Bunuel
If x is a positive integer, does the remainder, when (3^x + 2) is divided by 100, have 1 as the units digit?
(1) x = 2(2n + 1), where n is a positive integer.
(2) 10 > x > 4
This question can be simplified by analysing the question stem.
Remainder
when divided by 100 would be last two digits. So units digit of the remainder will be the units digit of \(3^x+2\), so \(3^x+2\) also has a units digit of 1.
This further tell us that \(3^x \) should have unit digit of x1-2 or 9.
The cyclicity of unit digit of successive powers of 3 is 3, 9, 7, 1, 3.....
So x has to be of the type 4K+2, where k is non negative integer.
The question, therefore, becomes:
Is x of type 4K+2?(1) x = 2(2n + 1), where n is a positive integer.
x=4n+2
Exactly what we were looking for.
Sufficient
(2) 10 > x > 4
If x is 6 or 4*1+2, then yes, otherwise no.
Insufficient
A