carcass wrote:
If m and n are positive integers, is m + n divisible by 4 ?
(1) m and n are each divisible by 2.
(2) Neither m nor n is divisible by 4.
Statement 1- Let's say m = (2*a*b*...), which a,b,... are prime numbers.
- Let's say n = (2*x*y*...), which x,y,... are prime numbers.
- \(\frac{(m+n)}{4}\) = \(\frac{(2*a*b*...) + (2*x*y*...)}{4}\) = \(\frac{2 ( (a*b*...) + (x*y*...))}{4}\)
- Here we can see that a,b,x,y can be anything prime numbers.
- #1 If (a*b) + (x*y) = EVEN, YES, \(\frac{m+n}{4}\) divisible by 4.
- #2 If (a*b) + (x*y) = ODD, NO, \(\frac{m+n}{4}\) cannot divisible by 4.
- Hence, INSUFFICIENT.
Statement 2- Each of m and n has only one maximum one factor of 2 so it cannot divisible by 4.
- Try to plug simple number : m = 3 and n = 5, divisible by 4, YES.
- Try to plug another random number : m = 5 and n = 9, NOT divisible by 4, NO.
- Hence, INSUFFICIENT.
Both statement- Go back to this equation :
- \(\frac{(m+n)}{4}\) = \(\frac{(2*a*b*...) + (2*x*y*...)}{4}\) = \(\frac{2 ( (a*b*...) + (x*y*...))}{4}\)
- If m & n divisible by 2 but not by 4, so a,b,x,y MUST BE ODD.
- (ODD*ODD) + (ODD*ODD) = ODD + ODD = EVEN.
- 2 * ANY even number MUST BE DIVISIBLE BY 4.
- Hence, SUFFICIENT.
_________________
There's an app for that - Steve Jobs.