benejo wrote:

if a and b are positive integers, what is the remainder when \(9^{2a+b}\) is divided by 20?

(1) a = 3

(2) b = 5

Source: ExpertsGlobal

\(9^{2a+b}\) to know the remainder of this number when it is divided by 20 we need to know the last two digits of this number, i.e. we need to know the units and tens digit of this number.

The cyclicity of \(9\) is as follows -

\(9^1=9; 9^2=81; 9^3=729; 9^4=6561; 9^5=59049\)

So units digit of the number will be either \(1\) or \(9\) and tens digit will always be an even number. Hence the remainder of \(9^{2a+b}\) when divided by \(20\) will depend on the UNITS digit of this number.

Now \(2a\) is always EVEN irrespective of the value of \(a\). Hence \(9^{{Even}+b}\). So we only need to know the value of \(b\) to get the units digit.

Statement 1: Nothing mentioned about \(b\).

InsufficientStatement 2: \(b\) is odd so \(9^{2a+b}=9^{{Even}+Odd}=9^{Odd}\). Hence the remainder will be \(9\).

SufficientOption

B