# NCERT solutions for class 8 Maths Chapter 16 Playing With Numbers

#### Solution for Exercise 16.1

Find the values of the letters in each of the following and give reasons for the steps involved.
1.
$$\begin{array} \hline \;\;\;3 & A\\ + 2 & 5\\ \hline \;\;B & 2\\ \hline \end{array}$$

We can observe that A+5 gives 2 as the unit's digit to the sum.So it's only possible if A=7.

Now, the sum becomes :

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

And we can clearly see that value of B=6.

Therefore we got the values as A=7 and B=6

2.
$$\begin{array} \hline \;\;\;4 & A\\ + 9 & 8\\ \hline C\;B & 3\\ \hline \end{array}$$

We can observe that A+8 gives 3 as the unit's digit to the sum.So it's only possible if A=5.

Now, the sum becomes :

$$\begin{array} \hline \;\;\;4 & 5\\ + 9 & 8\\ \hline 1\;\;4 & 3\\ \hline \end{array}$$

By this, we can clearly see that value of B and C as 4 and 1 respectively.

Therefore we got the values as A=5, B=4 and C=1

3.
$$\;\; \begin{array} \hline \;\;\;1 & A\\ \;\;×& A\\ \hline \;\;9 & A\\ \hline \end{array}$$

We can observe that A×A gives A as the unit's digit to the multiplication, that means it's only possible if A=1, 6 and 5.

Checking for all the case:

For A=1: $$\;\;\;\begin{array} \hline \;\;\;1 & 1\\ \;\;×& 1\\ \hline \;\;1 & 1\\ \hline \end{array}$$

We can see clearly that it is not similar with the given multiplication so A$$\neq$$1.

For A=5: $$\;\;\;\begin{array} \hline \;\;\;1 & 5\\ \;\;×& 5\\ \hline \;\;7 & 5\\ \hline \end{array}$$

We can see clearly that it is not similar with the given multiplication so A$$\neq$$5.

For A=6: $$\;\;\;\begin{array} \hline \;\;\;1 & 6\\ \;\;×& 6\\ \hline \;\;9 & 6\\ \hline \end{array}$$

We can see clearly that it is similar with the given multiplication so A=6.

4.
$$\;\;\begin{array} \hline \;\;\;A & B\\ +\;3& 7\\ \hline \;\;6 & A\\ \hline \end{array}$$

We have here two conditions :

(i) B+7=A

(ii)A+3=6

By solving the second equation we get A=6-3=3

Also if we keep A=3 directly in eq. (i) we get B=3-7=negative value. So it can be said that a carry was taken so let the carry be 1.Therefore, now A=2+1(carry) and let us check if eq. (i) satisfies it or not.

We have B+7=12[∵ 1(carry)2(value of A)]

So we get B=12-7=5.

So the sum becomes :

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

Hence, the assumption is correct and values are A=2 and B=5

5. $$\begin{array} \hline \;\;\;A & B\\ \times& 3\\ \hline C \;\;A & B\\ \hline \end{array}$$

We can observe that we have equations:

(i)B$$\times$$3=B

Solving it we can have B=0 or 5

5 is not taken because, for satisfying the condition A$$\times$$3=CA, we should not take carry. If carry is taken the ten’s value becomes A+1 not A.

$$∴B=0$$

(ii)A$$\times$$3=C A

we can satisfy this condition only if A=5 i.e., 5$$\times$$3=15.

So we have A=5, B=0 and C=1

6.
$$\;\;\; \begin{array} \hline \;\;\;A & B\\ \times& 5\\ \hline C \;\;A & B\\ \hline \end{array}$$

We can observe that we have equations:

(i)B$$\times$$5=B

Solving it we can have B= 0 or 5

5 is removed because for satisfying the condition A$$\times$$5=CA, we should not take carry. If carry is taken the ten’s value becomes A+1 not A.

$$∴B=0$$

(ii)A$$\times$$5=C A

we can satisfy this condition only if A=5 i.e., 5$$\times$$5=25.

So we have A=5, B=0 and C=2

7.
$$\;\; \begin{array} \hline \;\;\;A & B\\ \times& 6\\ \hline B\;\;B & B\\ \hline \end{array}$$

We can observe that: Possible values of BBB are 111, 222, 333, etc.
Let us divide these numbers by 6 and check the nummber which is completely divisible.

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

222 ÷ 6 = 37, remainder 0 but the quotient 37 is not of the form A 2. So,222 is rejected.

333 ÷ 6 = 55, remainder 3 so, 333 is rejected. 444 ÷ 6 = 74, remainder 0 and also the quotient 74 is of the form A 4.

So, we can say the number B=4 is satisfying.Let's check and find value of A

$$\begin{array} \hline \;\;\;A & 4\\ \times& 6\\ \hline 4\;\;4 & 4\\ \hline \end{array}$$

We know this is only possible if A=7

Hence, we have got A=7 and B=4

8.
$$\;\;\begin{array} \hline \;\;\;A & 1\\ + 1 & B\\ \hline \;\;B & 0\\ \hline \end{array}$$

We have condition as:

B+1=0 i.e, B must be 9 to get 0 at unit's digit of the addition.So, we got B=9

Also we have A+1+1(carry)=B

$$\Rightarrow$$ A+2=9

$$\Rightarrow$$ A=7

Hence, we have A=7 and B=9

9.
$$\;\;\begin{array} \hline \;\;\;2 & A &B\\ + A & B&1\\ \hline \;\;B&1 & 8\\ \hline \end{array}$$

We have condition as:

B+1=8.So, we got B=7

Also we have A+B=1

$$\Rightarrow$$ A+7=11

A must be 4 to get unit's digit of A+7 as 1.So, A=4

And the last condition we have 2+A+1(carry)=B

Checking the values of A and B if it satisfies or not:

We have A=4 and B=7.Substituting, we have:

LHS: 2+A+1=2+4+1=7

RHS: B=7

Hence the values are verified.So we have A=4 and B=7

10.
$$\;\; \begin{array} \hline \;\;\;1 & 2 &A\\ + 6 & A&B\\ \hline \;\;A&0 & 9\\ \hline \end{array}$$

We have condition as:

(i)A+B=9.

(ii)2+A=10.

(iii)1+6+1(carry)=A

We can say by equation (iii) and equation (ii) that A=8

Also, we have A+B=9. Sustituting A=8, we get B=1

Hence, we have A=8 and B=1

#### Solution for Exercise 16.2

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

For a number is divisible by 9 we know that the sum of its digits should also divisible by 9.
Here, Sum of the digits of 21y5 = 2 + 1 +y + 5 = 8 + y

So we should have (8 + y) ÷ 9 = 1

$$\Rightarrow 8 + y = 9$$

$$\Rightarrow y = 9 – 8 = 1$$

Hence, the required value of y = 1.

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

We know that a number is a multiple of 9 if the sum of its digits is also divisible by 9.

Here, Sum of the digits of 31z5 = 3 + 1 + z + 5

So we have , 3 + 1 + z + 5 = 9k where k is an integer.

Checking for different values of k

For k = 1:

$$\Rightarrow$$ 3 + 1 + z + 5 = 9

$$\Rightarrow$$ z = 9 – 9 = 0

For k = 2:

We have, 3 + 1 + z + 5 = 18

$$\Rightarrow$$ z = 18 – 9 = 9

k = 3 is not possible because 3 + 1 + z + 5 = 27

$$\Rightarrow$$ z = 27 – 9 = 18 which is not a digit.

Hence the required value of z is 0 or 9

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

We are given that 24 x is a multiple of 3. So, the sum of its digits 2 + 4 + x = (6 + x) should be a multiple of 3.
Therefore, (6 + x) is one of the numbers 0, 3, 6, 9, 12, 15, 18…so on but as per given x can only be 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

$$\Rightarrow$$ x = 0 or 3 or 6 or 9

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

4.31z5 is a multiple of 3, where z is a digit, what might be the values of z?

$$\Rightarrow$$ z = 0 or 3 or 6 or 9