Mathematics NCERT Grade 8, Chapter 16: Playing with Numbers- This chapter focusses on concepts like Numbers in General FormLetters for Digits and various divisibility tests.
• Numbers can be written in general form. Thus, a two-digit number ab will be written as ab = 10a + b.
• A three digit number abc can be written as abc = 100a + 10b + c
The chapter also contains an interesting topic- Games with Numbers.
• The general form of numbers are helpful in solving puzzles or number games
• The games involve reversing two digit and three digit numbers, forming two digit numbers with given three digit numbers.
To make the content more attractive and interesting some games and puzzles related to numbers are given. These games will be fun to play and will also clear the concepts of students.
In certain number games, letters are replaced with numbers to form a code and vice-versa. These are interesting trick number games based on the concept of general form of numbers.

Divisibility tests explained in this chapter are as follows:
• Divisibility by 10:  If the one's digit of a number is 0, then the number is a multiple of 10; and if the one's digit is not 0, then the number is not a multiple of 10.
• Divisibility by 5: If the one's digit of a number is 0 or 5, then it is divisible by 5.
• Divisibility by 2: If the one’s digit of a number is 0, 2, 4, 6 or 8 then the number is divisible by 2.
• Divisibility by 9: A number is divisible by 9 if the sum of its digits is divisible by 9. Otherwise, it is not divisible by 9.
• Divisibility by 3: A number is divisible by 3 if the sum of its digits is divisible by 3. Otherwise, it is not divisible by 3.
​Students will learn in this chapter about the divisibility by 11 as well and how a number divisible by m will also be divisible by each of the factors of m. In this chapter, emphasis will be laid on why certain tests are done to check the divisibility.

The entire chapter is summarized at the end with a focus on important points.

#### Question 1:

Find the values of the letters in the following and give reasons for the steps involved. The addition of A and 5 is giving 2 i.e., a number whose ones digit is 2. This is possible only when digit A is 7. In that case, the addition of A (7) and 5 will give 12 and thus, 1 will be the carry for the next step. In the next step,

1 + 3 + 2 = 6

Therefore, the addition is as follows. Clearly, B is 6.

Hence, A and B are 7 and 6 respectively.

#### Question 2:

Find the values of the letters in the following and give reasons for the steps involved. The addition of A and 8 is giving 3 i.e., a number whose ones digit is 3. This is possible only when digit A is 5. In that case, the addition of A and 8 will give 13 and thus, 1 will be the carry for the next step. In the next step,

1 + 4 + 9 = 14

Therefore, the addition is as follows. Clearly, B and C are 4 and 1 respectively.

Hence, A, B, and C are 5, 4, and 1 respectively.

#### Question 3:

Find the value of the letter in the following and give reasons for the steps involved. The multiplication of A with A itself gives a number whose ones digit is A again. This happens only when A = 1, 5, or 6.

If A = 1, then the multiplication will be 11 × 1 = 11. However, here the tens digit is given as 9. Therefore, A = 1 is not possible. Similarly, if A = 5, then the multiplication will be 15 × 5 = 75. Thus, A = 5 is also not possible.

If we take A = 6, then 16 × 6 = 96. Therefore, A should be 6.

The multiplication is as follows. Hence, the value of A is 6.

#### Question 4:

Find the values of the letters in the following and give reasons for the steps involved. The addition of A and 3 is giving 6. There can be two cases.

(1) First step is not producing a carry

In that case, A comes to be 3 as 3 + 3 = 6. Considering the first step in which the addition of B and 7 is giving A (i.e., 3), B should be a number such that the units digit of this addition comes to be 3. It is possible only when B = 6. In this case, A = 6 + 7 = 13. However, A is a single digit number. Hence, it is not possible.

(2) First step is producing a carry

In that case, A comes to be 2 as 1 + 2 + 3 = 6. Considering the first step in which the addition of B and 7 is giving A (i.e., 2), B should be a number such that the units digit of this addition comes to be 2. It is possible only when B = 5 and 5 + 7 = 12. Hence, the values of A and B are 2 and 5 respectively.

#### Question 5:

Find the values of the letters in the following and give reasons for the steps involved. The multiplication of 3 and B gives a number whose ones digit is B again.

Hence, B must be 0 or 5.

Let B is 5.

Multiplication of first step = 3 × 5 = 15

1 will be a carry for the next step.

We have, 3 × A + 1 = CA

This is not possible for any value of A.

Hence, B must be 0 only. If B = 0, then there will be no carry for the next step.

We should obtain, 3 × A = CA

That is, the one’s digit of 3 × A should be A. This is possible when A = 5 or 0.

However, A cannot be 0 as AB is a two-digit number.

Therefore, A must be 5 only. The multiplication is as follows. Hence, the values of A, B, and C are 5, 0, and 1 respectively.

#### Question 6:

Find the values of the letters in the following and give reasons for the steps involved. The multiplication of B and 5 is giving a number whose ones digit is B again. This is possible when B = 5 or B = 0 only.

In case of B = 5, the product, B × 5 = 5 × 5 = 25

2 will be a carry for the next step.

We have, 5 × A + 2 = CA, which is possible for A = 2 or 7

The multiplication is as follows.  If B = 0,

B × 5 = B ⇒ 0 × 5 = 0

There will not be any carry in this step.

In the next step, 5 × A = CA

It can happen only when A = 5 or A = 0

However, A cannot be 0 as AB is a two-digit number.

Hence, A can be 5 only. The multiplication is as follows. Hence, there are 3 possible values of A, B, and C.

(i) 5, 0, and 2 respectively

(ii) 2, 5, and 1 respectively

(iii) 7, 5, and 3 respectively

#### Question 7:

Find the values of the letters in the following and give reasons for the steps involved. The multiplication of 6 and B gives a number whose one’s digit is B again.

It is possible only when B = 0, 2, 4, 6, or 8

If B = 0, then the product will be 0. Therefore, this value of B is not possible.

If B = 2, then B × 6 = 12 and 1 will be a carry for the next step.

6A + 1 = BB = 22 ⇒ 6A = 21 and hence, any integer value of A is not possible.

If B = 6, then B × 6 = 36 and 3 will be a carry for the next step.

6A + 3 = BB = 66 ⇒ 6A = 63 and hence, any integer value of A is not possible.

If B = 8, then B × 6 = 48 and 4 will be a carry for the next step.

6A + 4 = BB = 88 ⇒ 6A = 84 and hence, A = 14. However, A is a single digit number. Therefore, this value of A is not possible.

If B = 4, then B × 6 = 24 and 2 will be a carry for the next step.

6A + 2 = BB = 44 ⇒ 6A = 42 and hence, A = 7

The multiplication is as follows. Hence, the values of A and B are 7 and 4 respectively.

#### Question 8:

Find the values of the letters in the following and give reasons for the steps involved. The addition of 1 and B is giving 0 i.e., a number whose ones digits is 0. This is possible only when digit B is 9. In that case, the addition of 1 and B will give 10 and thus, 1 will be the carry for the next step. In the next step,

1 + A + 1 = B

Clearly, A is 7 as 1 + 7 + 1 = 9 = B

Therefore, the addition is as follows. Hence, the values of A and B are 7 and 9 respectively.

#### Question 9:

Find the values of the letters in the following and give reasons for the steps involved. The addition of B and 1 is giving 8 i.e., a number whose ones digits is 8. This is possible only when digit B is 7. In that case, the addition of B and 1 will give 8. In the next step,

A + B = 1

Clearly, A is 4.

4 + 7 = 11 and 1 will be a carry for the next step. In the next step,

1 + 2 + A = B

1 + 2 + 4 = 7

Therefore, the addition is as follows. Hence, the values of A and B are 4 and 7 respectively.

#### Question 10:

Find the values of the letters in the following and give reasons for the steps involved. The addition of A and B is giving 9 i.e., a number whose ones digits is 9. The sum can be 9 only as the sum of two single digit numbers cannot be 19. Therefore, there will not be any carry in this step.

In the next step, 2 + A = 0

It is possible only when A = 8

2 + 8 = 10 and 1 will be the carry for the next step.

1 + 1 + 6 = A

Clearly, A is 8. We know that the addition of A and B is giving 9. As A is 8, therefore, B is 1.

Therefore, the addition is as follows. Hence, the values of A and B are 8 and 1 respectively.

#### Question 1:

If 21y5 is a multiple of 9, where y is a digit, what is the value of y?

If a number is a multiple of 9, then the sum of its digits will be divisible by 9.

Sum of digits of 21y5 = 2 + 1 + y + 5 = 8 + y

Hence, 8 + y should be a multiple of 9.

This is possible when 8 + y is any one of these numbers 0, 9, 18, 27, and so on …

However, since y is a single digit number, this sum can be 9 only. Therefore, y should be 1 only.

#### Question 2:

If 31z5 is a multiple of 9, where z is a digit, what is the value of z?

You will find that there are two answers for the last problem. Why is this so?

If a number is a multiple of 9, then the sum of its digits will be divisible by 9.

Sum of digits of 31z5 = 3 + 1 + z + 5 = 9 + z

Hence, 9 + z should be a multiple of 9.

This is possible when 9 + z is any one of these numbers 0, 9, 18, 27, and so on …

However, since z is a single digit number, this sum can be either 9 or 18. Therefore, z should be either 0 or 9.

#### Question 3:

If 24x is a multiple of 3, where x is a digit, what is the value of x?

(Since 24x is a multiple of 3, its sum of digits 6 + x is a multiple of 3; so 6 + x is one of these numbers: 0, 3, 6, 9, 12, 15, 18…. But since x is a digit, it can only be that 6 + x = 6 or 9 or 12 or 15. Therefore, x = 0 or 3 or 6 or 9. Thus, x can have any of four different values)

Since 24x is a multiple of 3, the sum of its digits is a multiple of 3.

Sum of digits of 24x = 2 + 4 + x = 6 + x

Hence, 6 + x is a multiple of 3.

This is possible when 6 + x is any one of these numbers 0, 3, 6, 9, and so on …

Since x is a single digit number, the sum of the digits can be 6 or 9 or 12 or 15 and thus, the value of x comes to 0 or 3 or 6 or 9 respectively.

Thus, x can have its value as any of the four different values 0, 3, 6, or 9.

#### Question 4:

If 31z5 is a multiple of 3, where z is a digit, what might be the values of z?

Since 31z5 is a multiple of 3, the sum of its digits will be a multiple of 3.

That is, 3 + 1 + z + 5 = 9 + z is a multiple of 3.

This is possible when 9 + z is any one of 0, 3, 6, 9, 12, 15, 18, and so on …

Since z is a single digit number, the value of 9 + z can only be 9 or 12 or 15 or 18 and thus, the value of x comes to 0 or 3 or 6 or 9 respectively.

Thus, z can have its value as any one of the four different values 0, 3, 6, or 9.

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