# NCERT Solutions for Class 8 Maths Chapter 16 Playing With Numbers  Written by Team Trustudies
Updated at 2021-05-07

## NCERT solutions for class 8 Maths Chapter 16 Playing With Numbers Exercise 16.1

Find the values of the letters in each of the following and give reasons for the steps involved.
1.
$\overline{)\begin{array}{ll}\phantom{\rule{0.278em}{0ex}}\phantom{\rule{0.278em}{0ex}}\phantom{\rule{0.278em}{0ex}}3& A\\ +2& 5\\ \phantom{\rule{0.278em}{0ex}}\phantom{\rule{0.278em}{0ex}}B& 2\end{array}}$

NCERT Solutions for Class 8 Maths Chapter 16 Playing With Numbers

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 :

$\overline{)\begin{array}{ll}\phantom{\rule{0.278em}{0ex}}\phantom{\rule{0.278em}{0ex}}\phantom{\rule{0.278em}{0ex}}3& 7\\ +2& 5\\ \phantom{\rule{0.278em}{0ex}}\phantom{\rule{0.278em}{0ex}}6& 2\end{array}}$

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

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

2.
$\overline{)\begin{array}{ll}\phantom{\rule{0.278em}{0ex}}\phantom{\rule{0.278em}{0ex}}\phantom{\rule{0.278em}{0ex}}4& A\\ +9& 8\\ C\phantom{\rule{0.278em}{0ex}}B& 3\end{array}}$

NCERT Solutions for Class 8 Maths Chapter 16 Playing With Numbers

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 :

$\overline{)\begin{array}{ll}\phantom{\rule{0.278em}{0ex}}\phantom{\rule{0.278em}{0ex}}\phantom{\rule{0.278em}{0ex}}4& 5\\ +9& 8\\ 1\phantom{\rule{0.278em}{0ex}}\phantom{\rule{0.278em}{0ex}}4& 3\end{array}}$

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.
$\phantom{\rule{0.278em}{0ex}}\phantom{\rule{0.278em}{0ex}}\overline{)\begin{array}{ll}\phantom{\rule{0.278em}{0ex}}\phantom{\rule{0.278em}{0ex}}\phantom{\rule{0.278em}{0ex}}1& A\\ \phantom{\rule{0.278em}{0ex}}\phantom{\rule{0.278em}{0ex}}×& A\\ \phantom{\rule{0.278em}{0ex}}\phantom{\rule{0.278em}{0ex}}9& A\end{array}}$

NCERT Solutions for Class 8 Maths Chapter 16 Playing With Numbers

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: $\phantom{\rule{0.278em}{0ex}}\phantom{\rule{0.278em}{0ex}}\phantom{\rule{0.278em}{0ex}}\overline{)\begin{array}{ll}\phantom{\rule{0.278em}{0ex}}\phantom{\rule{0.278em}{0ex}}\phantom{\rule{0.278em}{0ex}}1& 1\\ \phantom{\rule{0.278em}{0ex}}\phantom{\rule{0.278em}{0ex}}×& 1\\ \phantom{\rule{0.278em}{0ex}}\phantom{\rule{0.278em}{0ex}}1& 1\end{array}}$

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

For A=5: $\phantom{\rule{0.278em}{0ex}}\phantom{\rule{0.278em}{0ex}}\phantom{\rule{0.278em}{0ex}}\overline{)\begin{array}{ll}\phantom{\rule{0.278em}{0ex}}\phantom{\rule{0.278em}{0ex}}\phantom{\rule{0.278em}{0ex}}1& 5\\ \phantom{\rule{0.278em}{0ex}}\phantom{\rule{0.278em}{0ex}}×& 5\\ \phantom{\rule{0.278em}{0ex}}\phantom{\rule{0.278em}{0ex}}7& 5\end{array}}$

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

For A=6: $\phantom{\rule{0.278em}{0ex}}\phantom{\rule{0.278em}{0ex}}\phantom{\rule{0.278em}{0ex}}\overline{)\begin{array}{ll}\phantom{\rule{0.278em}{0ex}}\phantom{\rule{0.278em}{0ex}}\phantom{\rule{0.278em}{0ex}}1& 6\\ \phantom{\rule{0.278em}{0ex}}\phantom{\rule{0.278em}{0ex}}×& 6\\ \phantom{\rule{0.278em}{0ex}}\phantom{\rule{0.278em}{0ex}}9& 6\end{array}}$

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

4.
$\phantom{\rule{0.278em}{0ex}}\phantom{\rule{0.278em}{0ex}}\overline{)\begin{array}{ll}\phantom{\rule{0.278em}{0ex}}\phantom{\rule{0.278em}{0ex}}\phantom{\rule{0.278em}{0ex}}A& B\\ +\phantom{\rule{0.278em}{0ex}}3& 7\\ \phantom{\rule{0.278em}{0ex}}\phantom{\rule{0.278em}{0ex}}6& A\end{array}}$

NCERT Solutions for Class 8 Maths Chapter 16 Playing With Numbers

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 :

$\overline{)\begin{array}{ll}\phantom{\rule{0.278em}{0ex}}\phantom{\rule{0.278em}{0ex}}\phantom{\rule{0.278em}{0ex}}2& 5\\ +\phantom{\rule{0.278em}{0ex}}3& 7\\ \phantom{\rule{0.278em}{0ex}}\phantom{\rule{0.278em}{0ex}}6& 2\end{array}}$

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

5. $\overline{)\begin{array}{ll}\phantom{\rule{0.278em}{0ex}}\phantom{\rule{0.278em}{0ex}}\phantom{\rule{0.278em}{0ex}}A& B\\ ×& 3\\ C\phantom{\rule{0.278em}{0ex}}\phantom{\rule{0.278em}{0ex}}A& B\end{array}}$

NCERT Solutions for Class 8 Maths Chapter 16 Playing With Numbers

We can observe that we have equations:

(i)B$×$3=B

Solving it we can have B=0 or 5

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

$?B=0$

(ii)A$×$3=C A

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

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

6.
$\phantom{\rule{0.278em}{0ex}}\phantom{\rule{0.278em}{0ex}}\phantom{\rule{0.278em}{0ex}}\overline{)\begin{array}{ll}\phantom{\rule{0.278em}{0ex}}\phantom{\rule{0.278em}{0ex}}\phantom{\rule{0.278em}{0ex}}A& B\\ ×& 5\\ C\phantom{\rule{0.278em}{0ex}}\phantom{\rule{0.278em}{0ex}}A& B\end{array}}$

NCERT Solutions for Class 8 Maths Chapter 16 Playing With Numbers

We can observe that we have equations:

(i)B$×$5=B

Solving it we can have B= 0 or 5

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

$?B=0$

(ii)A$×$5=C A

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

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

7.
$\phantom{\rule{0.278em}{0ex}}\phantom{\rule{0.278em}{0ex}}\overline{)\begin{array}{ll}\phantom{\rule{0.278em}{0ex}}\phantom{\rule{0.278em}{0ex}}\phantom{\rule{0.278em}{0ex}}A& B\\ ×& 6\\ B\phantom{\rule{0.278em}{0ex}}\phantom{\rule{0.278em}{0ex}}B& B\end{array}}$

NCERT Solutions for Class 8 Maths Chapter 16 Playing With Numbers

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

$\overline{)\begin{array}{ll}\phantom{\rule{0.278em}{0ex}}\phantom{\rule{0.278em}{0ex}}\phantom{\rule{0.278em}{0ex}}A& 4\\ ×& 6\\ 4\phantom{\rule{0.278em}{0ex}}\phantom{\rule{0.278em}{0ex}}4& 4\end{array}}$

We know this is only possible if A=7

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

8.
$\phantom{\rule{0.278em}{0ex}}\phantom{\rule{0.278em}{0ex}}\overline{)\begin{array}{ll}\phantom{\rule{0.278em}{0ex}}\phantom{\rule{0.278em}{0ex}}\phantom{\rule{0.278em}{0ex}}A& 1\\ +1& B\\ \phantom{\rule{0.278em}{0ex}}\phantom{\rule{0.278em}{0ex}}B& 0\end{array}}$

NCERT Solutions for Class 8 Maths Chapter 16 Playing With Numbers

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

$?$ A+2=9

$?$ A=7

Hence, we have A=7 and B=9

9.
$\phantom{\rule{0.278em}{0ex}}\phantom{\rule{0.278em}{0ex}}\overline{)\begin{array}{lll}\phantom{\rule{0.278em}{0ex}}\phantom{\rule{0.278em}{0ex}}\phantom{\rule{0.278em}{0ex}}2& A& B\\ +A& B& 1\\ \phantom{\rule{0.278em}{0ex}}\phantom{\rule{0.278em}{0ex}}B& 1& 8\end{array}}$

NCERT Solutions for Class 8 Maths Chapter 16 Playing With Numbers

We have condition as:

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

Also we have A+B=1

$?$ 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.
$\phantom{\rule{0.278em}{0ex}}\phantom{\rule{0.278em}{0ex}}\overline{)\begin{array}{lll}\phantom{\rule{0.278em}{0ex}}\phantom{\rule{0.278em}{0ex}}\phantom{\rule{0.278em}{0ex}}1& 2& A\\ +6& A& B\\ \phantom{\rule{0.278em}{0ex}}\phantom{\rule{0.278em}{0ex}}A& 0& 9\end{array}}$

NCERT Solutions for Class 8 Maths Chapter 16 Playing With Numbers

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

## NCERT solutions for class 8 Maths Chapter 16 Playing With Numbers Exercise 16.2

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

NCERT Solutions for Class 8 Maths Chapter 16 Playing With Numbers

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

$?8+y=9$

$?y=9–8=1$

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?

NCERT Solutions for Class 8 Maths Chapter 16 Playing With Numbers

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:

$?$ 3 + 1 + z + 5 = 9

$?$ z = 9 – 9 = 0

For k = 2:

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

$?$ z = 18 – 9 = 9

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

$?$ 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?

NCERT Solutions for Class 8 Maths Chapter 16 Playing With Numbers

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

$?$ 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?

NCERT Solutions for Class 8 Maths Chapter 16 Playing With Numbers

We are given that 31z5 is a multiple of 3. So, the sum of its digits 3 + 1 + z + 5 = (9 + z) should be multiple of 3.
Therefore, (9 + z) is one of the numbers 0, 3, 6, 9, 12, 15, 18, …so on but given that z is a digit. Therefore (9 + z) must be equal to 9 or 12 or 15 or 18.

So we have, 9 + z = 9 or 12 or 15 or 18

$?$ z = 0 or 3 or 6 or 9

Hence, z can have any of the four different values, which are 0, 3, 6 or 9.

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