DS: If p, x, and y are positive integers, y is odd, and p

[cesar.rodriguez.blanco]

I do not know how to solve this problem. Please help:

If p, x, and y are positive integers, y is odd, and p = x^2 + y^2, is x divisible by 4?

(1) When p is divided by 8, the remainder is 5.
(2) x - y = 3

[nitin_prakash_khanna]

Hi,

This is how i did it. Hope it helps.

1. p can be represented as p = 8n+5
so p can take values , 13, 21, 29,37, 45.....just plugging in various values of integer n.

Now the next task is to represent all these odd numbers as sum of 2 perfect squares.

13 = 4+9 = 2^2+3^2 this implies x=2, y=3 . As question already told us that y is odd.
21 = cant represnt as sum of two +ve integers
29 = 4+ 25 = 2^2+ 5^2 ....implies x=2, y=5.
37 = 1+36 ....implies y=1 , x=6
45 = 9+36....implies y=3, x=6

So it tells us that x = 2, 6...not divisble by 4.

Hence SUFFICIENT

2. Condition B tells us
x-y= 3 and y is odd

so y=1 , x=4 Div by 4
y= 3 , x= 6 Not Div by 4
y= 5, x= 8 Div by 4
y= 7, x=10, Not DIv by 4
So INSUFFICIENT

Whats the OA?

thanks
nk.

[vishalsongra]

Agree IMO A

OA pls ?

[Ben Ku]

Please post problems in the correct folder, and not in multiple folders! Here is the response in the General Folder:

http://www.manhattangmat.com/forums/divisible-by-4-t7787.html