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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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16 Sep 2014, 00:16
Official Solution: 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 \(ab=3\), 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 \(ab=1\), 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 \(ab=1\), is not divisible by 3 OR: \(a=6\) and \(b=3\), neither is divisible by 5: \(a+b=9\) and \(ab=3\), neither is divisible by 5 OR: \(a=2\) and \(b=2\), neither is divisible by 4: \(a+b=4\) and \(ab=0\), 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 M0202
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07 Jan 2015, 09:28
I think this question is good and helpful.
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10 Feb 2016, 05:55
I think this is a highquality question and I agree with explanation.



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Re: M0202
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09 Mar 2016, 00:45
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: M0202
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23 May 2016, 18:08
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: M0202
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09 Jan 2018, 22:24
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)(x8y4) (x8y4)=(x4+y2)(x4y2) (x4y2)=(x2+y)(x2y) you see my point?



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Re: M0202
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09 Jan 2018, 23:10
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)(x8y4) (x8y4)=(x4+y2)(x4y2) (x4y2)=(x2+y)(x2y) 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 reread the solution above carefully, it explains in detail (with examples) why the first statement is not sufficient.
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Re: M0202
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12 May 2018, 15:54
Hello Bunuel, I approached this question using the remainder theorem. According to the theorem, the remainder of the expression [AxBxC]/M is the same as the remainder of expression [ArxBrxCr]/M (where Ar, Br, and Cr are the respective remainders when A, B, and C are divided by M. (For example, remainder of [16x19]/15 == remainder [1x3]/15 = 3). Please let me know if you agree with this. If so, then taking x=25 and y=40, we get : Rem [(25^16 – 40^8)]/15 == [10^16 – 10^8]/15 (because 25/15 gives a remainder of 10, and 40/15 gives a remainder of 10) = [100^8 – 10^8]/15 (rewriting 10^16 as 100^8 ) = [10^8 – 10^8]/15 (remainder of 100/15 = 10) = 0 Therefore, if we take x=25, and y =40, the expression IS divisible by 15. Similarly, I checked the value of expression \(x^16  y^8\) for different values of (x,y) such as (50,40), (25,20), (50,20), etc. and got the remainder to be 0. Please let me know where am I going wrong. Thank you in advance.



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Re: M0202
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17 May 2018, 05:49
Hello Bunuel, or other experts, any thoughts on my post above? Thank you for your help. Regards, Paresh



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Re: M0202
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05 Jun 2018, 19:23
prajpurohit wrote: I approached this question using the remainder theorem. According to the theorem, the remainder of the expression [AxBxC]/M is the same as the remainder of expression [ArxBrxCr]/M (where Ar, Br, and Cr are the respective remainders when A, B, and C are divided by M. (For example, remainder of [16x19]/15 == remainder [1x3]/15 = 3). Please let me know if you agree with this.
If so, then taking x=25 and y=40, we get :
Rem [(25^16 – 40^8)]/15 == [10^16 – 10^8]/15 (because 25/15 gives a remainder of 10, and 40/15 gives a remainder of 10) = [100^8 – 10^8]/15 (rewriting 10^16 as 100^8 ) = [10^8 – 10^8]/15 (remainder of 100/15 = 10) = 0
Therefore, if we take x=25, and y =40, the expression IS divisible by 15.
Similarly, I checked the value of expression \(x^16  y^8\) for different values of (x,y) such as (50,40), (25,20), (50,20), etc. and got the remainder to be 0.
Please let me know where am I going wrong.
Thank you in advance. prajpurohit I think this is an interesting thought. To use the theorem as you outlined, however, wouldn't the expression need to be in the form A*B*C instead of the form ABC? It does not seem that the theorem would be applicable to the expression in the question because it is subtraction, not multiplication.



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Bunuel wrote: 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) \(y = x^2\) Another way of solving this question. Following are a few divisibility rules.
1 a^n  b^n is divisible by (a+b) if n = even
2 a^n  b^n is divisible by (ab) if n = even/odd
3 a^n + b^n is divisible by (a+b) if n = oddNow, \(x^{16}  y^8 + 345y^2\) can be written as (x^2)^8  y^8. Im not including 345y^2 as it is divisible by 15. (1) \(x\) is a multiple of 25, and \(y\) is a multiple of 20 If a= 25; b= 20. Then a+b = 45. Then x^{16}  y^8 is divisible by 15 But if a = 50; b = 60, Then a+b = 110, So x^{16}  y^8 is NOT divisible by 15. Hence, NOT SUFFICIENT (2) Substituting y = x^2. We get x^16  x^16 = 0. 0 is divisible by 15. Hence answer is B Hope this helps.



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28 Sep 2018, 21:25
I think this is a highquality question and I don't agree with the explanation. using x^2Y^2 = (x+y)(xy) multiple times we get 25^322^16 as 5^8((5^3+2^2)(5^32^2)(5^6+2^4)(5^12+2^8))
that gives us a multiple of 5 multipled with 129 which is a multiple of 3
hence a multiple of 15. thus 1 is also sufficient



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Re: M0202
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29 Sep 2018, 00:14
Chinmoy25 wrote: I think this is a highquality question and I don't agree with the explanation. using x^2Y^2 = (x+y)(xy) multiple times we get 25^322^16 as 5^8((5^3+2^2)(5^32^2)(5^6+2^4)(5^12+2^8))
that gives us a multiple of 5 multipled with 129 which is a multiple of 3
hence a multiple of 15. thus 1 is also sufficient If x=25∗15 and y=20, then x^16 −y^8 + 345y^2 = 4304672099999999999999999999974400138000, which is NOT a multiple of 15. So, (1) is not sufficient.
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Bunuel is this sound logic? (1) \(x\) is a multiple of 25, and \(y\) is a multiple of 20If we rewrite ... x=25^n and y=20^m (25^n)^16  (20^m)^8 + 345y*(20^m)^2 We know all 3 are divisible by 5. We only know for sure the 3rd term is divisible by 3, so it's not sufficient. This depends on n and m (e.g. n=1, m=1 is no but n=3, m=3 is yes) (2) \(y = x^2\)x^16  x^16 + 345x^4 Since 345 is divisible by 15, the product of 345 and x^4 is as well; sufficient.










