thanatoz wrote:

So I know the basic rule is that, if a is a factor of both b and c, then a is a factor of (b + c)

But anything else?

Specifically, for a question like this on the math challenge. Is there a clear rule that will show quickly that since x and y are not multiples of 3, their addition cannot have it either? Thanks.

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. \(y = x^2\)

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.

thanatoz wrote:

So I know the basic rule is that, if a is a factor of both b and c, then a is a factor of (b + c)

But anything else?

Specifically, for a question like this on the math challenge. Is there a clear rule that will show quickly that since x and y are not multiples of 3, their addition cannot have it either? Thanks.

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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