NCERT Solutions Class 8 Mathematics Chapter 16 Playing with Numbers

Exercise 16.1

Q.1) Find the values of the letters in the following and give reasons for the steps involved.

Sol.1) i) 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.
∴ 𝐴 = 7
And putting 2 and carry over 1, we get B = 6
Hence, A = 7 and B = 6

ii) 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.
∴ A = 5
And putting 3 and carry over 1,
we get B = 4 and C = 1
Hence, A = 5, B = 4 and C = 1

iii) 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.
∴ A = 6 Hence, A = 6

iv) 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

v) Here on putting B = 0,
We get 0 x 3 = 0.
And A = 5,
then 5 × 3 = 15
⇒ A = 5 and C = 1
Hence, A = 5, B = 0 and C = 1

vi) On putting B = 0,
we get 0 x 5 = 0 and A = 5,
then 5 × 5 = 25
⇒ 𝐴 = 5, 𝐶 = 2
Hence, A = 5, B = 0 and C = 2

vii) 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.
∴ For 6 × 7 = 42 + 2 = 44
Hence, A = 7 and B = 4

viii) 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

ix) 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

x) 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

Q.1) If 21y5 is a multiple of 9, where y is a digit, what is the value of y
Sol.1) 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.
∴ 2 + 1 + 𝑦 + 5 = 8 + 𝑦
⇒ 8 + 𝑦 = 9
⇒ 𝑦 = 1

Q.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?
Sol.2) 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.
∴ 3 + 1 + 𝑧 + 5 = 9 + 𝑧
⇒ 9 + 𝑧 = 9
⇒ 𝑧 = 0
⇒ If 3 + 1 + 𝑧 + 5 = 9 + 𝑧
⇒ 9 + 𝑧 = 18
⇒ 𝑧 = 9
Hence, 0 and 9 are two possible answers.

Q.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.)
Sol.3) 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.
⇒ 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.

Q.4) If 31𝑧5 is a multiple of 3, where z is a digit, what might be the values of 𝑧?
Sol.4) Since 31 5𝑧 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. Since 𝑧 is a digit.
∴ 3 + 1 + 𝑧 + 5 = 9 + 𝑧
⇒ 9 + 𝑧 = 9 ⇒ 𝑧 = 0
If 3 + 1 + 𝑧 + 5 = 9 + 𝑧
⇒ 9 + 𝑧 = 12 ⇒ 𝑧 = 3
If 3 + 1 + 𝑧 + 5 = 9 + 𝑧
⇒ 9 + 𝑧 = 15 ⇒ 𝑧 = 6
If 3 + 1 + 𝑧 + 5 = 9 + 𝑧
⇒ 9 + 𝑧 = 18 ⇒ 𝑧 = 9
Hence, 0, 3, 6 and 9 are four possible answers.