Playing with Numbers

EXERCISE-16.1

P1. Find the values of the letters in the following and give reasons for the steps involved.

$\begin{array}{r}\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{3}\mathbf{ }\mathbf{A}\\ \mathbf{+}\mathbf{2}\mathbf{ }\mathbf{5}\\ \mathbf{B}\mathbf{ }\mathbf{2}\end{array}$

Sol. 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.

$\begin{array}{r}3 7\\ +2 5\\ 6 2\end{array}$

Clearly, B is 6.

Hence, A and B are 7 and 6 respectively.

P2. Find the values of the letters in the following and give reasons for the steps involved.

$\begin{array}{r}\mathbf{4}\mathbf{ }\mathbf{A}\\ \mathbf{+}\mathbf{9}\mathbf{ }\mathbf{8}\\ \mathbf{C}\mathbf{ }\mathbf{B}\mathbf{ }\mathbf{3}\end{array}$

Sol. 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.

$\begin{array}{r}4 5\\ +9 8\\ 1 4 3\end{array}$

Clearly, B and C are 4 and 1 respectively.

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

P3. Find the values of the letters in the following and give reasons for the steps involved.

$\begin{array}{r}\mathbf{ }\mathbf{A}\mathbf{B}\\ \mathbf{+}\mathbf{3}\mathbf{ }\mathbf{7}\\ \mathbf{6}\mathbf{ }\mathbf{A}\end{array}$

Sol. 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.

$\begin{array}{rr}2& 5\\ + 3& 7\\ 6& 2\end{array}$

Hence, the values of A and B are 2 and 5 respectively.

P4. Find the values of the letters in the following and give reasons for the steps involved.

$\begin{array}{r}\mathbf{A}\mathbf{ }\mathbf{ }\mathbf{B}\\ \mathbf{\times }\mathbf{ }\mathbf{3}\\ \mathbf{C}\mathbf{ }\mathbf{A}\mathbf{ }\mathbf{B}\end{array}$

Sol. 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 = C A

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.

$\begin{array}{r}50\\ \times 3\\ 150\end{array}$

Hence, the values of A, B, and C are 5, 0, and 1 respectively.

P5.Find the values of the letters in the following and give reasons for the steps involved.

$\begin{array}{r}\mathbf{A}\mathbf{ }\mathbf{B}\\ \mathbf{\times }\mathbf{ }\mathbf{5}\\ \mathbf{C}\mathbf{ }\mathbf{A}\mathbf{ }\mathbf{B}\end{array}$

Sol. 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.

$\begin{array}{r}25\\ \times 5\\ 125\end{array}$    $\begin{array}{r}75\\ \times 5\\ 375\end{array}$

If B = 0,

B × 5 = B $\Rightarrow$ 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.

$\begin{array}{r}50\\ \times 5\\ 250\end{array}$

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

P6. Find the values of the letters in the following and give reasons for the steps involved.

$\begin{array}{r}\mathrm{A} \mathrm{B}\\ \times 6\\ B B B\end{array}$

Sol. 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 $\Rightarrow$ 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 $\Rightarrow$ 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 $\Rightarrow$ 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 $\Rightarrow$ 6A = 42 and hence, A = 7

The multiplication is as follows.

$\begin{array}{r}7 4\\ \times 6\\ 444\end{array}$

Hence, the values of A and B are 7 and 4 respectively.

P8. Find the values of the letters in the following and give reasons for the steps involved.

$\begin{array}{cc}\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{A}& \mathbf{1}\\ \mathbf{+}\mathbf{1}& \mathbf{ }\mathbf{B}\\ \mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{B}& \mathbf{0}\end{array}$

Sol. 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.

$\begin{array}{rr}7& 1\\ +1& 9\\ 9& 0\end{array}$

Hence, the values of A and B are 7 and 9 respectively.

P8. Find the values of the letters in the following and give reasons for the steps involved.

$\begin{array}{rrr}\mathbf{2}& \mathbf{7}& \mathbf{B}\\ \mathbf{+}\mathbf{A}& \mathbf{B}& \mathbf{1}\\ \mathbf{B}& \mathbf{1}& \mathbf{8}\end{array}$

Sol. 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.

$\begin{array}{r}2 4 7\\ +4 7 1\\ 7 1 8\end{array}$

Hence, the values of A and B are 4 and 7 respectively.

P9. Find the values of the letters in the following and give reasons for the steps involved.

$\begin{array}{ccc}\mathbf{ }\mathbf{ }\mathbf{1}& \mathbf{2}& \mathbf{A}\\ \mathbf{+}\mathbf{6}& \mathbf{A}& \mathbf{B}\\ \mathbf{ }\mathbf{ }\mathbf{ }\mathbf{A}& \mathbf{0}& \mathbf{9}\end{array}$

Sol. 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.

$\begin{array}{r}1 2 8\\ +6 8 1\\ 8 0 9\end{array}$

Hence, the values of A and B are 8 and 1 respectively.

EXERCISE-16.2

P1. If 21y5 is a multiple of 9, where y is a digit, what is the value of y?

Sol. 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.

P2. 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?

Sol. 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.

P3. 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 xis 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)

Sol. 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.

P4. If 31z5 is a multiple of 3, where z is a digit, what might be the values of z?

Sol. 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.