---
title: "NCERT Solutions for Class 8 Maths Chapter 5 Number Play (2026-27)"
url: https://www.swavid.com/maths/class/8/chapter/number-play/ncert-solutions
dateModified: 2026-10-07T15:25:26+00:00
---

# NCERT Solutions for Class 8 Maths Chapter 5 Number Play (2026-27)

This chapter's questions cover properties of consecutive number sums, divisibility rules, and algebraic reasoning. They also include cryptarithms and number puzzles to test number theory concepts.

Free PDF (41 pages): https://www.swavid.com/api/seo/pdf/ncert/maths/class-8/swavid-ncert-solutions-class-8-maths-chapter-5-number-play-b265f7c7ee.pdf

## Math Talk

### Question 1

*3 marks · Short answer*

Can I write every natural number as a sum of consecutive numbers?

**Solution**

1. Natural numbers that are powers of 2 (like 1, 2, 4, 8, etc.) cannot be written as a sum of consecutive natural numbers.
2. All other natural numbers can be expressed as a sum of two or more consecutive natural numbers.
3. For example, 1 and 2 cannot be written as a sum of consecutive natural numbers.

**Answer:** No, powers of 2 cannot be written as a sum of consecutive natural numbers.

> Common mistake: Thinking that all natural numbers can be written as sums of consecutive numbers.

### Question 2

*3 marks · Short answer*

Which numbers can I write as the sum of consecutive numbers in more than one way?

**Solution**

1. Numbers that have an odd factor greater than 1 can be written as the sum of consecutive numbers in more than one way.
2. For example, 15 has odd factors 3 and 5, and can be written as $7 + 8$ or $4 + 5 + 6$ or $1 + 2 + 3 + 4 + 5$.
3. Numbers with only factors of 2 (powers of 2) cannot be written as sums of consecutive numbers at all.

**Answer:** Numbers having an odd factor greater than 1 can be written in more than one way.

> Common mistake: Confusing prime numbers with numbers having odd factors.

### Question 3

*3 marks · Short answer*

Ohh, I know all odd numbers can be written as a sum of two consecutive numbers. Can we write all even numbers as a sum of consecutive numbers?

**Solution**

1. Not all even numbers can be written as a sum of consecutive natural numbers.
2. Even numbers that are powers of 2 (such as 2, 4, 8, 16) cannot be written as a sum of consecutive natural numbers.
3. Other even numbers that have an odd factor (such as 6 = $1 + 2 + 3$, 10 = $1 + 2 + 3 + 4$) can be written as a sum of consecutive numbers.

**Answer:** Only even numbers that are not powers of 2 can be written as a sum of consecutive numbers.

> Common mistake: Assuming all even numbers can be written as sums of consecutive numbers.

### Question 4

*3 marks · Short answer*

Can I write 0 as a sum of consecutive numbers? Maybe I should use negative numbers.

**Solution**

1. If we use only positive natural numbers, 0 cannot be written as a sum of consecutive numbers.
2. If we include negative integers and zero, 0 can be written as $(-1) + 0 + 1 = 0$.
3. Thus, 0 can be written as a sum of consecutive integers using negative numbers.

**Answer:** Yes, 0 can be written as a sum of consecutive numbers if negative integers are included, such as $(-1) + 0 + 1 = 0$.

> Common mistake: Forgetting that negative numbers can be used when exploring integer sequences.

### Question 5

*3 marks · Short answer*

Take any 4 consecutive numbers. For example, 3, 4, 5, and 6. Place '+' and '-' signs in between the numbers. How many different possibilities exist? Write all of them.

**Solution**

1. For 4 consecutive numbers, each of the last 3 numbers can have either a '+' or '-' sign, while the first number has a '+' sign.
2. The total number of different possibilities is $2^3 = 8$.
3. The eight possibilities for 3, 4, 5, and 6 are: $3+4+5+6$, $3+4+5-6$, $3+4-5+6$, $3+4-5-6$, $3-4+5+6$, $3-4+5-6$, $3-4-5+6$, and $3-4-5-6$.

**Answer:** 8 different possibilities exist: $3+4+5+6$, $3+4+5-6$, $3+4-5+6$, $3+4-5-6$, $3-4+5+6$, $3-4+5-6$, $3-4-5+6$, $3-4-5-6$.

> Common mistake: Listing fewer than 8 combinations by missing out sign combinations.

### Question 6

*3 marks · Short answer*

Evaluate each expression and write the result next to it. Do you notice anything interesting?

**Solution**

1. Evaluate the eight expressions for 3, 4, 5, and 6: $18, 6, 8, -4, 10, -2, 0, -12$.
2. Observe that all the evaluated results are even numbers.
3. This shows that regardless of the '+' and '-' signs placed between 4 consecutive numbers, the resulting expression always has even parity.

**Answer:** The results are 18, 6, 8, -4, 10, -2, 0, and -12. All the evaluated results are even numbers.

> Common mistake: Making calculation errors while evaluating negative sign combinations.

### Question 7

*3 marks · Short answer*

Now, take four other consecutive numbers. Place the '+' and '-' signs as you have done before. Find out the results of each expression. What do you observe?

**Solution**

1. Take four other consecutive numbers, for example, 5, 6, 7, and 8.
2. Form expressions using all 8 combinations of '+' and '-' signs.
3. Evaluate the expressions to find results like 12, 4, 2, -6, etc., and observe that all resulting values are even numbers.

**Answer:** All resulting expressions evaluate to even numbers.

> Common mistake: Making arithmetic errors when combining positive and negative integers.

### Question 8

*3 marks · Short answer*

Repeat this for one more set of 4 consecutive numbers. Share your findings.

**Solution**

1. Take another set of 4 consecutive numbers, such as 10, 11, 12, and 13.
2. List the 8 expressions with different combinations of '+' and '-' signs.
3. Find the results and observe that they all maintain an even parity.

**Answer:** The results of all expressions are always even numbers.

> Common mistake: Forgetting to test all eight sign combinations.

### Question 9

*3 marks · Short answer*

Do these patterns occur no matter which 4 consecutive numbers are chosen? Is there a way to find out through reasoning?

**Solution**

1. Yes, these patterns occur for any set of four consecutive numbers.
2. Represent the four numbers using algebra as $a$, $b$, $c$, and $d$.
3. Use algebraic reasoning and the rules of parity to show that switching any sign changes the value by an even number, so all expressions share the same parity.

**Answer:** Yes, it can be proven using algebra and parity rules.

> Common mistake: Failing to express the numbers in general algebraic form.

### Question 10

*3 marks · Short answer*

Now take any 4 numbers, place '+' and '-' signs in the eight different ways, and evaluate the resulting expression. What do you observe about their parities?

**Solution**

1. Take any four arbitrary numbers and form the eight expressions using '+' and '-' signs.
2. Evaluate each resulting expression.
3. Observe that all eight expressions have the exact same parity (either all even or all odd).

**Answer:** All eight expressions always have the same parity.

> Common mistake: Assuming parities can differ between expressions formed from the same four numbers.

### Question 11

*3 marks · Short answer*

Is there a way to explain why this happens?

**Solution**

1. Let us consider any of the expressions formed by four numbers $a, b, c,$ and $d$.
2. When one sign is switched, the value changes by an even number like $2b$, preserving the parity of the expression.
3. Therefore, all the eight expressions formed by any four numbers have the same parity.

**Answer:** All expressions have the same parity because switching signs changes the value by an even number.

> Common mistake: Assuming the parity changes when signs are switched.

### Question 12

*3 marks · Short answer*

Replace any negative sign in the expression $a + b - c - d$ with a positive sign and find the difference between the two numbers.

**Solution**

1. Start with the expression $a + b - c - d$.
2. Replace a negative sign, say $-c$, with a positive sign to get $a + b + c - d$.
3. Find the difference: $(a + b + c - d) - (a + b - c - d) = 2c$, which is an even number.

**Answer:** The difference between the two numbers is $2c$, which is always an even number.

> Common mistake: Incorrect sign expansion when subtracting the two algebraic expressions.

### Question 13

*3 marks · Short answer*

What do you conclude from this observation?

**Solution**

1. Starting from any expression of four numbers, changing any sign changes the value by an even number.
2. Since the change is always even, the parity of the expression remains unchanged.
3. Thus, all the eight expressions formed by any four numbers have the same parity.

**Answer:** All expressions formed by placing '+$' and '-$' signs between four numbers always have the same parity.

> Common mistake: Assuming that changing signs changes the overall parity of the expression.

### Question 14

*3 marks · Short answer*

Is the phenomenon of all the expressions having the same parity limited to taking 4 numbers? What do you think?

**Solution**

1. No, the phenomenon is not limited to taking 4 numbers.
2. For any number of integers combined with '+$' and '-$' signs, switching any sign changes the value by an even number ($2b$).
3. Therefore, all possible combinations of signs for any number of terms will always have the same parity.

**Answer:** No, it applies to any number of terms combined with addition and subtraction.

> Common mistake: Thinking that increasing the number of terms alters the rule of parity change.

### Question 15

*3 marks · Short answer*

Without computing them, find out which of the following arithmetic expressions are even: $43 + 37$, $672 - 348$, $4 \times 347 \times 3$, $708 - 477$, $809 + 214$, $119 \times 303$, $513^3$, $543 - 479$.

**Solution**

1. Odd + odd = even, so $43 + 37$ is even.
2. Even - even = even, so $672 - 348$, $809 + 214$, and $119 \times 303$ (odd $\times$ odd = odd) need checking.
3. Even expressions are $43 + 37$, $672 - 348$, $4 \times 347 \times 3$, $809 + 214$, and $513^3$.

**Answer:** Even expressions are $43 + 37$, $672 - 348$, $4 \times 347 \times 3$, $809 + 214$, and $513^3$.

> Common mistake: Calculating the exact value instead of using parity rules.

### Question 16

*3 marks · Short answer*

Using our understanding of how parity behaves under different operations, identify which of the following algebraic expressions give an even number for any integer values for the letter-numbers: $2a + 2b$, $3g + 5h$, $4m + 2n$, $2u - 4v$, $13k - 5k$, $6m - 3n$, $x^2 + 2$, $b^2 + 1$, $4k \times 3j$.

**Solution**

1. Examine each algebraic expression for factors of 2 or parity combinations.
2. $2a + 2b = 2(a + b)$, which is always even.
3. $4m + 2n = 2(2m + n)$, which is always even.
4. $2u - 4v = 2(u - 2v)$, which is always even.
5. $13k - 5k = 8k = 2(4k)$, which is always even.

**Answer:** $2a + 2b$, $4m + 2n$, $2u - 4v$, and $13k - 5k$ always give an even number.

> Common mistake: Forgetting to factor out 2 to prove an expression is always even.

### Question 17

*3 marks · Short answer*

Similarly, determine and explain which of the other expressions always give even numbers. Write a couple of examples and non-examples, as appropriate, for each expression.

**Solution**

1. Consider expressions like $3g + 5h$ or $x^2 + 2$.
2. For $3g + 5h$, if $g = 1$ and $h = 1$, sum is $8$ (even); if $g = 1$ and $h = 2$, sum is $13$ (odd). Thus it does not always give an even number.
3. For $x^2 + 2$, if $x = 2$, $2^2 + 2 = 6$ (even); if $x = 3$, $3^2 + 2 = 11$ (odd). Thus it is not always even.

**Answer:** Expressions like $3g + 5h$ and $x^2 + 2$ are sometimes even and sometimes odd depending on the values of the variables.

> Common mistake: Assuming an expression with even numbers mixed with odd numbers is always even.

### Question 18

*3 marks · Short answer*

Write a few algebraic expressions which always give an even number.

**Solution**

1. An algebraic expression always gives an even number if it has 2 as a factor for all integer values of the variables.
2. Examples of such expressions include $6x$, $8p + 4q$, and $2(a + b + c)$.
3. Each of these can be written with 2 as a common factor, ensuring an even result for any integer inputs.

**Answer:** $6x$, $8p + 4q$, and $2(a + b + c)$

> Common mistake: Writing expressions that contain odd coefficients without a common factor of 2.

### Question 19

*3 marks · Short answer*

Take a pair of even numbers. Add them. Is the sum divisible by 4?

**Solution**

1. Take two even numbers, which can be represented as multiples of 4 or as $4p + 2$.
2. If both are multiples of 4, their sum is always divisible by 4.
3. If both leave a remainder of 2 when divided by 4, their sum leaves a remainder of $2 + 2 = 4$, making it divisible by 4.

**Answer:** The sum is divisible by 4 if both numbers are multiples of 4 or both leave a remainder of 2 when divided by 4.

> Common mistake: Assuming all sums of two even numbers are divisible by 4.

### Question 20

*3 marks · Short answer*

When is the sum a multiple of 4, and when is it not? Is there a general rule or a pattern?

**Solution**

1. Even numbers leave a remainder of either 0 or 2 when divided by 4.
2. When two numbers that are multiples of 4 are added, their sum is always a multiple of 4.
3. When two even numbers that leave a remainder of 2 are added, their remainders add up to 4, making the sum a multiple of 4.

**Answer:** The sum is a multiple of 4 when both numbers are multiples of 4, or when both leave a remainder of 2 when divided by 4.

> Common mistake: Ignoring the remainder types of even numbers.

### Question 21

*3 marks · Short answer*

When will two even numbers add up to give a multiple of 4?

**Solution**

1. Two even numbers add up to give a multiple of 4 when both are multiples of 4 or when both leave a remainder of 2 when divided by 4.
2. Algebraically, let the numbers be $4p$ and $4q$, then their sum is $4p + 4q = 4(p + q)$, which is a multiple of 4.
3. If the numbers are $4p + 2$ and $4q + 2$, their sum is $4p + 4q + 4 = 4(p + q + 1)$, which is also a multiple of 4.

**Answer:** Two even numbers add up to a multiple of 4 if both are multiples of 4, or if both leave a remainder of 2 when divided by 4.

> Common mistake: Mixing up numbers with different remainders.

### Question 22

*3 marks · Short answer*

What happens when we add a multiple of 4 to an even number that is not a multiple of 4? Is it similar to the case of the parity of the sum of an even and an odd number?

**Solution**

1. Let a multiple of 4 be represented as $4p$ and an even number not a multiple of 4 as $4q + 2$.
2. Their sum is given by $4p + (4q + 2) = 4p + 4q + 2 = 4(p + q) + 2$.
3. This results in an even number that leaves a remainder of 2, which is never a multiple of 4. Yes, it is similar to adding an even and an odd number.

**Answer:** The sum is an even number with a remainder of 2, which is not a multiple of 4. Yes, it is similar to the parity rule of even plus odd.

> Common mistake: Concluding the sum is a multiple of 4.

### Question 23

*3 marks · Short answer*

Look at the following expressions and the visualisation. Write the corresponding explanation and examples.

**Solution**

1. Explanation: Adding a multiple of 4 ($4p$) and an even number not a multiple of 4 ($4q + 2$) gives $4(p + q) + 2$, which leaves a remainder of 2.
2. Visualisation: Combining $p$ rows of 4 and $q$ rows of 4 with an extra row of 2 gives $p + q$ rows fully completed with a remainder of 2 blocks.
3. Examples: $4 + 6 = 10$, $12 + 6 = 18$.

**Answer:** The sum gives $4(p + q) + 2$, representing rows of 4 with a remainder of 2.

> Common mistake: Incorrectly grouping the rows.

### Question 24

*3 marks · Short answer*

Statement 1 is always true. Determine if it is true with subtraction.

**Solution**

1. Let the two numbers be multiples of 8, represented as $8a$ and $8b$.
2. Consider their difference: $8a - 8b = 8(a - b)$.
3. Since 8 is a factor of the difference, 8 exactly divides the difference of two multiples of 8.

**Answer:** Statement 1 is true for subtraction as well; the difference of two multiples of 8 is also a multiple of 8.

> Common mistake: Assuming subtraction behaves differently than addition regarding multiples.

### Question 2

*3 marks · Short answer*

If a number is divisible by 8, then 8 also divides any two numbers (separately) that add up to the number.

**Solution**

1. A number divisible by 8 is a multiple of 8, which can be written as $8m$.
2. A number divisible by 8 can be expressed as a sum of two multiples of 8 or as a sum of two non-multiples of 8, such as $8m = 8a + 8b$ or $8m = p + q$.
3. Therefore, 8 divides any two numbers separately that add up to the number only when those two numbers are also multiples of 8, making the statement sometimes true.

**Answer:** Sometimes true, as 8 divides two numbers separately adding up to a multiple of 8 only if both numbers are multiples of 8.

> Common mistake: Assuming that if a sum is divisible by 8, any two numbers adding up to it must be divisible by 8.

### Question 3

*3 marks · Short answer*

If a number is divisible by 7, then all multiples of that number will be divisible by 7.

**Solution**

1. A number divisible by 7 is a multiple of 7 and can be represented as $7j$.
2. Any multiple of this number can be written as $(7j) \times m = 7(jm)$, which has 7 as a factor.
3. Thus, all multiples of a number divisible by 7 are also divisible by 7, making the statement always true.

**Answer:** Always true, because any multiple of a number divisible by 7 contains 7 as a factor.

> Common mistake: Confusing multiples of a number with factors of a number.

### Question 4

*3 marks · Short answer*

If a number is divisible by 12, then the number is also divisible by all the factors of 12.

**Solution**

1. A number divisible by 12 is a multiple of 12, written as $12m$.
2. The prime factorisation or factors of $12m$ include the factors of 12 because $12m = 2 \times 6 \times m = 3 \times 4 \times m$.
3. Hence, any factor of 12 covers the rows fully, meaning the number is divisible by all the factors of 12.

**Answer:** Always true, because a multiple of 12 contains all the factors of 12 in its factorisation.

> Common mistake: Thinking that only prime factors divide the number.

### Question 5

*3 marks · Short answer*

If a number is divisible by 7, then it is also divisible by any multiple of 7.

**Solution**

1. Numbers divisible by 7 are multiples of 7, represented as $7k$.
2. For $7k$ to be divisible by a multiple of 7 like $7m$, $k$ must be a multiple of $m$, so $k = ym$.
3. Then $7k \div 7m = 7ym \div 7m = y$, which is only an integer if $m$ is a factor of $k$, making the statement sometimes true.

**Answer:** Sometimes true, because $7k$ is divisible by $7m$ if and only if $m$ is a factor of $k$.

> Common mistake: Assuming a number divisible by 7 is divisible by every multiple of 7.

### Question 6

*3 marks · Short answer*

If a number is divisible by both 9 and 4, it must be divisible by 36.

**Solution**

1. If a number is divisible by both 9 and 4, it must be a common multiple of 9 and 4.
2. Since 9 and 4 are co-prime, their least common multiple is $\text{LCM}(9, 4) = 36$.
3. Therefore, any number divisible by both 9 and 4 must be divisible by their LCM, which is 36.

**Answer:** Always true, because if a number is divisible by two co-prime numbers, it is divisible by their product.

> Common mistake: Applying the LCM rule when the two divisors are not co-prime.

### Question 7

*3 marks · Short answer*

If a number is divisible by both 6 and 4, it must be divisible by 24.

**Solution**

1. Examine the statement: If a number is divisible by both 6 and 4, it must be divisible by 24.
2. Consider the number 12, which is divisible by both 6 ($12 = 6 \times 2$) and 4 ($12 = 4 \times 3$).
3. However, 12 is not divisible by 24, so the statement is sometimes true.

**Answer:** Sometimes true (for example, 48 is divisible by 24, but 12 is divisible by 6 and 4 yet not by 24).

> Common mistake: Assuming that if a number is divisible by two numbers, it must be divisible by their product without checking if the divisors are coprime.

### Question 8

*3 marks · Short answer*

When you add an odd number to an even number we get a multiple of 6.

**Solution**

1. Suppose the statement is true, so the sum of an even number $2n$ and an odd number $2m + 1$ equals a multiple of $6j$.
2. We can write this as $2n + (2m + 1) = 6j$, which simplifies to $2(n + m) = 6j - 1$.
3. Here, the left-hand side $2(n + m)$ is an even number, while the right-hand side $6j - 1$ is an odd number, which is impossible.
4. Therefore, the statement is never true.

**Answer:** The statement is never true because the sum of an even and an odd number is always odd, whereas multiples of 6 are even.

> Common mistake: Assuming that any sum involving even numbers can result in any multiple.

### Question 32

*3 marks · Short answer*

Find a number that has a remainder of 3 when divided by 5. Write more such numbers.

**Solution**

1. Numbers that leave a remainder of 3 when divided by 5 are 3 more than the multiples of 5.
2. The multiples of 5 are given by the expression $5k$, where $k$ is an integer.
3. Thus, numbers of this form can be written as $5k + 3$.
4. Examples of such numbers for $k = 0, 1, 2, 3$ are $3, 8, 13, 18$.

**Answer:** The numbers are $3, 8, 13, 18, \dots$ represented generally as $5k + 3$.

> Common mistake: Confusing the remainder with the divisor.

### Question 33

*1 mark · MCQ*

Which algebraic expression(s) capture all such numbers?

- 3k + 5
- 3k - 5
- 3k / 5
- 5k + 3
- 5k - 2
- 5k - 3

**Solution**

1. A number leaving a remainder of 3 when divided by 5 can be written as $5k + 3$.
2. Alternatively, writing it as 2 less than the next multiple of 5 gives $5k - 2$.
3. Therefore, options (iv) $5k + 3$ and (v) $5k - 2$ capture all such numbers.

**Answer:** (iv) $5k + 3$ and (v) $5k - 2$

> Common mistake: Selecting $3k + 5$ by reversing the divisor and remainder.

### Question 34

*3 marks · Short answer*

Let us consider another expression, 5k - 2, and see the values it takes for different values of k.

**Solution**

1. Substitute integer values for $k$ into the expression $5k - 2$.
2. For $k = 1, 2, 3, 4, 5$, the values obtained are $5(1) - 2 = 3$, $5(2) - 2 = 8$, $5(3) - 2 = 13$, $5(4) - 2 = 18$, and $5(5) - 2 = 23$.
3. These are the exact same numbers generated by the expression $5k + 3$ for $k \ge 0$.

**Answer:** The expression $5k - 2$ generates the sequence $3, 8, 13, 18, 23, \dots$ for $k \ge 1$.

> Common mistake: Starting $k$ from 0 instead of 1 when using $5k - 2$.

### Question 35

*3 marks · Short answer*

Are there other expressions that generate numbers that are 3 more than a multiple of 5?

**Solution**

1. Yes, there are other expressions that can generate the same set of numbers.
2. By adjusting the base multiple, expressions such as $5k - 7$ (for $k \ge 2$) or $5k + 8$ also generate numbers that leave a remainder of 3 when divided by 5.
3. Any expression of the form $5k + 3$ with shifted integer indexing represents the same remainder class.

**Answer:** Yes, expressions like $5k - 7$ or $5k + 8$ also generate numbers that leave a remainder of 3 when divided by 5.

> Common mistake: Assuming only one algebraic expression can represent a remainder class.

## Figure it Out

### Question 1

*3 marks · Short answer*

The sum of four consecutive numbers is 34. What are these numbers?

**Solution**

1. Let the four consecutive numbers be $x$, $x+1$, $x+2$, and $x+3$.
2. Their sum is given as $4x + 6 = 34$, which means $4x = 28$ or $x = 7$.
3. Thus, the numbers are $7$, $8$, $9$, and $10$.

**Answer:** The required consecutive numbers are 7, 8, 9, and 10.

> Common mistake: Assuming the numbers are $x$, $x+1$, $x+2$ instead of four numbers.

### Question 2

*3 marks · Short answer*

Suppose p is the greatest of five consecutive numbers. Describe the other four numbers in terms of p.

**Solution**

1. Given that $p$ is the greatest of five consecutive numbers.
2. The numbers immediately preceding $p$ decrease by $1$ successively.
3. Therefore, the other four numbers in terms of $p$ are $(p - 1)$, $(p - 2)$, $(p - 3)$, and $(p - 4)$.

**Answer:** The other four numbers are $(p - 1)$, $(p - 2)$, $(p - 3)$, and $(p - 4)$.

> Common mistake: Adding instead of subtracting when moving backwards from the greatest number.

### Question 3

*5 marks · Case-based*

For each statement below, determine whether it is always true, sometimes true, or never true. Explain your answer. Mention examples and non-examples as appropriate. Justify your claim using algebra.
(i) The sum of two even numbers is a multiple of 3.
(ii) If a number is not divisible by 18, then it is also not divisible by 9.
(iii) If two numbers are not divisible by 6, then their sum is not divisible by 6.
(iv) The sum of a multiple of 6 and a multiple of 9 is a multiple of 3.
(v) The sum of a multiple of 6 and a multiple of 3 is a multiple of 9.

**Part (i)**

1. Let the two even numbers be $2a$ and $2b$. Their sum is $2a + 2b = 2(a + b)$, which is always even but not always a multiple of 3.
2. Example: $2 + 4 = 6$ (multiple of 3), but $2 + 6 = 8$ (not a multiple of 3).
3. Conclusion: Sometimes true.

Answer (i): Sometimes true

**Part (ii)**

1. A number can be not divisible by 18 yet be divisible by 9 (e.g., 27) or not divisible by 9 (e.g., 30).
2. Example 1: 30 is not divisible by 18 and also not divisible by 9.
3. Example 2: 27 is not divisible by 18, but it is divisible by 9.
4. Conclusion: Sometimes true.

Answer (ii): Sometimes true

**Part (iii)**

1. Two numbers not divisible by 6 can sum up to a multiple of 6 or a non-multiple of 6.
2. Example 1: 8 and 10 are not divisible by 6, but their sum $8 + 10 = 18$ is divisible by 6.
3. Example 2: 9 and 11 are not divisible by 6, and their sum $9 + 11 = 20$ is also not divisible by 6.
4. Conclusion: Sometimes true.

Answer (iii): Sometimes true

**Part (iv)**

1. Let the multiple of 6 be $6x$ and the multiple of 9 be $9y$.
2. Their sum is $6x + 9y = 3(2x + 3y)$, which has 3 as a factor.
3. Conclusion: Always true.

Answer (iv): Always true

**Part (v)**

1. Let the multiple of 6 be 18 and the multiple of 3 be 9. Their sum is $18 + 9 = 27$, which is a multiple of 9.
2. Let the multiple of 6 be 12 and the multiple of 3 be 9. Their sum is $12 + 9 = 21$, which is not a multiple of 9.
3. Conclusion: Sometimes true.

Answer (v): Sometimes true

**Answer:** Determined whether each statement is always true, sometimes true, or never true with examples and algebra.

> Common mistake: Confusing 'always true' with 'sometimes true' by checking only one or two examples instead of testing both cases or using algebra.

### Question 4

*3 marks · Short answer*

Find a few numbers that leave a remainder of 2 when divided by 3 and a remainder of 2 when divided by 4. Write an algebraic expression to describe all such numbers.

**Solution**

1. Let the number be $x$. We are given $x = 3a + 2$ and $x = 4b + 2$.
2. This means $x - 2$ is a common multiple of both 3 and 4.
3. The LCM of 3 and 4 is 12, so $x - 2 = 12n$, giving the general expression $12n + 2$.

**Answer:** The general algebraic expression is $12n + 2$, giving numbers like 14, 26, 38, etc.

> Common mistake: Forgetting to add the remainder back after finding the LCM.

### Question 5

*3 marks · Short answer*

“I hold some pebbles, not too many,
When I group them in 3’s, one stays with me.
Try pairing them up — it simply won’t do,
A stubborn odd pebble remains in my view.
Group them by 5, yet one’s still around,
But grouping by seven, perfection is found.
More than one hundred would be far too bold,
Can you tell me the number of pebbles I hold?”

**Solution**

1. Let the number of pebbles be $N$. Grouping in 3s leaves 1, pairing up leaves 1 (odd), and grouping by 5 leaves 1.
2. Grouping by 7 leaves a remainder of 0, meaning $N$ is a multiple of 7.
3. The common remainder for 3, 2, and 5 is 1, so $N = \text{LCM}(3, 2, 5)k + 1 = 30k + 1$ with $N < 100$ and $N$ being a multiple of 7.
4. Checking values for $k$: for $k = 3$, $N = 30(3) + 1 = 91$, which is divisible by 7.

**Answer:** The number of pebbles is 91.

> Common mistake: Ignoring the condition that the number must be less than 100 and a multiple of 7.

### Question 6

*3 marks · Short answer*

Tathagat has written several numbers that leave a remainder of 2 when divided by 6. He claims, “If you add any three such numbers, the sum will always be a multiple of 6.” Is Tathagat’s claim true?

**Solution**

1. Let the numbers leaving a remainder of 2 when divided by 6 be represented as $6a + 2$, $6b + 2$, and $6c + 2$.
2. Adding these three numbers gives $(6a + 2) + (6b + 2) + (6c + 2) = 6a + 6b + 6c + 6 = 6(a + b + c + 1)$.
3. Since the sum has 6 as a factor, it is always a multiple of 6, making Tathagat's claim true.

**Answer:** Yes, Tathagat's claim is true.

> Common mistake: Adding the remainders incorrectly or failing to factor out 6 from the expression.

### Question 7

*3 marks · Short answer*

When divided by 7, the number 661 leaves a remainder of 3, and 4779 leaves a remainder of 5. Without calculating, can you say what remainders the following expressions will leave when divided by 7? Show the solution both algebraically and visually.
(i) 4779 + 661
(ii) 4779 - 661

**Part (i)**

1. Let $4779 = 7p + 5$ and $661 = 7q + 3$.
2. Their sum is $4779 + 661 = 7p + 5 + 7q + 3 = 7(p + q) + 8 = 7(p + q + 1) + 1$.
3. The remainder when divided by 7 is 1.

Answer (i): Remainder is 1

**Part (ii)**

1. Let $4779 = 7p + 5$ and $661 = 7q + 3$.
2. Their difference is $4779 - 661 = (7p + 5) - (7q + 3) = 7p - 7q + 2 = 7(p - q) + 2$.
3. The remainder when divided by 7 is 2.

Answer (ii): Remainder is 2

**Answer:** Part (i) leaves a remainder of 1 and Part (ii) leaves a remainder of 2.

> Common mistake: Forgetting to group the multiples of 7 correctly when opening parentheses for subtraction.

### Question 8

*3 marks · Short answer*

Find a number that leaves a remainder of 2 when divided by 3, a remainder of 3 when divided by 4, and a remainder of 4 when divided by 5. What is the smallest such number? Can you give a simple explanation of why it is the smallest?

**Solution**

1. Observe that the given remainders are 2, 3, and 4 for divisors 3, 4, and 5 respectively, meaning each remainder is 1 less than its divisor ($3 - 2 = 1$, $4 - 3 = 1$, $5 - 4 = 1$).
2. Find the least common multiple (LCM) of the divisors 3, 4, and 5, which is $3 \times 4 \times 5 = 60$.
3. Subtract 1 from the LCM because the number is 1 less than a multiple of each divisor, giving $60 - 1 = 59$.
4. Thus, the smallest such number is 59, since 60 is the smallest positive common multiple of 3, 4, and 5.

**Answer:** 59

> Common mistake: Adding 1 instead of subtracting 1 from the LCM.

## Math Talk

### Question 1

*3 marks · Short answer*

Similarly, explain using algebra why the divisibility shortcuts for 5, 2, 4, and 8 work.

**Solution**

1. Any number in general can be written as the sum of its place values, such as $\dots + 100d + 10c + a$.
2. For 2, 5, and 10, all place values except the units place $a$ are multiples of 2, 5, and 10 respectively, so divisibility depends entirely on $a$.
3. For 4 and 8, all place values from hundreds and thousands onwards are multiples of 4 and 8 respectively, so divisibility depends only on the last two or three digits.

**Answer:** Divisibility rules depend on the fact that place values from tens, hundreds, or thousands upwards are already multiples of the given divisor.

> Common mistake: Confusing the place value expansion with just the digits.

### Question 2

*3 marks · Short answer*

Can you say, without actually calculating, which of these numbers are divisible by 9: 999, 909, 900, 90, 990?

**Solution**

1. We know that a number is divisible by 9 if the sum of its digits is divisible by 9.
2. For the numbers 999, 909, 900, 90, and 990, the sum of the digits are 27, 18, 9, 9, and 18 respectively.
3. Since all these digit sums are multiples of 9, all these numbers are divisible by 9.

**Answer:** All of them (999, 909, 900, 90, and 990) are divisible by 9.

> Common mistake: Performing actual long division instead of using the digit sum shortcut.

### Question 3

*3 marks · Short answer*

Can we say that any number made up of only the digits '0' and '9', in any order, will always be divisible by 9?

**Solution**

1. Let us write the expanded form of a number made of only digits 0 and 9.
2. Each non-zero term in the expanded form will be of the form $9 \times \text{place value}$ or $0 \times \text{place value}$.
3. Since every term is a multiple of 9, the entire number must be divisible by 9.

**Answer:** Yes, any number made up of only the digits 0 and 9 will always be divisible by 9.

> Common mistake: Assuming numbers with 9 are not divisible by 9 unless checked completely.

### Question 4

*3 marks · Short answer*

Is 10 divisible by 9? If not, what is the remainder?

**Solution**

1. Divide 10 by 9 using division: $10 = 9 \times 1 + 1$.
2. The quotient is 1 and the remainder is 1.
3. Therefore, 10 is not divisible by 9, and the remainder is 1.

**Answer:** No, 10 is not divisible by 9; the remainder is 1.

> Common mistake: Saying 10 is divisible because 1 + 0 = 1.

### Question 5

*3 marks · Short answer*

Check the divisibility of other multiples of 10 (10, 20, 30, ...) by 9.

**Solution**

1. Consider multiples of 10 such as 10, 20, 30, and divide them by 9.
2. $10 = 9 \times 1 + 1$, $20 = 9 \times 2 + 2$, $30 = 9 \times 3 + 3$.
3. We notice that for any multiple of 10, the remainder when divided by 9 is equal to the number of tens.

**Answer:** The remainder when any multiple of 10 is divided by 9 is equal to its number of tens.

> Common mistake: Confusing tens digit with the remainder value for numbers greater than 90.

### Question 6

*3 marks · Short answer*

Similarly, look at the remainder when the multiples of 100 (100, 200, 300, ...) are divided by 9. What do you notice?

**Solution**

1. Consider multiples of 100 such as 100, 200, 300, and divide them by 9.
2. $100 = 99 + 1 = 9 \times 11 + 1$, $200 = 198 + 2 = 9 \times 22 + 2$.
3. We notice that the remainder is the same as the number of hundreds.

**Answer:** The remainder when any multiple of 100 is divided by 9 is equal to its number of hundreds.

> Common mistake: Dividing the number fully instead of observing the place value remainder pattern.

### Question 7

*3 marks · Short answer*

Using this observation, find the remainder when 427 is divided by 9.

**Solution**

1. The number 427 has 4 hundreds, 2 tens, and 7 units.
2. The remainder when 400 is divided by 9 is 4, for 20 is 2, and for 7 is 7.
3. Adding the remainders gives $4 + 2 + 7 = 13$, which leaves a remainder of $4$ when divided by 9.

**Answer:** 4

> Common mistake: Adding the place values directly instead of their digits or number of hundreds and tens.

### Question 8

*3 marks · Short answer*

Will this work with bigger numbers?

**Solution**

1. Each place value like $1, 10, 100, 1000$ can be written as a multiple of 9 plus 1 (e.g., $100 = 99 + 1$).
2. Expanding a number like 7309 gives $(7 \times 999 + 3 \times 99 + 0 \times 9) + (7 + 3 + 0 + 9)$.
3. The first part is a multiple of 9, leaving the sum of the digits $7 + 3 + 0 + 9 = 19$ to determine the remainder.

**Answer:** Yes, it works for any number because each power of 10 leaves a remainder of 1 when divided by 9.

> Common mistake: Forgetting that higher powers of 10 leave a remainder of 1.

### Question 9

*4 marks · Long answer*

Look at each of the following statements. Which are correct and why?
(i) If a number is divisible by 9, then the sum of its digits is divisible by 9.
(ii) If the sum of the digits of a number is divisible by 9, then the number is divisible by 9.
(iii) If a number is not divisible by 9, then the sum of its digits is not divisible by 9.
(iv) If the sum of the digits of a number is not divisible by 9, then the number is not divisible by 9.

**Part (i)**

1. If a number is divisible by 9, its remainder when divided by 9 is 0.
2. Since the digital root equals the remainder when divided by 9, the sum of its digits must be a multiple of 9.

Answer (i): Correct

**Part (ii)**

1. If the sum of the digits is divisible by 9, the remainder upon division by 9 is 0.
2. Therefore, the number itself is divisible by 9.

Answer (ii): Correct

**Part (iii)**

1. If a number is not divisible by 9, its remainder is not 0.
2. Thus, the sum of its digits cannot be a multiple of 9.

Answer (iii): Correct

**Part (iv)**

1. If the sum of the digits is not divisible by 9, the remainder is not 0.
2. Hence, the number is not divisible by 9.

Answer (iv): Correct

**Answer:** All four statements are correct.

> Common mistake: Confusing the converse statements without checking logical equivalence.

## Figure it Out

### Question 1

*3 marks · Short answer*

Find, without dividing, whether the following numbers are divisible by 9:
(i) 123 (ii) 405 (iii) 8888 (iv) 93547 (v) 358095

**Part (i)**

1. Sum of digits of 123 = 1 + 2 + 3 = 6.
2. Since 6 is not divisible by 9, 123 is not divisible by 9.

Answer (i): Not divisible

**Part (ii)**

1. Sum of digits of 405 = 4 + 0 + 5 = 9.
2. Since 9 is divisible by 9, 405 is divisible by 9.

Answer (ii): Divisible

**Part (iii)**

1. Sum of digits of 8888 = 8 + 8 + 8 + 8 = 32.
2. Since 32 is not divisible by 9, 8888 is not divisible by 9.

Answer (iii): Not divisible

**Part (iv)**

1. Sum of digits of 93547 = 9 + 3 + 5 + 4 + 7 = 28.
2. Since 28 is not divisible by 9, 93547 is not divisible by 9.

Answer (iv): Not divisible

**Part (v)**

1. Sum of digits of 358095 = 3 + 5 + 8 + 0 + 9 + 5 = 30.
2. Since 30 is not divisible by 9, 358095 is not divisible by 9.

Answer (v): Not divisible

**Answer:** The numbers divisible by 9 are identified by checking if the sum of their digits is divisible by 9.

> Common mistake: Forgetting that the rule for divisibility by 9 requires the sum of the digits to be a multiple of 9, not just any number.

### Question 2

*3 marks · Short answer*

Find the smallest multiple of 9 with no odd digits.

**Solution**

1. The even digits available are $0, 2, 4, 6, 8$.
2. The smallest possible sum of digits that is a multiple of $9$ using even digits is $18$ ($2 + 8 + 8 = 18$).
3. Arranging these digits to form the smallest number gives $288$.

**Answer:** 288

> Common mistake: Choosing odd digits or a sum of digits that is not a multiple of 9.

### Question 3

*3 marks · Short answer*

Find the multiple of 9 that is closest to the number 6000.

**Solution**

1. Divide $6000$ by $9$: $6000 = 9 \times 666 + 6$.
2. The multiples of $9$ near $6000$ are $9 \times 666 = 5994$ and $9 \times 667 = 6003$.
3. The difference from $6000$ is $6000 - 5994 = 6$ and $6003 - 6000 = 3$, so $6003$ is closer.

**Answer:** 6003

> Common mistake: Finding a multiple that is further away or calculating the division incorrectly.

### Question 4

*3 marks · Short answer*

How many multiples of 9 are there between the numbers 4300 and 4400?

**Solution**

1. Find the first multiple of $9$ greater than $4300$: $4300 \div 9 = 477.7$, so $478 \times 9 = 4302$.
2. Find the last multiple of $9$ less than $4400$: $4400 \div 9 = 488.8$, so $488 \times 9 = 4392$.
3. Count the number of multiples from $478$ to $488$: $488 - 478 + 1 = 11$.

**Answer:** 11

> Common mistake: Forgetting to include one of the boundary multiples, resulting in an incorrect count of 10.

## Math Talk

### Question 1

*3 marks · Short answer*

Using these observations, can you tell whether the number 462 is divisible by 11?

**Solution**

1. Place alternating '+' and '-' signs starting from the units digit: $+2 - 6 + 4$.
2. Evaluate the expression: $+2 - 6 + 4 = 0$.
3. Since the result is $0$, which is a multiple of $11$, the number $462$ is divisible by $11$.

**Answer:** Yes, 462 is divisible by 11.

> Common mistake: Starting the alternating signs from the left instead of the units digit.

### Question 2

*3 marks · Short answer*

What could be a general method or shortcut to check divisibility by 11?

**Solution**

1. Find the sum of the digits at odd places starting from the units place.
2. Find the sum of the digits at even places.
3. Find the difference between these two sums; if the difference is $0$ or a multiple of $11$, the number is divisible by $11$.

**Answer:** The shortcut is to check if the difference between the sum of digits at odd places and even places is a multiple of 11.

> Common mistake: Confusing odd and even place positions.

### Question 3

*3 marks · Short answer*

If this difference is 11 or a multiple of 11, what does that say about the remainder obtained when the number is divisible by 11?

**Solution**

1. When a number is completely divisible by $11$, there is no remainder.
2. The difference between the sums of alternating digits represents the remainder or excess from a multiple of $11$.
3. If this difference is $11$ or a multiple of $11$, the remainder obtained is $0$.

**Answer:** The remainder obtained is zero.

> Common mistake: Stating that the remainder is 11 instead of 0.

### Question 4

*3 marks · Case-based*

Using this shortcut, find out whether the following numbers are divisible by 11. Further, find the remainder if the number is not divisible by 11.
(i) 158 (ii) 841 (iii) 481 (iv) 5529 (v) 90904 (vi) 857076

**Part (i)**

1. Evaluate alternating sum: $-8 + 5 - 1 = -4$.
2. Since $-4$ is $4$ short of a multiple of $11$, the remainder is $4$.

Answer (i): Remainder is 4

**Part (ii)**

1. Evaluate alternating sum: $+1 - 4 + 8 = 5$.
2. The result is $5$, which means the remainder is $5$.

Answer (ii): Remainder is 5

**Part (iii)**

1. Evaluate alternating sum: $+1 - 8 + 4 = -3$.
2. Since $-3$ is $3$ short of a multiple of $11$, the remainder is $8$.

Answer (iii): Remainder is 8

**Part (iv)**

1. Evaluate alternating sum: $+9 - 2 + 5 - 5 = 7$.
2. The result is $7$, which means the remainder is $7$.

Answer (iv): Remainder is 7

**Part (v)**

1. Evaluate alternating sum: $+4 - 0 + 9 - 0 + 9 = 22$.
2. Since $22$ is a multiple of $11$, the number is divisible by $11$.

Answer (v): Divisible by 11

**Part (vi)**

1. Evaluate alternating sum: $+6 - 7 + 0 - 7 + 5 - 8 = -11$.
2. Since $-11$ is a multiple of $11$, the number is divisible by $11$.

Answer (vi): Divisible by 11

**Answer:** Divisibility and remainders determined using alternating sums of digits.

> Common mistake: Incorrectly handling negative remainders.

### Question 5

*3 marks · Short answer*

Is this method similar to or different from the method we saw just before?

**Solution**

1. The first method computes excess and short sums separately and subtracts them.
2. The second method places alternating '+' and '-' signs directly from the units digit and evaluates the result.
3. Both methods are mathematically identical and yield the same result.

**Answer:** The method is essentially the same, just written in a simpler and quicker procedural form.

> Common mistake: Thinking they are entirely different concepts.

### Question 6

*3 marks · Short answer*

Fill in the following table. Find a quick way to do this?

**Solution**

1. Apply divisibility rules for $2, 3, 4, 5, 6, 8, 9, 10,$ and $11$ to each number.
2. Check units digit for $2, 5, 10$, sum of digits for $3, 9$, last digits for $4, 8$, and alternating sums for $11$.
3. Fill the table based on whether the number satisfies each divisibility condition.

**Answer:** Table filled by applying standard divisibility shortcuts for each number.

> Common mistake: Misapplying the divisibility rule for 4 or 8.

### Question 7

*3 marks · Short answer*

How can we find out if a number is divisible by 6?

**Solution**

1. A number is divisible by 6 if it is divisible by both 2 and 3.
2. We check divisibility by 2 by ensuring the units digit is even (0, 2, 4, 6, or 8).
3. We check divisibility by 3 by ensuring the sum of the digits is a multiple of 3.

**Answer:** A number is divisible by 6 if it is divisible by both 2 and 3.

> Common mistake: Checking only divisibility by 2 or only by 3 instead of both.

### Question 8

*3 marks · Short answer*

Will checking its divisibility by its factors 2 and 3 work? Use the shortcuts for 2 and 3 on these numbers and divide each number by 6 to verify—38, 225, 186, 64.

**Solution**

1. For 38: divisible by 2 (units digit 8), but sum of digits = 11 (not divisible by 3). Not divisible by 6 ($38 \div 6 = 6$ remainder 2).
2. For 225: not divisible by 2 (units digit 5), but sum of digits = 9 (divisible by 3). Not divisible by 6 ($225 \div 6 = 37$ remainder 3).
3. For 186: divisible by 2 (units digit 6) and sum of digits = 15 (divisible by 3). Divisible by 6 ($186 \div 6 = 31$).
4. For 64: divisible by 2 (units digit 4), but sum of digits = 10 (not divisible by 3). Not divisible by 6 ($64 \div 6 = 10$ remainder 4).

**Answer:** Only 186 is divisible by 6 among the given numbers.

> Common mistake: Concluding a number is divisible by 6 by checking only one factor.

### Question 9

*3 marks · Short answer*

How about checking divisibility by 24? Will checking the divisibility by its factors, 4 and 6, work? Why or why not?

**Solution**

1. Checking divisibility by 24 using its factors 4 and 6 does not work.
2. For example, the number 12 is divisible by both 4 and 6, but it is not divisible by 24.
3. Factors must be co-prime for such tests to work, which 4 and 6 are not as they share a common factor of 2.

**Answer:** Checking divisibility by 4 and 6 does not work because 4 and 6 are not co-prime.

> Common mistake: Assuming any set of factors of a number can be used to test divisibility.

### Question 10

*3 marks · Short answer*

Explain using prime factorisation why checking divisibility by 3 and 8 works for checking divisibility by 24, but checking divisibility by 4 and 6 is not sufficient for checking divisibility by 24.

**Solution**

1. The prime factorisation of 24 is $2^3 \times 3$, which gives the co-prime factors 3 and 8.
2. If a number is divisible by both 3 and 8, its prime factorisation must contain the prime factors of both, ensuring divisibility by 24.
3. For 4 and 6, they are not co-prime, so a number can be divisible by both 4 and 6 (like 12) without being divisible by their product or LCM.

**Answer:** Divisibility by 3 and 8 works because 3 and 8 are co-prime and their product is 24, whereas 4 and 6 share a common factor.

> Common mistake: Forgetting that test factors for composite numbers must be co-prime.

### Question 11

*3 marks · Short answer*

What property do you think this digital root will have? Recall that we did this while finding the divisibility shortcut for 9.

**Solution**

1. The digital root of a number is the remainder obtained when the number is divided by 9.
2. If the number is exactly divisible by 9, its digital root will be 9.
3. This property arises because each place value in the decimal number system leaves a remainder of 1 when divided by 9.

**Answer:** The digital root represents the remainder when divided by 9, with 9 representing a remainder of 0.

> Common mistake: Confusing the digital root of a multiple of 9 as 0 instead of 9.

### Question 12

*3 marks · Case-based*

Between the numbers 600 and 700, which numbers have the digital root: (i) 5, (ii) 7, (iii) 3?

**Part (i)**

1. We look for numbers between 600 and 700 whose digits sum to 5, 14, or 23.
2. The numbers are 608, 617, 626, 635, 644, 653, 662, 671, 680, 698.

Answer (i): 608, 617, 626, 635, 644, 653, 662, 671, 680, 689, 698

**Part (ii)**

1. We look for numbers between 600 and 700 whose digits sum to 7 or 16.
2. The numbers are 601, 610, 619, 628, 637, 646, 655, 664, 673, 682, 691.

Answer (ii): 601, 610, 619, 628, 637, 646, 655, 664, 673, 682, 691

**Part (iii)**

1. We look for numbers between 600 and 700 whose digits sum to 3, 12, or 21.
2. The numbers are 606, 615, 624, 633, 642, 651, 660, 669, 678, 687, 696.

Answer (iii): 606, 615, 624, 633, 642, 651, 660, 669, 678, 687, 696

**Answer:** The numbers between 600 and 700 with digital roots 5, 7, and 3 are listed in the parts.

> Common mistake: Missing some numbers by not checking all possible digit sums that reduce to the given digital root.

### Question 13

*3 marks · Short answer*

Write the digital roots of any 12 consecutive numbers. What do you observe?

**Solution**

1. Take any 12 consecutive numbers, for example, 1 to 12.
2. Find their digital roots by repeatedly adding digits until a single digit is obtained.
3. The digital roots form a repeating sequence from 1 to 9, such as $1, 2, 3, 4, 5, 6, 7, 8, 9, 1, 2, 3$.

**Answer:** The digital roots of 12 consecutive numbers form a repeating sequence from 1 to 9.

> Common mistake: Stopping at a two-digit sum instead of continuing until a single-digit digital root is found.

### Question 14

*3 marks · Short answer*

Now, find the digital roots of some consecutive multiples of (i) 3, (ii) 4, and (iii) 6.

**Solution**

1. (i) For consecutive multiples of 3 ($3, 6, 9, 12, 15, 18, \dots$), the digital roots are $3, 6, 9, 3, 6, 9, \dots$
2. (ii) For consecutive multiples of 4 ($4, 8, 12, 16, 20, 24, \dots$), the digital roots are $4, 8, 3, 7, 2, 6, 1, 5, 9, 4, 8, 3, \dots$
3. (iii) For consecutive multiples of 6 ($6, 12, 18, 24, 30, 36, \dots$), the digital roots are $6, 3, 9, 6, 3, 9, \dots$.

**Answer:** Digital roots repeat in specific patterns: {3, 6, 9} for multiples of 3 and 6, and a 9-step cycle for multiples of 4.

> Common mistake: Miscalculating the digital root of numbers like 18 or 24.

### Question 15

*3 marks · Short answer*

What are the digital roots of numbers that are 1 more than a multiple of 6? What do you notice?

**Solution**

1. Numbers that are 1 more than a multiple of 6 are of the form $6k + 1$ (e.g., $7, 13, 19, 25, 31, 37, \dots$).
2. Find their digital roots by repeatedly adding digits.
3. The digital roots obtained are $7, 4, 1, 7, 4, 1, \dots$, which follow a repeating decreasing pattern of $7, 4, 1$.

**Answer:** The digital roots of numbers that are 1 more than a multiple of 6 follow a repeating pattern of 7, 4, and 1.

> Common mistake: Listing incorrect digital roots by making addition errors.

### Question 16

*3 marks · Short answer*

Try to explain the patterns noticed.

**Solution**

1. The digital root of a number is congruent to the number modulo 9.
2. Adding a fixed value to a number shifts its remainder when divided by 9 by the same fixed value.
3. This modular arithmetic property explains why sequences formed by adding constants like 6 or 11 produce repeating patterns in their digital roots.

**Answer:** The patterns occur because digital roots correspond to remainders modulo 9, which cycle predictably when numbers increase by a constant amount.

> Common mistake: Explaining patterns through mere observation without connecting them to place value or multiples of 9.

## Figure it Out

### Question 1

*3 marks · Short answer*

The digital root of an 8-digit number is 5. What will be the digital root of 10 more than that number?

**Solution**

1. Let the 8-digit number be $x$, so its digital root is 5.
2. When we add 10 to the number, the sum is $x + 10$.
3. Adding 10 increases the sum of the digits by $1 + 0 = 1$.
4. Therefore, the digital root of the new number is $5 + 1 = 6$.

**Answer:** 6

> Common mistake: Adding 10 directly to the digital root without considering how adding 10 affects the sum of digits.

### Question 2

*3 marks · Short answer*

Write any number. Generate a sequence of numbers by repeatedly adding 11. What would be the digital roots of this sequence of numbers? Share your observations.

**Solution**

1. Take any starting number, for example, 10, and generate a sequence by repeatedly adding 11: 10, 21, 32, 43, 54, 65, 76, 87, 98, 109, 120, ...
2. Find the digital roots of these numbers by repeatedly adding their digits until a single-digit number is obtained.
3. The digital roots for this sequence are 1, 3, 5, 7, 9, 2, 4, 6, 8, 1, 3, ... which form a repeating cycle of nine numbers from 1 to 9.

**Answer:** The digital roots form a repeating sequence from 1 to 9 that cycles continuously.

> Common mistake: Stopping the digit addition before obtaining a single-digit number.

### Question 3

*3 marks · Short answer*

What will be the digital root of the number 9a + 36b + 13?

**Solution**

1. Write the given expression: $9a + 36b + 13$.
2. Split $13$ as $9 + 4$ to separate multiples of $9$: $9a + 36b + 9 + 4$.
3. Take $9$ common from the terms: $9(a + 4b + 1) + 4$.
4. Since $9(a + 4b + 1)$ is a multiple of $9$, its digital root is $9$.
5. Adding $4$ to $9$ gives $13$, and its digital root is $1 + 3 = 4$.

**Answer:** 4

> Common mistake: Forgetting to compute the digital root of the final sum $9 + 4 = 13$, leading to an incorrect single digit.

### Question 4

*4 marks · Case-based*

Make conjectures by examining if there are any patterns or relations between
(i) the parity of a number and its digital root.
(ii) the digital root of a number and the remainder obtained when the number is divided by 3 or 9.

**Part (i)**

1. Examine numbers and their digital roots to check for parity relationships.
2. An even number can have an odd or even digital root, and an odd number can also have an odd or even digital root.
3. Therefore, there is no consistent pattern between the parity of a number and its digital root.

Answer (i): There is no consistent pattern between the parity of a number and its digital root.

**Part (ii)**

1. For division by 3: if the digital root is 1, 4, or 7, the remainder is 1; if 2, 5, or 8, the remainder is 2; if 3, 6, or 9, the remainder is 0.
2. For division by 9: if the digital root is 9, the remainder is 0; otherwise, the remainder is equal to the digital root of the number.

Answer (ii): The digital root determines the remainder when divided by 3 or 9.

**Answer:** Parity and digital roots show no consistent pattern, but digital roots directly determine remainders upon division by 3 or 9.

> Common mistake: Assuming that even numbers always have even digital roots.

## Math Talk

### Question 1

*3 marks · Short answer*

Solve the cryptarithms given below:
(i) A1 + 1B = B0
(ii) AB + 37 = 6A
(iii) ON + ON + ON = PO
(iv) QR + QR + QR = PRR

**Part (i)**

1. Consider the units column: $1 + B = 0$, so $B = 9$ with a carry of $1$.
2. Consider the tens column with the carry: $A + 1 + 1 = B$, which gives $A + 2 = 9$, so $A = 7$.
3. Thus, $A = 7$ and $B = 9$.

Answer (i): $A = 7, B = 9$

**Part (ii)**

1. Consider the units column: $B + 7 = A$ or $B + 7 = 10 + A$.
2. Consider the tens column: $A + 3 + 1 = 6$, which gives $A + 4 = 6$, so $A = 2$.
3. Substitute $A = 2$ into the units equation: $B + 7 = 12$, so $B = 5$.
4. Thus, $A = 2$ and $B = 5$.

Answer (ii): $A = 2, B = 5$

**Part (iii)**

1. The cryptarithm is $3 \times \text{ON} = \text{PO}$.
2. Since $3 \times \text{ON}$ results in a 2-digit number starting with $P$, $O$ must be a small digit.
3. Testing values, if $O = 3$ and $N = 1$, then $3 \times 31 = 93$, which matches $PO = 93$.
4. Thus, $N = 1, O = 3, P = 9$.

Answer (iii): $N = 1, O = 3, P = 9$

**Part (iv)**

1. The cryptarithm is $3 \times \text{QR} = \text{PRR}$, where a two-digit number multiplied by 3 gives a three-digit number with identical last two digits.
2. Testing possible values for two-digit numbers multiplied by 3, we find $3 \times 85 = 255$.
3. Thus, $Q = 8, R = 5, P = 2$.

Answer (iv): $Q = 8, R = 5, P = 2$

**Answer:** Solutions for all four cryptarithms are given in parts.

> Common mistake: Ignoring the carry from the units column.

### Question 2

*3 marks · Short answer*

(v) PQ * 8 = RS.

**Solution**

1. The equation is $\text{PQ} \times 8 = \text{RS}$.
2. We need to find a 2-digit number $\text{PQ}$ which when multiplied by 8 yields another 2-digit number $\text{RS}$.
3. Testing small 2-digit values, $12 \times 8 = 96$, which satisfies all conditions since $P = 1, Q = 2, R = 9, S = 6$ are distinct digits.

**Answer:** $PQ = 12$ ($P = 1, Q = 2, R = 9, S = 6$)

> Common mistake: Choosing a number like 10 or 11 where digits repeat or units digit is the same.

### Question 3

*3 marks · Short answer*

Can PQ be 13? Think.

**Solution**

1. Calculate $13 \times 8$.
2. $13 \times 8 = 104$.
3. Since 104 is a 3-digit number, PQ cannot be 13 because the product must be a 2-digit number $\text{RS}$.

**Answer:** No, PQ cannot be 13 because $13 \times 8 = 104$, which is a 3-digit number.

> Common mistake: Assuming 2-digit numbers multiplied by a single digit always yield a 2-digit number.

### Question 4

*3 marks · Short answer*

(vi) Try this now: GH * H = 9K.

**Solution**

1. The cryptarithm is $\text{GH} \times \text{H} = 9\text{K}$.
2. We test the given options for a 2-digit number multiplied by its units digit resulting in the 90s.
3. Checking $12 \times 8 = 96$, which is in the 90s and has distinct letters representing distinct digits.
4. Thus, the correct solution is $12 \times 8 = 96$.

**Answer:** $12 \times 8 = 96$

> Common mistake: Picking combinations that do not result in the 90s range.

### Question 5

*3 marks · Short answer*

(vii) Here is one more: BYE * 6 = RAY.

**Solution**

1. The cryptarithm is $\text{BYE} \times 6 = \text{RAY}$.
2. Since the product is a 3-digit number starting with $R$, and $B$ is the hundreds digit of the 3-digit number $\text{BYE}$, $B$ cannot be 2 or more (as $200 \times 6 = 1200$, a 4-digit number).
3. Therefore, $B = 1$.

**Answer:** $B = 1$

> Common mistake: Assuming $B$ can be greater than 1 without checking the scale of the product.

### Question 6

*3 marks · Short answer*

What can you say about 'Y'? What digits are possible/not possible?

**Solution**

1. Analyze the digit $Y$ in $\text{BYE} \times 6 = \text{RAY}$.
2. $Y$ cannot be 7 or more because if $Y \ge 7$, then $170 \times 6 = 1020$, which exceeds 3 digits.
3. Also, since the number is multiplied by 6 (an even number), the product ends in an even digit, so $Y$ must be even.

**Answer:** $Y$ must be an even digit less than 7 (possible values are 0, 2, 4, 6).

> Common mistake: Forgetting that multiplying by an even number always results in an even units digit.

### Question 7

*5 marks · Case-based*

Solve the following:
(i) UT * 3 = PUT
(ii) AB * 5 = BC
(iii) L2N * 2 = 2NP
(iv) XY * 4 = ZX
(v) PP * QQ = PRP
(vi) JK * 6 = KKK

**Part (i)**

1. Given the cryptarithm $\text{UT} \times 3 = \text{PUT}$.
2. Testing values for T and U, if $T = 0$ and $U = 5$, we get $50 \times 3 = 150$.
3. Thus, $U = 5$, $T = 0$, and $P = 1$.

Answer (i): $U = 5, T = 0, P = 1$

**Part (ii)**

1. Given the cryptarithm $\text{AB} \times 5 = \text{BC}$.
2. Testing values for B and A, if $B = 9$ and $A = 1$, we get $19 \times 5 = 95$.
3. Thus, $A = 1$, $B = 9$, and $C = 5$.

Answer (ii): $A = 1, B = 9, C = 5$

**Part (iii)**

1. Given the cryptarithm $\text{L}2\text{N} \times 2 = 2\text{NP}$.
2. Testing values, if $L = 1$, $N = 5$, we get $125 \times 2 = 250$.
3. Thus, $L = 1$, $N = 5$, and $P = 0$.

Answer (iii): $L = 1, N = 5, P = 0$

**Part (iv)**

1. Given the cryptarithm $\text{XY} \times 4 = \text{ZX}$.
2. Testing values, if $X = 2$, $Y = 3$, we get $23 \times 4 = 92$.
3. Thus, $X = 2$, $Y = 3$, and $Z = 9$.

Answer (iv): $X = 2, Y = 3, Z = 9$

**Part (v)**

1. Given the cryptarithm $\text{PP} \times \text{QQ} = \text{PRP}$.
2. Testing values, if $P = 2$ and $Q = 1$, we get $22 \times 11 = 242$.
3. Thus, $P = 2$, $Q = 1$, and $R = 4$.

Answer (v): $P = 2, Q = 1, R = 4$

**Part (vi)**

1. Given the cryptarithm $\text{JK} \times 6 = \text{KKK}$.
2. Testing values, if $J = 7$ and $K = 4$, we get $74 \times 6 = 444$.
3. Thus, $J = 7$ and $K = 4$.

Answer (vi): $J = 7, K = 4$

**Answer:** The solved values for the variables in the given cryptarithms are provided in the respective parts.

> Common mistake: Confusing digits assigned to letters or failing to check unique digit constraints for each letter.

## Figure it Out

### Question 1

*3 marks · Short answer*

If 31z5 is a multiple of 9, where z is a digit, what is the value of z? Explain why there are two answers to this problem.

**Solution**

1. The sum of the digits of the number $31z5$ is $3 + 1 + z + 5 = 9 + z$.
2. For $31z5$ to be a multiple of 9, the sum of its digits must be a multiple of 9.
3. Since $z$ is a single digit from 0 to 9, the possible values for $9 + z$ that are multiples of 9 are 9 and 18, giving $z = 0$ or $z = 9$.
4. There are two answers because both $9 + 0 = 9$ and $9 + 9 = 18$ are multiples of 9.

**Answer:** $z = 0$ or $z = 9$

> Common mistake: Forgetting that $z = 9$ is a valid digit value along with $z = 0$.

### Question 2

*3 marks · Short answer*

“I take a number that leaves a remainder of 8 when divided by 12. I take another number which is 4 short of a multiple of 12. Their sum will always be a multiple of 8”, claims Snehal. Examine his claim and justify your conclusion.

**Solution**

1. Let the first number be $a = 12n + 8$ and the second number be $b = 12m - 4 = 12m - 12 + 8$, which also leaves a remainder of 8 or can be written directly as $12m - 4$.
2. Adding the two numbers gives $a + b = (12n + 8) + (12m - 4) = 12n + 12m + 4 = 12(n + m) + 4$.
3. Let $k = n + m$, so the sum is $12k + 4$, which is 4 more than a multiple of 12.
4. Snehal's claim is false because $12k + 4$ is not always a multiple of 8 (for example, if $k = 2$, the sum is $28$, which is not divisible by 8).

**Answer:** Snehal's claim is false.

> Common mistake: Assuming that a combination of multiples and remainders will always result in a multiple of 8 without checking algebraically.

### Question 3

*3 marks · Short answer*

When is the sum of two multiples of 3, a multiple of 6 and when is it not? Explain the different possible cases, and generalise the pattern.

**Solution**

1. Let the two multiples of 3 be $3m$ and $3n$, where $m$ and $n$ are integers.
2. Their sum is $3m + 3n = 3(m + n)$.
3. For this sum to be a multiple of 6, $3(m + n)$ must be divisible by 6, which requires $(m + n)$ to be a multiple of 2 (i.e., an even number).
4. Thus, the sum is a multiple of 6 if $m$ and $n$ are both even or both odd, and it is not a multiple of 6 if one is even and the other is odd.

**Answer:** The sum is a multiple of 6 when the sum of their factors $(m + n)$ is even.

> Common mistake: Concluding the sum is always a multiple of 6 just because both terms are multiples of 3.

### Question 4

*3 marks · Short answer*

Sreelatha says, “I have a number that is divisible by 9. If I reverse its digits, it will still be divisible by 9”.
(i) Examine if her conjecture is true for any multiple of 9.
(ii) Are any other digit shuffles possible such that the number formed is still a multiple of 9?

**Part (i)**

1. A number is divisible by 9 if and only if the sum of its digits is divisible by 9.
2. When the digits of a number are reversed, the set of digits remains exactly the same, so the sum of the digits does not change.
3. Therefore, if the original number is divisible by 9, the reversed number is also divisible by 9.

Answer (i): True, because reversing digits does not change their sum.

**Part (ii)**

1. Any shuffle or rearrangement of the digits of a number changes only the order of the digits, not the digits themselves.
2. The sum of the digits remains unchanged under any rearrangement.
3. Thus, any digit shuffle of a multiple of 9 remains a multiple of 9.

Answer (ii): Yes, any shuffle of digits keeps the number divisible by 9.

**Answer:** Yes, both conjectures are true.

> Common mistake: Thinking that changing the order of digits alters the sum of the digits.

### Question 5

*3 marks · Short answer*

If 48a23b is a multiple of 18, list all possible pairs of values for a and b.

**Solution**

1. A number is divisible by 18 if it is divisible by both 2 and 9.
2. For divisibility by 2, the units digit $b$ must be an even number, so $b \in \{0, 2, 4, 6, 8\}$.
3. For divisibility by 9, the sum of the digits $4 + 8 + a + 2 + 3 + b = 17 + a + b$ must be a multiple of 9.
4. Testing each value of $b$ from the allowed set to find integer digits $a$ ($0 \le a \le 9$) yields the possible pairs $(1, 0), (8, 2), (6, 4), (4, 6), (2, 8)$.

**Answer:** (1, 0), (8, 2), (6, 4), (4, 6), (2, 8)

> Common mistake: Missing some values of $b$ or obtaining values of $a$ greater than 9.

### Question 6

*3 marks · Short answer*

If 3p7q8 is divisible by 44, list all possible pairs of values for p and q.

**Solution**

1. A number is divisible by 44 if it is divisible by both 4 and 11.
2. For divisibility by 4, the last two digits $q8$ must be divisible by 4, which means $q$ can be any digit from 0 to 9.
3. For divisibility by 11, the difference between the sum of digits at odd places and even places, $18 - (p + q)$, must be 0 or a multiple of 11.
4. Solving the conditions for $p$ and $q$ gives the possible pairs: $(7, 0), (5, 2), (3, 4), (1, 6)$.

**Answer:** (7, 0), (5, 2), (3, 4), (1, 6)

> Common mistake: Incorrectly calculating the alternating sum of digits for divisibility by 11.

### Question 7

*3 marks · Short answer*

Find three consecutive numbers such that the first number is a multiple of 2, the second number is a multiple of 3, and the third number is a multiple of 4. Are there more such numbers? How often do they occur?

**Solution**

1. Let the three consecutive numbers be $n$, $n + 1$, and $n + 2$.
2. We are given that $n$ is a multiple of 2, $n + 1$ is a multiple of 3, and $n + 2$ is a multiple of 4.
3. One such set of numbers is $2, 3, 4$ where $2$ is a multiple of 2, $3$ is a multiple of 3, and $4$ is a multiple of 4.
4. The pattern repeats every 12 numbers since the LCM of 2, 3, and 4 is 12, so the next set is $14, 15, 16$.

**Answer:** The first set is 2, 3, 4 and they occur repeatedly after every 12 numbers, with the next set being 14, 15, 16.

> Common mistake: Forgetting that consecutive numbers are spaced by 1.

### Question 8

*3 marks · Short answer*

Write five multiples of 36 between 45,000 and 47,000. Share your approach with the class.

**Solution**

1. A number is a multiple of 36 if it is a multiple of both 4 and 9.
2. For a number to be a multiple of 4, its last two digits must form a number divisible by 4.
3. For a number to be a multiple of 9, the sum of its digits must be a multiple of 9.
4. Five multiples of 36 between 45,000 and 47,000 are 45036, 45072, 45108, 45144, and 45180.

**Answer:** 45036, 45072, 45108, 45144, 45180

> Common mistake: Checking only divisibility by 4 or 9 instead of both.

### Question 9

*3 marks · Short answer*

The middle number in the sequence of 5 consecutive even numbers is 5p. Express the other four numbers in sequence in terms of p.

**Solution**

1. Let the middle number of the five consecutive even numbers be $5p$.
2. Consecutive even numbers differ by 2.
3. The two even numbers preceding $5p$ are $5p - 2$ and $5p - 4$.
4. The two even numbers succeeding $5p$ are $5p + 2$ and $5p + 4$.

**Answer:** The other four numbers are $5p - 4$, $5p - 2$, $5p + 2$, and $5p + 4$.

> Common mistake: Subtracting 1 instead of 2 for consecutive even numbers.

### Question 10

*3 marks · Short answer*

Write a 6-digit number that it is divisible by 15, such that when the digits are reversed, it is divisible by 6.

**Solution**

1. A 6-digit number divisible by 15 must end in 0 or 5 and have a digit sum divisible by 3.
2. If it ends in 0, reversing the digits gives a number starting with 0, which is not a 6-digit number. Thus, the units digit must be 5.
3. When reversed, the number becomes divisible by 2 and 3, so its first digit must be an even number ($2, 4, 6, \text{or } 8$).
4. Possible numbers include 200025, 200055, and 200085.

**Answer:** 200025, 200055, 200085 (any valid 6-digit number starting with an even digit and ending in 5 with digit sum divisible by 3)

> Common mistake: Ignoring the constraint that the reversed number must remain a 6-digit number.

### Question 11

*3 marks · Short answer*

Deepak claims, “There are some multiples of 11 which, when doubled, are still multiples of 11. But other multiples of 11 don’t remain multiples of 11 when doubled”. Examine if his conjecture is true; explain your conclusion.

**Solution**

1. Let $n$ be any multiple of 11, so we can write $n = 11k$ for some integer $k$.
2. When doubled, the number becomes $2n = 2(11k) = 11(2k)$.
3. Since $2k$ is an integer, $2n$ is always a multiple of 11.
4. Therefore, Deepak's conjecture is false because all multiples of 11 remain multiples of 11 when doubled.

**Answer:** Deepak's conjecture is false; every multiple of 11 remains a multiple of 11 when doubled.

> Common mistake: Assuming doubling a multiple changes its prime factors regarding 11.

### Question 12

*3 marks · Short answer*

Determine whether the statements below are ‘Always True’, ‘Sometimes True’, or ‘Never True’. Explain your reasoning.
(i) The product of a multiple of 6 and a multiple of 3 is a multiple of 9.
(ii) The sum of three consecutive even numbers will be divisible by 6.
(iii) If abcdef is a multiple of 6, then badcef will be a multiple of 6.
(iv) 8 (7b - 3) - 4 (11b + 1) is a multiple of 12.

**Part (i)**

1. Let the multiple of 6 be $6a$ and the multiple of 3 be $3b$.
2. Their product is $(6a)(3b) = 18ab = 9(2ab)$, which is always a multiple of 9.

Answer (i): Always true

**Part (ii)**

1. Let three consecutive even numbers be $2n, 2n+2, 2n+4$.
2. Their sum is $2n + 2n + 2 + 2n + 4 = 6n + 6 = 6(n + 1)$, which is always divisible by 6.

Answer (ii): Always true

**Part (iii)**

1. If $abcdef$ is a multiple of 6, it is divisible by 2 and 3.
2. Rearranging the digits to $badcef$ does not change the sum of the digits (divisible by 3) nor the units digit parity (divisible by 2), so it remains a multiple of 6.

Answer (iii): Always true

**Part (iv)**

1. Simplify the expression: $8(7b - 3) - 4(11b + 1) = 56b - 24 - 44b - 4 = 12b - 28$.
2. This expression cannot be a multiple of 12 for all integers $b$ since $-28$ is not divisible by 12.

Answer (iv): Never true

**Answer:** (i) Always true (ii) Always true (iii) Always true (iv) Never true

> Common mistake: Failing to simplify algebraic expressions fully before checking for divisibility.

### Question 13

*3 marks · Short answer*

Choose any 3 numbers. When is their sum divisible by 3? Explore all possible cases and generalise.

**Solution**

1. Let the three numbers be $n_1$, $n_2$, and $n_3$, and let their sum be $S = n_1 + n_2 + n_3$.
2. When any number is divided by 3, the possible remainders are 0, 1, or 2.
3. The sum $S$ is divisible by 3 when the sum of the remainders of the three numbers upon division by 3 is equal to 0, 3, or 6.

**Answer:** The sum of three numbers is divisible by 3 if and only if the sum of their remainders when divided by 3 is 0, 3, or 6.

> Common mistake: Forgetting that remainders can combine in different ways to give a multiple of 3.

### Question 14

*3 marks · Short answer*

Is the product of two consecutive integers always multiple of 2? Why? What about the product of these consecutive integers? Is it always a multiple of 6? Why or why not? What can you say about the product of 4 consecutive integers? What about the product of five consecutive integers?

**Part (i)**

1. The product of two consecutive integers is always a multiple of 2 because one of them must be even.

Answer (i): Yes, always a multiple of 2.

**Part (ii)**

1. The product of three consecutive integers is always a multiple of 6 because there is at least one multiple of 2 and one multiple of 3.

Answer (ii): Yes, always a multiple of 6.

**Part (iii)**

1. The product of 4 consecutive integers is a multiple of 24 ($2 \times 3 \times 4$), and the product of 5 consecutive integers is a multiple of 120 ($2 \times 3 \times 4 \times 5$).

Answer (iii): Multiple of 24 for 4 integers and multiple of 120 for 5 integers.

**Answer:** Products of consecutive integers are always multiples of factorials corresponding to the number of terms, due to alternating parity and multiples.

> Common mistake: Assuming the product of three consecutive integers is only divisible by 3 and missing the factor of 2.

### Question 15

*3 marks · Short answer*

Solve the cryptarithms —
(i) EF * E = GGG
(ii) WOW * 5 = MEOW

**Part (i)**

1. Consider $EF \times E = GGG$, where $GGG$ is a three-digit number with identical digits.
2. Testing values for $E$, if $E = 3$, we get $37 \times 3 = 111$, which satisfies the condition.

Answer (i): $E = 3, F = 7, G = 1$

**Part (ii)**

1. Consider $WOW \times 5 = MEOW$.
2. Testing values, $575 \times 5 = 2875$, which matches the letter pattern for $W = 5, O = 7, M = 2, E = 8$.

Answer (ii): $W = 5, O = 7, M = 2, E = 8$

**Answer:** (i) $E = 3, F = 7, G = 1$ (ii) $W = 5, O = 7, M = 2, E = 8$

> Common mistake: Not checking if letters represent unique digits.

### Question 16

*1 mark · MCQ*

Which of the following Venn diagrams captures the relationship between the multiples of 4, 8, and 32?

**Solution**

1. Every multiple of 32 is also a multiple of 8, and every multiple of 8 is a multiple of 4.
2. Therefore, the set of multiples of 32 is completely contained inside the set of multiples of 8, which is completely contained inside the set of multiples of 4 (nested circles).

**Answer:** (iv)

> Common mistake: Confusing subset containment direction between larger and smaller multiples.

## Frequently asked questions

### How many questions are there in NCERT Class 8 Maths Chapter 5 Number Play?

This chapter is divided into multiple sections featuring a total of 79 questions across Math Talk and Figure it Out exercises. You can find step-by-step solutions for all of them in SwaVid's free PDF available on this page.

### What topics are covered in Class 8 Maths Chapter 5 Number Play?

The questions cover interesting number patterns, algebraic explanations of even numbers, divisibility rules for 9, 11, and 24, digital roots, and cryptarithmetic puzzles. SwaVid's free PDF on this page provides clear explanations for each of these concepts.

### Are cryptarithms and divisibility proofs asked in Class 8 Maths Chapter 5?

Yes, students often find cryptarithms and algebraic proofs for divisibility and remainders to be the trickiest question types. To score full marks, you should break down puzzles step by step using logical deduction, and you can study SwaVid's detailed solutions on this page to master the approach.

### How should I write answers to get full marks in Class 8 Maths Chapter 5?

To secure full marks, you need to clearly show every step of your algebraic reasoning, especially when proving divisibility or writing expressions for consecutive numbers. SwaVid's free PDF on this page demonstrates the exact presentation style required for school exams.

### Is the free PDF for Class 8 Maths Chapter 5 Number Play based on the new NCERT book?

Yes, the solutions are strictly based on the new NCERT book as per the NCF 2023 guidelines for the 2026-27 session. You can access SwaVid's comprehensive free PDF and step-by-step solutions exclusively on this page.

## Related pages

- [Class 8 Maths chapters](https://www.swavid.com/maths/class/8)

Solutions written by SwaVid, a personal AI tutor for Class 6 to 10 Maths and Science. Practise this chapter free: https://www.swavid.com/start/student