NCERT Solutions For Class 8 MATH Chapter – 16 Playing with numbers Exercise – 16.2
Q1. If 21y5 is a multiple of 9, where y is a digit, what is the value of y ?
Solution:-
Suppose 21y5 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.
That is , 2 + 1 + y + 5 = 8 + y
Therefore , 8 + y is a factor 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 , the value of should be 1 only I.e. 8 + y = 8 + 1 = 9
Q2. If 31z5 is 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?
Solution:-
Since 31z5 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
Therefore, 9 + z is a multiple of 9.
This is only possible when 9 + z is any one of these numbers : 0,,18,27 and so on.
This implies , 9 + 0 = 9 and 9 + 9 = 18
Hence 0 and 9 are two possible answers.
Q3. If 24x is a multiple of 3, where x is a digit, what is the value of x ?
Solution:-
Let’s say 24x is a multiple of 3.
Then, according to the divisibility rule of 3, the sum of all the digits should be a multiple of 3.
2 + 4 + x = 6 + x
So, 6 + x is a multiple of 3, and 6 + x is one of these numbers : 0,3,6,9,12,15,18 and so on.
Since x is a digit the value of x will either 0 or 3 or 6 or 9 and the sum of the digits can be 6 or 9 or 12 or 15 respectively.
Thus , x can have any of the four different values: 0 or 3 or 6 or 9.
Q4. If 31z5 is a multiple of 3, where z is a digit, what might be the values of z?
Solution:-
Since 31z5 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.
That is 3 + 1+ z + 5 = 9 + z
Therefore , 9 + z is a multiple of 3.
This is possible when the value of 9 + z is any of the values; 0,3,6,9,12,15 and so on.
At z = 0 , 9 + z = 9 + 0 = 9
At z = 3, 9 + z = 9 + 3 = 12
At z = 6 9 + z = 9 + 6 = 15
At z = 9, 9 + z = 9 + 9 = 18
The value of 9 + z can be 9 or 12 or 15 or 18.
Hence 0,3,6 or 9 are four possible answers for z.