Contents

NCERT Solutions for Class 8 Maths Chapter 16 Playing with Numbers are part of NCERT Solutions for Class 8 Maths. Here we have given NCERT Solutions for Class 8 Maths Chapter 16 Playing with Numbers.

Board |
CBSE |

Textbook |
NCERT |

Class |
Class 8 |

Subject |
Maths |

Chapter |
Chapter 16 |

Chapter Name |
Playing with Numbers |

Exercise |
Ex 16.1, Ex 16.2 |

Number of Questions Solved |
5 |

Category |
NCERT Solutions |

## NCERT Solutions for Class 8 Maths Chapter 16 Playing with Numbers

### Chapter 16 Playing with Numbers Exercise 16.1

**Question 1.**

Find the values of the letters in each of the following and give reasons for the steps involved.

**Solution:**

**1.** There are two letters whose values are to be found out.

Study the addition in the ones column, i. e., from A + 5 we get 2, i. e., the number whose ones digit is 2.

For this to happen, A must be 7 (∵ A + 5 = 7 + 5 = 12).

So for the addition in tens column, we have

1 + 3 + 2 = B ⇒ B = 6

So, the puzzle has been decoded as

∴ The possible values of A and B are 7 and 6 respectively.

**2.** There are three letters whose values are to be found out.

Study the addition in the ones column, i. e., from A + 8 we get 3, i. e., the number whose ones digit is 3.

For this to happen, A must be 5 (∵ A + 8 = 5 + 8 = 13).

So for the addition in tens column, we have 1 + 4 + 9 = B ⇒ 14 = B

∴Clearly, B is 4 and C is 1.

So the puzzle has been decoded as

∴ The possible values of A, B and C are 5, 4 and 1 respectively.

**3.** Since the ones digit of A x A is A, it must be that A = 1 or A = 5 or A = 6.

**4.** There are two letters whose values are to be found : A and B.

Study the addition in the given puzzle. .

i. e., from B + 7 we get A and from A + 3 we get 6

Possible values can be

**5.** This has three letters whose values are to be found : A, B and C.

Since the ones digit of 3 x B = B, it must be B = 0. So, the puzzle becomes

**6.** This has three letters whose values are to be found : A, B and C. Since the ones digit of 5 x B is B, it must be that B = 0 or B = 5 If B = 0, then the puzzle becomes

**7.** This has two letters whose values are to be found : A and B.

Possible values of BBB are 111, 222, 333, etc.

Let us divide these numbers by 6.

111 ÷ 6 =18, remainder 3. So, 111 is rejected.

222 ÷ 6 = 37, remainder 0. So, the quotient 37 is not of the form A2. Thus,

222 is rejected.

333 ÷ 6 = 55, remainder 3. So, 333 is rejected.

444 ÷ 6 = 74, remainder 0. The quotient 74 is of the form A4.

So, the puzzle has been decoded as

This one clearly works out correctly.

So, the answer is A = 7 and B = 4.

**8.** This has two letters whose values are to be found : A and B.

Study the addition in the ones column : from 1 + B we get 0, that is, a number whose ones digit is 0.

For this to happen, the ones digit of B should be 9. And since B itself is a digit, we get B =9. So, the puzzle becomes

But 90 – 19 = 71 so, A1 = 71 ⇒ A = 7

Thus, A = 7 and B = 9 is the answer.

**9.** There are two letters whose values are to be found out.

Study the addition in the ones column, i. e., from B + 1 we get 8, that is, a number whose ones digit is 8.

For this to happen, the ones digit of B should be 7. As B itself is a digit, we get B = 7. So the puzzle becomes

Now study the addition in the tens column, i. e., from A+ 7 we get 1, that is, a number whose ones digit is 1.

For this to happen, the ones digit of A should be 4. As A itself is a digit, we get A = 4. So, the puzzle has been decoded as

Therefore, possible values of A and B are 4 and 7 respectively.

**10.** This has two letters whose values are to be found : A and B.

Study the addition in the tens column : from 2 + A we get 0, that is, a number whose ones digit is 0.

For this to happen, the ones digit of A should be 8. And since A is a digit, we get A = 8. So, the puzzle becomes

Now, study the addition in the ones column: from 8 + B we get 9. For this to happen, we must have B = 1. Therefore, the puzzle has been decoded as

### Chapter 16 Playing with Numbers Exercise 16.2

**Question 1.**

If 21 y 5 is a multiple of 9, where y is a digit, what is the value of y?

**Solution:**

Since the number 21y5 is a multiple of 9.

So, the sum of its digits 2 + 1 + y + 5 = 8 + y is a multiple of 9.

∴ (8 + y) is either 0 or 9 or 18 or 27 …

But since y is a digit, so (8 + y) must be equal to 9.

i.e.,8 + y = 9 ⇒ y = 9- 8= l

**Question 2.**

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

**Solution:**

Since the number 31z5 is a multiple of 9.

So, the sum of its digits 3 + 1 + z + 5 = 9+ z is a multiple of 9.

∴ (9 + z) is either 0 or 9 or 18 or 27 …

But since z is a digit, so (9 + z) must be equal to 9 or 18…

i.e., 9 + z = 9 ⇒ z = 9 + z = 18 ⇒ z = 9

**Question 3.**

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

**Solution:**

Since 24 x is a multiple of 3. So, the sum of its digits 2 + 4 + x = (6 + x) is

a multiple of 3.

∴ (6 + x) is one of the numbers 0, 3, 6, 9, 12, 15, 18…

But x is a digit. Therefore, (6 + x) must be equal to 6 or 9 or 12 or 15.

i.e., 6 + x = 6 or 9 or 12 or 15

⇒ x = 0 or 3 or 6 or 9

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

**Question 4.**

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

**Solution:**

Since 31z5 is a multiple of 3. So, the sum of its digits 3 + 1 + z + 5 = (9 + z) is a multiple of 3.

∴ (9 + z) is one of the numbers 0, 3, 6, 9, 12, 15, 18, …

But z is a digit. Therefore (9 + z) must be equal to 9 or 12 or 15 or 18.

i.e., 9 + z = 9 or 12 or 15 or 18

⇒ z = 0 or 3 or 6 or 9

Thus, z can have any of the four different values, namely, 0, 3, 6 or 9.

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