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is x^16-y^8+345y^2 divisible by 15

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by bhumika.k.shah » Tue Jan 26, 2010 4:32 am
If x and y are positive integers, is x^16-y^8+345y^2 divisible by 15?

1. x is a multiple of 25, and y is a multiple of 20
2.y=x^2

OA B

A quicker approach is needed!
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Source: — Data Sufficiency |
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by bhumika.k.shah » Tue Jan 26, 2010 4:34 am
Also is there a way in which these quadratics can be broken down to a simple question?
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by dmitriyaleyev » Tue Jan 26, 2010 9:14 am
Bhumika,

The easiest way to look at statement 1 is to imagine multiple of 25 - say 25 - and a multiple of 20.
Multiple of 25 in 16th power will always end on "0" or "5". since 345 itself is a multiple of 15, and we don't know the result of the first 2 numbers, statement 1 is insufficient.

Statement 2 on the other hand, written as y^8 - y^8 + 345y^2 = 345(blablabla)/15 will def be divisible by 15.

let me know if you ahve any questions.
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