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M02-02

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M02-02 [#permalink]

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

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. Now, both \(x\) and \(y\) 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 (e.g. \(x=25*15\) and \(y=20\)) and in this case \(x^{16}-y^8\) won't be divisible by 15 (as we cannot factor out 15 from \(x^{16}-y^8\)). Not sufficient.

(2) \(y = x^2\). Substitute \(y\) with \(x^2\): \(x^{16}-y^8=x^{16}-(x^2)^8=x^{16}-x^{16}=0\). 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 \(k \gt 1\) (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 \(k \gt 1\) 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 \(k \gt 1\) (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.


Answer: B
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Re M02-02 [#permalink]

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New post 07 Jan 2015, 08:28
I think this question is good and helpful.
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Re M02-02 [#permalink]

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New post 10 Feb 2016, 04:55
I think this is a high-quality question and I agree with explanation.
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Re: M02-02 [#permalink]

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New post 08 Mar 2016, 23:45
This is an Excellent Question
here is my approach
here in statement 1 => 1st is a non multiple of 15 and the rest two terms are
hence the result is a non multiple of 15 as we know that a multiple + non multiple is a non multiple.
statement 2 is sufficient as 345 is a multiple of 15 and rest two terms cancel out
So B
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Re: M02-02 [#permalink]

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New post 23 May 2016, 17:08
Thanks for the question, got caught by it :) - was struggling with the (1) condition


stonecold:
stonecold wrote:
This is an Excellent Question
here is my approach
here in statement 1 => 1st is a non multiple of 15 and the rest two terms are
hence the result is a non multiple of 15 as we know that a multiple + non multiple is a non multiple.
statement 2 is sufficient as 345 is a multiple of 15 and rest two terms cancel out
So B


I quite not understand your reason for y^8 is divisible by 15 - the statement only says it's a multiple of 20 - could you explain a little bit more?
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Re: M02-02 [#permalink]

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New post 09 Jan 2018, 21:24
I think the right answer is D.
x16−y8 has the x2+y as a factor which is 645 and divisible by 15 ==> D is correct.
x16−y8 = (x8+y4)(x8-y4)
(x8-y4)=(x4+y2)(x4-y2)
(x4-y2)=(x2+y)(x2-y)
you see my point?
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Re: M02-02 [#permalink]

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New post 09 Jan 2018, 22:10
Java85 wrote:
I think the right answer is D.
x16−y8 has the x2+y as a factor which is 645 and divisible by 15 ==> D is correct.
x16−y8 = (x8+y4)(x8-y4)
(x8-y4)=(x4+y2)(x4-y2)
(x4-y2)=(x2+y)(x2-y)
you see my point?


You are not right. x is a multiple of 25, and y is a multiple of 20 does NOT mean that x^2 + y is a multiple of 645. You could check it VERY easily by picking some numbers: x = 25 any y = 40 --> x^2 + y = 665, which is NOT a multiple of 645. Please re-read the solution above carefully, it explains in detail (with examples) why the first statement is not sufficient.
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Re: M02-02   [#permalink] 09 Jan 2018, 22:10
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