Class 8 - Maths - Playing with Numbers

Exercise 16.1

Question 1:

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

On putting A = 1, 2, 3, 4, 5, 6, 7 and so on and we get,

7 + 5 = 12 in which ones place is 2.

So, A = 7

And putting 2 and carry over 1, we get

B = 6

Hence, A = 7 and B = 6

Question 2:

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

On putting A = 1, 2, 3, 4, 5, 6, 7 and so on and we get,

8 + 5 = 13 in which ones place is 3.

So, A = 5

And putting 3 and carry over 1, we get B = 4 and C = 1

Hence, A = 5, B = 4 and C = 1

Question 3:

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

On putting A = 1, 2, 3, 4, 5, 6, 7 and so on and we get,

A * A = 6 * 6 = 36 in which ones place is 6.

So, A = 6

Hence, A = 6

Question 4:

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

Here, we observe that B = 5 so that 7 + 5 = 12.

Putting 2 at ones place and carry over 1 and A = 2, we get

2 + 3 + 1 = 6

Hence, A = 2 and B = 5

Question 5:

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

Here on putting B = 0, we get 0 * 3 = 0.

And A = 5, then 5 * 3 = 15

So, A = 5 and C = 1

Hence, A = 5, B = 0 and C = 1

Question 6:

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

On putting B = 0, we get

0 * 5 = 0 and A = 5, then 5 * 5 = 25

So, A = 5, C = 2

Hence, A = 5, B = 0 and C = 2

Question 7:

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

Here product of B and 6 must be same as ones place digit as B.

6 * 1 = 6, 6 * 2 = 12, 6 * 3 = 18, 6 * 4 = 24

On putting B = 4, we get the ones digit 4 and remaining two B’s value should be 44.

So, For 6 * 7 = 42 + 2 = 44

Hence, A = 7 and B = 4

Question 8:

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

On putting B = 9, we get 9 + 1 = 10

Putting 0 at ones place and carry over 1, we get

For A = 7

=> 7 + 1 + 1 = 9

Hence, A = 7 and B = 9

Question 9:

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

On putting B = 7,

=> 7 + 1 = 8

Now A = 4, then 4 + 7 = 11

Putting 1 at tens place and carry over 1, we get

2 + 4 + 1 = 7

Hence, A = 4 and B = 7

Question 10:

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

Putting A = 8 and B = 1, we get

8 + 1 = 9

Now again we add 2 + 8 = 10

Tens place digit is 0 and carry over 1.

Now 1 + 6 + 1 = 8 = A

Hence, A = 8 and B = 1

Exercise 16.2

Question 1:

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

Since

21 5y is a multiple of 9.

Therefore according to the divisibility rule of 9, the sum of all the digits should be a

multiple of 9. Since y is a single digit number, this sum can be 9 only.

So, 2 + 1 + y + 5 = 9

Now, 8 + y = 9

=> y = 9 – 8

=> y = 1

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?

Since 31 5z is a multiple of 9.

Therefore, according to the divisibility rule of 9, the sum of all the digits should be a

multiple of 9. Since z is a single digit number, this sum can be either 9 or 18.

So, 3 + 1 + 5 + z = 9 or 3 + 1 + 5 + z = 18

=>  9 + z = 9 or 9 + z = 18

=> z = 9 – 9 or z = 18 – 9

=> z = 0 or 9

Therefore, z should be 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.

Therefore according to the divisibility rule of 3, the sum of all the digits should be a

multiple of 3.

2 + 4 + x = 6 + x

Since x is a digit.

So, 6 + x = 6 => x = 0

6 + x = 9 => x = 3

6 + x = 12 => x = 6

6 + x = 15 => x = 9

Thus, x can have any of four different values.

Question 4:

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

Since 315z is a multiple of 3.

Therefore according to the divisibility rule of 3, the sum of all the digits should be a

multiple of 3.

3 + 1 + 5 + z = 9 + z

Since z is a digit.

So, 9 + z = 9 => x = 0

9 + z = 12 => x = 3

9 + z = 15 => x = 6

9 + z = 18 => x = 9

Hence, 0, 3, 6 and 9 are four possible answers.