Bunuel wrote:
If x and y are positive integers, is x^{16} - y^8 + 345y^2 divisible by 15?
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 (zero is divisible by every integer except zero itself). Sufficient.
Answer: B.
Notes for statement (1):
If integers a and b are both multiples of some integer k>1 (divisible by k), then their sum and difference will also be a multiple of k (divisible by k):
Example: a=6 and b=9, both divisible by 3 ---> a+b=15 and a-b=-3, again both divisible by 3.
If out of integers a and b one is a multiple of some integer k>1 and another is not, then their sum and difference will NOT be a multiple of k (divisible by k):
Example: a=6, divisible by 3 and b=5, not divisible by 3 ---> a+b=11 and a-b=1, neither is divisible by 3.
If integers a and b both are NOT multiples of some integer k>1 (divisible by k), then their sum and difference may or may not be a multiple of k (divisible by k):
Example: a=5 and b=4, neither is divisible by 3 ---> a+b=9, is divisible by 3 and a-b=1, is not divisible by 3;
OR: a=6 and b=3, neither is divisible by 5 ---> a+b=9 and a-b=3, neither is divisible by 5;
OR: a=2 and b=2, neither is divisible by 4 ---> a+b=4 and a-b=0, both are divisible by 4.
So according to above info that x is a multiple of 25, and y is a multiple of 20 tells us nothing whether x^{16}-y^8 is divisible by 15.
Hope it helps.
Definitely helps! Thanks!
close
so the answer is B
