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# M02 02 : Not convinced with method used to explaing the ans

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Re: M02 02 : Not convinced with method used to explaing the ans [#permalink]  27 Nov 2011, 12:42
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.

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!
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Re: M02 02 : Not convinced with method used to explaing the ans [#permalink]  26 Nov 2012, 12:17
pleonasm wrote:
Actually I came up with a slightly different explanation. I may be biased because I saw the answer accidentally before solving the problem.

Is x^16 -y^8+345 y^2 divisible by 15 - This can be broken down to :

x^16 -y^8+345 y^2 is divisible by 5 and 3 .

Statement 1: x is a multiple of 25 and y is a multiple of 20. So lets say
x=25k y =20l
now we have :

is (25k)^16 - (20l)^8 + 345*(20l)^2 divisible by 5 and 3.

This term is definitely divisible by 5 because all the individual terms end up in a 0 in the units place. However it cannot be determined if they are divisible by 3. So this is not sufficient.

Statement 2; y=x^2. So the question now becomes

Is x^16 - x^16 + 345 * x^4 divisible by 5 and 3

Is 345 * x^4 divisible by 5 and 3

Yes, because 345 is divisible by both 3 and 5 and any multiple of 345 has to be divisible by 3 and 5. Hence sufficient.

Though the explanation looks long, it took me under 2 minutes to solve.

This is really a lovely explanation however couldn't understand how a multiple of 25 raised to any degree of power end up with ZERO at the units digit place?
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Re: M02 02 : Not convinced with method used to explaing the ans [#permalink]  26 Nov 2012, 12:33
Hi there,

To me there question is very straight forward. We've to find out whether each term of expression X^16 - Y^8+ 345y^2 is divisible by 15 or not i.e. X^16/15 then Y^8/15 and 345Y^2/15
With help of 1 we can see it's not possible because 25*25*..../15 having reminder then 20*20*.../15 is also having reminder however the last term is divisible by 15
With help of 2 we can see the first and second term got vanished and the third is divisible by 15 so the answer is B

Now I need Kudos...
Re: M02 02 : Not convinced with method used to explaing the ans   [#permalink] 26 Nov 2012, 12:33
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