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

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.