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.
\( \begin{array} \hline \;\;\;3 & A\\ + 2 & 5\\ \hline \;\;B & 2\\ \hline \end{array} \)


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Answer :

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


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Answer :

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.
\(\;\; \begin{array} \hline \;\;\;1 & A\\ \;\;×& A\\ \hline \;\;9 & A\\ \hline \end{array} \)


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Answer :

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


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Answer :

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. \( \begin{array} \hline \;\;\;A & B\\ \times& 3\\ \hline C \;\;A & B\\ \hline \end{array} \)


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Answer :

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.
\(\;\;\; \begin{array} \hline \;\;\;A & B\\ \times& 5\\ \hline C \;\;A & B\\ \hline \end{array} \)


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Answer :

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.
\(\;\; \begin{array} \hline \;\;\;A & B\\ \times& 6\\ \hline B\;\;B & B\\ \hline \end{array} \)


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Answer :

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.
\( \;\;\begin{array} \hline \;\;\;A & 1\\ + 1 & B\\ \hline \;\;B & 0\\ \hline \end{array} \)


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Answer :

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


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Answer :

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


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Answer :

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?


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Answer :

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?


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Answer :

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?


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Answer :

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?


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Answer :

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

\(\Rightarrow\) 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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