a is a factor of both b and c then a is a factor of b-c, b*c...
if x and y are not multiples of 3, when x^2n and y^2n are divided by 3, the remainder is 1.
1. That x is a multiple of 25 and y is a multiple of 20 doesn't give any clue of remainder of x,y when divided by 3 so insuf
2. suf

First of all as \(345y^2\) is divisible by 15 (this term won't affect the remainder), we can drop it.

The question becomes: is \(x^{16}-y^8\) divisible by 15?

(1) \(x\) is a multiple of 25, and \(y\) is a multiple of 20 --> they both could be multiples of 15 as well (eg x=25*15 and y=20*15) and in this case \(x^{16}-y^8=15*(...)\) will be divisible by 15 OR one could be multiple of 15 and another not (eg x=25*15 and y=20) and in this case \(x^{16}-y^8\) won't be divisible by 15 (as we can not factor out 15 from \(x^{16}-y^8\)). Not sufficient.

(2) \(y = x^2\) --> \(x^{16}-y^8=x^{16}-(x^2)^8=x^{16}-x^{16}=0\). 0 is divisible by 15. Sufficient.

Answer: B.



If x and y both are not multiples of 3, then their sum or difference may or may not be multiple of 3:

x=2 and y=1 --> x+y=3 --> sum is multiple of 3;
x=2 and y=2 --> x-y=0 --> difference is multiple of 3;
x=2 and y=3 --> x+y=5 --> sum is not multiple of 3;
x=2 and y=0 --> x-y=2 --> difference is not multiple of 3.

BUT, if x is multiple of 3 and y is not (or vise-versa), then their sum or difference won't be multiple of 3.

Hope it helps.
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REM(x^16-y^8 +345*y^2)/15

REM(345/15)=0

So REM(x^16-y^8/15)

(1).

It may have different remainders for different values of x/y.
Insufficient

(2).

y=x^2
=> REM(x^16-x^16)/15
=>REM=0

=>Sufficient

(B) it is !

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