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.