GMAT Question of the Day - Daily to your Mailbox; hard ones only

 It is currently 22 Jan 2019, 07:41

### 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

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 January
PrevNext
SuMoTuWeThFrSa
303112345
6789101112
13141516171819
20212223242526
272829303112
Open Detailed Calendar
• ### The winners of the GMAT game show

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.
• ### Key Strategies to Master GMAT SC

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.

# M02-02

Author Message
TAGS:

### Hide Tags

Math Expert
Joined: 02 Sep 2009
Posts: 52385

### Show Tags

15 Sep 2014, 23:16
1
12
00:00

Difficulty:

25% (medium)

Question Stats:

73% (01:15) correct 27% (01:21) wrong based on 167 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) $$y = x^2$$

_________________
Math Expert
Joined: 02 Sep 2009
Posts: 52385

### Show Tags

15 Sep 2014, 23:16
1
6
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.

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

### Show Tags

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

Hasan Mahmud

Intern
Joined: 17 Oct 2015
Posts: 21

### Show Tags

10 Feb 2016, 04:55
I think this is a high-quality question and I agree with explanation.
Current Student
Joined: 12 Aug 2015
Posts: 2626
Schools: Boston U '20 (M)
GRE 1: Q169 V154

### 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 GMAT-Quant(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)

### 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

### 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: 52385

### 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.
_________________
Intern
Joined: 31 Jan 2016
Posts: 2

### 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.

Intern
Joined: 31 Jan 2016
Posts: 2

### Show Tags

17 May 2018, 04:49
Hello Bunuel, or other experts, any thoughts on my post above?

Regards,
Paresh
Intern
Joined: 03 Jul 2017
Posts: 1

### 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.

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 A-B-C? 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

### Show Tags

04 Jul 2018, 01:55
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 (a-b) if n = even/odd

3 a^n + b^n is divisible by (a+b) if n = odd

Now, $$x^{16} - y^8 + 345y^2$$ 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.

Hope this helps.
Intern
Joined: 11 Feb 2018
Posts: 2

### Show Tags

28 Sep 2018, 20:25
I think this is a high-quality question and I don't agree with the explanation. using x^2-Y^2 = (x+y)(x-y) multiple times we get 25^32-2^16 as 5^8((5^3+2^2)(5^3-2^2)(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

### Show Tags

28 Sep 2018, 23:14
Chinmoy25 wrote:
I think this is a high-quality question and I don't agree with the explanation. using x^2-Y^2 = (x+y)(x-y) multiple times we get 25^32-2^16 as 5^8((5^3+2^2)(5^3-2^2)(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.
_________________
Re: M02-02 &nbs [#permalink] 28 Sep 2018, 23:14
Display posts from previous: Sort by

# M02-02

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®.