number of odd numbers = odd (the odd numbers pair up to give an even sum, leaving one odd number over).

What is the parity of the sum of the numbers from 1 to 100?

Answer

From 1 to 100 there are 50 odd numbers (1, 3, 5, ..., 99) and 50 even numbers (2, 4, 6, ..., 100).

The sum of the 50 even numbers is even.

The sum of the 50 odd numbers is a sum of an even count of odd numbers, which is even.

So the total is even + even = even.

We can also confirm this directly:

$1 + 2 + 3 + \cdots + 100 = \dfrac{100 \times 101}{2} = 5050$

which is an even number.

Hence, the parity of the sum of the numbers from 1 to 100 is even.

Two consecutive numbers in the Virahāṅka 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 Virahāṅka sequence, each number is the sum of the two numbers before it.

So to move forward we add, and to move backward we subtract.

Next two numbers (adding):

987 + 1597 = 2584

1597 + 2584 = 4181

Previous two numbers (subtracting):

1597 – 987 = 610

987 – 610 = 377

So the sequence around these terms reads:

..., 377, 610, 987, 1597, 2584, 4181, ...

Hence, the next two numbers are 2584 and 4181, and the previous two numbers are 610 and 377.

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 can count directly by grouping the paths according to how many 2-steps are used:

No 2-step (eight 1-steps): 1 way

One 2-step (with six 1-steps): 7 ways

Two 2-steps (with four 1-steps): 15 ways

Three 2-steps (with two 1-steps): 10 ways

Four 2-steps (no 1-step): 1 way

Total = 1 + 7 + 15 + 10 + 1 = 34 ways.

Hence, Angaan can reach the top of the 8-step staircase in 34 different ways.

What is the parity of the 20th term of the Virahāṅka sequence?

Answer

The Virahāṅka sequence is

1, 2, 3, 5, 8, 13, 21, 34, 55, 89, …

where each term (from the third onward) is the sum of the two terms before it.

We do not need to calculate the actual 20th term — the parity of each term is fixed by the parities of the two terms before it, using the rules:

odd + odd = even
odd + even = odd
even + odd = odd

Starting with the parities of the first two terms (1 is odd, 2 is even) and applying these rules term by term:

O, E, O, O, E, O, O, E, O, O, E, O, O, E, O, O, E, O, O, E, …

The parities repeat in a block of three — O, O, E — so an even term occurs at every position that is a multiple of 3 more than position 2, i.e. at positions 2, 5, 8, 11, 14, 17, 20, …

The 20th position fits this list (20 = 3 × 6 + 2), so the 20th term is even.

⇒ Checking directly, the 20th term is

1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, 610, 987, 1597, 2584, 4181, 6765, 10946,

and 10946 is indeed an even number.

∴ The 20th term of the Virahāṅka sequence is even.

Hence, the parity of the 20th term of the Virahāṅka sequence is even.

Identify the statements that are true.

(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

We use the parity rules and the chapter's understanding that a letter-number stands for a counting number (1, 2, 3, …), just as 2n – 1 gives the nth odd number.

(a) 4m – 1

4m is a multiple of 4, so it is always even.

⇒ 4m – 1 = even – 1 = odd, for every value of m.

So 4m – 1 always gives odd numbers.

Hence, this statement is true.

(b) 6j – 4

For j = 1, 2, 3, … the expression gives:

6(1) – 4 = 2

6(2) – 4 = 8

6(3) – 4 = 14

6(4) – 4 = 20

These jump by 6 each time, so even numbers such as 4, 6, 10 and 12 never appear.

So not every even number can be written as 6j – 4.

Hence, this statement is false.

(c) 2p + 1 and 2q – 1

For q = 1, 2, 3, … the expression 2q – 1 gives 1, 3, 5, 7, … — this is exactly the list of all odd numbers.

For p = 1, 2, 3, … the expression 2p + 1 gives 3, 5, 7, 9, … — this leaves out the odd number 1.

So 2q – 1 describes all odd numbers, but 2p + 1 does not.

Since both expressions do not describe all odd numbers, this statement is false.

(d) 2f + 3

2f is always even.

⇒ 2f + 3 = even + odd = odd, for every value of f.

So 2f + 3 is always odd and never even; it does not give both even and odd numbers.

This statement is false.

Solve this cryptarithm:

$\begin{array}{r} UT \\ +\ TA \\ \hline TAT \end{array}$

Answer

Here each letter stands for a single digit. Writing the numbers by place value:

UT = 10U + T

TA = 10T + A

TAT = 100T + 10A + T = 101T + 10A

Step 1 — Find T.

UT and TA are both 2-digit numbers, so their sum is at most 99 + 99 = 198, which is less than 200.

But TAT is a 3-digit number whose first digit is T.

⇒ T must be 1 (the sum cannot reach 200).

Step 2 — Find A and U.

Substituting T = 1 into UT + TA = TAT:

(10U + 1) + (10 + A) = 101 + 10A

10U + 11 + A = 101 + 10A

10U = 90 + 9A

10U = 9(10 + A)

Now U is a single digit, so 10U is at most 90.

But 10U = 90 + 9A is at least 90.

⇒ 10U = 90 exactly, which forces 9A = 0.

⇒ A = 0 and U = 9.

∴ U = 9, T = 1 and A = 0.

Verification:

UT = 91, TA = 10 and TAT = 101, and indeed 91 + 10 = 101.

Hence, the solution of the cryptarithm is U = 9, T = 1 and A = 0, giving 91 + 10 = 101.