Hi,
I tried trial-and-error to solve this. Can anyone suggest a better approach.
Thanks
Amit
Power prep divisibility
This topic has expert replies
(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
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
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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
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