Is +ve Integer n is divisible by 4 ?

1) n^2 is divisible by 8
2)sqrt n is an even number



for 1 - n^2 has at least 3 two's in it's prime box so there must be at least 2 two's in each of n. Try an example of this

if n= 1, 2, 3, 4, 5, 6, 8, 12
then n^2 = 1, 4, 9, 16, 25, 36, 64, 144

notice that 4, 8 and 12 are each divisible by 4 and their square is divisible by 8.

The math side of this says that if a number has 2 two's as primes then it's square will have 4 two's, and hence will be divisible by any combination of two's from 1-4.

2) in this is really the inverse of what we did in 1. So we are told that the sqrt yields an even number - or a number divisible by 2. If we start from the sqrt and sqare it, we get a number that is divisible by 2 two's.

so n = 1, 4, 9, 16, 25, 36, 49, 64...
and sqrt(n) = 1, 2, 3, 4, 5, 6, 7, 8

even n yields even sqrt of n. we can derive this by knowing the following:

assume sqrt(n) = 2*whatever integer
so if we square we get n, so
2*whatever integer = 4*whatever^2
so we n is divisible by 4

Ans should be D

Hope this helps

Nicely done, Thanks M&M