Power prep divisibility

erjamit · Posted: Sun Jun 29, 2008 3:39 am Post subject: Power prep divisibility

Hi,

I tried trial-and-error to solve this. Can anyone suggest a better approach.

Thanks
Amit

gxliu · Posted: Mon Jun 30, 2008 9:53 am Post subject:

(p^2-n^2)/15 can be rewritten as (P+N)(P-N)/3*5.
C is the because you get see that (P-N)/3 gives you a remainder of 1 and (P+N)/5 gives you a remainder of 1. Both of these are factors in the question. 1+1=2

gogetter08 · Posted: Mon Jun 30, 2008 12:13 pm Post subject:

Trial and error works well but here is a systematic approach.

Lets assume that when x is divided by a, reminder is r1 and when y is divided by b, reminder is r2

x = N1*a + r1 ; N1- some integer
y = N2*b + r2 ; N2 - some integer

You are asked that given, r1, r2,a, b: can we find out reminder of x*y when divided by a*b:

x*y = N1N2ab + r1r2 + N1ar2 + N2br1

N1N2ab is divisible by ab

But we cannot tell what the fraction (N1ar2+N2br1)/ab will resolve to

we already know the value of the fraction r1r2/ab

Since it is not possible to arrive at a reminder independent of N1, N2, the ans should be E

atlantic · Posted: Tue Jul 01, 2008 12:55 am Post subject:

Oops! I'll go with E also.

But since I also got it by trial and error I doubt was is the final answer, C or D.

Anyone else?

beeparoo · Posted: Tue Jul 01, 2008 4:41 pm Post subject:

I got E as well. My solution looks similar to gogetter08, but I used the number values provided in the question stem to arrive at the same logic.

What is the OA?

erjamit · Posted: Sun Jul 06, 2008 1:27 am Post subject:

OA is E.