January 22, 2019 January 22, 2019 10:00 PM PST 11:00 PM PST In case you didn’t notice, we recently held the 1st ever GMAT game show and it was awesome! See who won a full GMAT course, and register to the next one. January 26, 2019 January 26, 2019 07:00 AM PST 09:00 AM PST Attend this webinar to learn how to leverage Meaning and Logic to solve the most challenging Sentence Correction Questions.
Author 
Message 
TAGS:

Hide Tags

Math Expert
Joined: 02 Sep 2009
Posts: 52385

Question Stats:
73% (01:15) correct 27% (01:21) wrong based on 167 sessions
HideShow timer Statistics



Math Expert
Joined: 02 Sep 2009
Posts: 52385

Re M0202
[#permalink]
Show Tags
15 Sep 2014, 23:16
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
_________________
New to the Math Forum? Please read this: Ultimate GMAT Quantitative Megathread  All You Need for Quant  PLEASE READ AND FOLLOW: 12 Rules for Posting!!! Resources: GMAT Math Book  Triangles  Polygons  Coordinate Geometry  Factorials  Circles  Number Theory  Remainders; 8. Overlapping Sets  PDF of Math Book; 10. Remainders  GMAT Prep Software Analysis  SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS)  Tricky questions from previous years.
Collection of Questions: PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.
What are GMAT Club Tests? Extrahard Quant Tests with Brilliant Analytics



Retired Moderator
Status: The best is yet to come.....
Joined: 10 Mar 2013
Posts: 498

Re M0202
[#permalink]
Show Tags
07 Jan 2015, 08:28
I think this question is good and helpful.
_________________
Hasan Mahmud



Intern
Joined: 17 Oct 2015
Posts: 21
Concentration: Technology, Leadership

Re M0202
[#permalink]
Show Tags
10 Feb 2016, 04:55
I think this is a highquality question and I agree with explanation.



Current Student
Joined: 12 Aug 2015
Posts: 2626

Re: M0202
[#permalink]
Show Tags
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
_________________
MBA Financing: INDIAN PUBLIC BANKS vs PRODIGY FINANCE! Getting into HOLLYWOOD with an MBA! The MOST AFFORDABLE MBA programs!STONECOLD's BRUTAL Mock Tests for GMATQuant(700+)AVERAGE GRE Scores At The Top Business Schools!



Manager
Joined: 08 Jul 2015
Posts: 56
GPA: 3.8
WE: Project Management (Energy and Utilities)

Re: M0202
[#permalink]
Show Tags
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?
_________________
[4.33] In the end, what would you gain from everlasting remembrance? Absolutely nothing. So what is left worth living for? This alone: justice in thought, goodness in action, speech that cannot deceive, and a disposition glad of whatever comes, welcoming it as necessary, as familiar, as flowing from the same source and fountain as yourself. (Marcus Aurelius)



Intern
Joined: 21 Aug 2013
Posts: 2

Re: M0202
[#permalink]
Show Tags
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)(x8y4) (x8y4)=(x4+y2)(x4y2) (x4y2)=(x2+y)(x2y) you see my point?



Math Expert
Joined: 02 Sep 2009
Posts: 52385

Re: M0202
[#permalink]
Show Tags
09 Jan 2018, 22:10



Intern
Joined: 31 Jan 2016
Posts: 2

Re: M0202
[#permalink]
Show Tags
12 May 2018, 14:54
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.



Intern
Joined: 31 Jan 2016
Posts: 2

Re: M0202
[#permalink]
Show Tags
17 May 2018, 04:49
Hello Bunuel, or other experts, any thoughts on my post above? Thank you for your help. Regards, Paresh



Intern
Joined: 03 Jul 2017
Posts: 1

Re: M0202
[#permalink]
Show Tags
05 Jun 2018, 18:23
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.



Intern
Joined: 18 Jul 2017
Posts: 2

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.



Intern
Joined: 11 Feb 2018
Posts: 2

Re M0202
[#permalink]
Show Tags
28 Sep 2018, 20:25
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



Math Expert
Joined: 02 Sep 2009
Posts: 52385

Re: M0202
[#permalink]
Show Tags
28 Sep 2018, 23:14










