It is currently 22 Feb 2018, 22:52

### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track
Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

# Events & Promotions

###### Events & Promotions in June
Open Detailed Calendar

# M02-02

 new topic post reply Question banks Downloads My Bookmarks Reviews Important topics
Author Message
TAGS:

### Hide Tags

Math Expert
Joined: 02 Sep 2009
Posts: 43889
M02-02 [#permalink]

### Show Tags

15 Sep 2014, 23:16
Expert's post
21
This post was
BOOKMARKED
00:00

Difficulty:

25% (medium)

Question Stats:

73% (01:15) correct 27% (01:21) wrong based on 153 sessions

### HideShow timer Statistics

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) $$y = x^2$$
[Reveal] Spoiler: OA

_________________
Math Expert
Joined: 02 Sep 2009
Posts: 43889
Re M02-02 [#permalink]

### Show Tags

15 Sep 2014, 23:16
Expert's post
5
This post was
BOOKMARKED
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 $$a-b=-3$$, 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
_________________
CR Forum Moderator
Status: The best is yet to come.....
Joined: 10 Mar 2013
Posts: 529
Re M02-02 [#permalink]

### Show Tags

07 Jan 2015, 08:28
I think this question is good and helpful.
_________________

Hasan Mahmud

Current Student
Joined: 17 Oct 2015
Posts: 30
Concentration: Technology, Leadership
Re M02-02 [#permalink]

### Show Tags

10 Feb 2016, 04:55
I think this is a high-quality question and I agree with explanation.
Retired Moderator
Joined: 12 Aug 2015
Posts: 2425
GRE 1: 323 Q169 V154
Re: M02-02 [#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
_________________

Getting into HOLLYWOOD with an MBA

Stone Cold's Mock Tests for GMAT-Quant(700+)

Manager
Joined: 08 Jul 2015
Posts: 58
GPA: 3.8
WE: Project Management (Energy and Utilities)
Re: M02-02 [#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: M02-02 [#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)(x8-y4)
(x8-y4)=(x4+y2)(x4-y2)
(x4-y2)=(x2+y)(x2-y)
you see my point?
Math Expert
Joined: 02 Sep 2009
Posts: 43889
Re: M02-02 [#permalink]

### Show Tags

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.
_________________
Re: M02-02   [#permalink] 09 Jan 2018, 22:10
Display posts from previous: Sort by

# M02-02

 new topic post reply Question banks Downloads My Bookmarks Reviews Important topics

Moderators: chetan2u, Bunuel

 Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®.