Number Play l Exercise 5.5 l Class 8 l NCERT Solutions l Ganita Prakash

Chapter: Number Play

Exercise 5.5 – Stepwise Solutions

Question 1

If 31z5 is a multiple of 9, where z is a digit, what is the value of z? Explain why there are two answers to this problem.

Solution

A number is divisible by 9 if the sum of its digits is divisible by 9.

Given number:

31z5

Sum of digits:

3+1+z+5=9+z

For divisibility by 9,

9+z

must be a multiple of 9.

Possible values:

• If z=0,

9+0=9

which is divisible by 9.

• If z=9,

9+9=18

which is also divisible by 9.

Therefore,

z=0 or z=9​

Why are there two answers?

Because both 9 and 18 are multiples of 9, and both are possible sums of digits.

Question 2

“I take a number that leaves a remainder of 8 when divided by 12. I take another number which is 4 short of a multiple of 12. Their sum will always be a multiple of 8,” claims Snehal. Examine his claim and justify your conclusion.

Solution

First number:

A number leaving remainder 8 when divided by 12 can be written as:

12a+8

Second number:

A number 4 less than a multiple of 12 can be written as:

12b−4

Their sum:

(12a+8)+ (12b−4) =12a+12b+4=12(a+b) +4

Now check whether this is always divisible by 8.

Take an example:

Let

a=1, b=1

Then,

First number:

12(1)+8=20

Second number:

12(1)−4=8

Sum:

20+8=28

But

28÷8=remainder 4

So 28 is not divisible by 8.

Therefore, Snehal’s claim is false.

The sum is not always a multiple of 8.

Question 3

When is the sum of two multiples of 3, a multiple of 6 and when is it not? Explain the different possible cases, and generalise the pattern.

Solution

Multiples of 3 are of two types:

• Even multiples of 3:
  6,12,18,…
• Odd multiples of 3:
  3,9,15,…

Case 1: Even multiple + Even multiple

Example:

12+18=30

30 is divisible by 6.

So, the sum is a multiple of 6.

Case 2: Odd multiple + Odd multiple

Example:

9+15=24

24 is divisible by 6.

So, the sum is a multiple of 6.

Case 3: Even multiple + Odd multiple

Example:

12+9=21

21 is not divisible by 6.

So, the sum is not a multiple of 6.

Generalisation

The sum of two multiples of 3 is a multiple of 6 when both multiples are:

• Both even, or
• Both odd.

It is not a multiple of 6 when one is even and the other is odd.

Same parity ⇒ multiple of 6​

Different parity ⇒ not a multiple of 6

Question 4

Sreelatha says, “I have a number that is divisible by 9. If I reverse its digits, it will still be divisible by 9.”

(i) Examine if her conjecture is true for any multiple of 9.

Solution

A number is divisible by 9 if the sum of its digits is divisible by 9.

Reversing digits does not change the sum of digits.

Example:

729

Sum of digits:

7+2+9=18

18 is divisible by 9.

Reverse the number:

927

Sum of digits:

9+2+7=18

Again divisible by 9.

Therefore, the reversed number is also divisible by 9.

Hence, Sreelatha’s conjecture is true.

Reversing digits of a multiple of 9 still gives a multiple of 9​

(ii) Are any other digit shuffles possible such that the number formed is still a multiple of 9?

Solution

Yes.

Any rearrangement (shuffle) of the digits keeps the digit sum unchanged.

Since divisibility by 9 depends only on the sum of digits, every shuffle will also be divisible by 9.

Example:

243→324, 432, 234

All have digit sum:

2+4+3=9

Hence all are divisible by 9.

Any digit arrangement remains divisible by 9​

Question 5

If 48a23b is a multiple of 18, list all possible pairs of values for a and b.

Solution

A number is divisible by 18 if it is divisible by:

• 2, and 9

Step 1: Divisibility by 2

The last digit b must be even.

So,

b=0,2,4,6,8

Step 2: Divisibility by 9

Sum of digits must be divisible by 9.

Digits are:

4+8+a+2+3+b=17+a+b

For divisibility by 9:

17+a+b

must be a multiple of 9.

Possible multiples near 17 are:

18 or 27

Case 1

17+a+b=18 where a+b=1

Possible even values of b:

• b=0⇒a=1

So pair:

(1,0)

Case 2

17+a+b=27 where a+b=10

Possible even values of b:

• b=0⇒a=10 ✗ not a digit
• b=2⇒a=8
• b=4⇒a=6
• b=6⇒a=4
• b=8⇒a=2

Possible pairs:

(8,2), (6,4), (4,6), (2,8)

Final Answer

(a,b)=(1,0), (8,2), (6,4), (4,6), (2,8)