It is currently 23 Sep 2017, 09:55

# Happening Now:

Alleviate MBA app anxiety! Come to Chat Room #2

### 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 June
Open Detailed Calendar

# If x and y are positive integers, is x^16 - y^8 + 345y^2

 post reply Question banks Downloads My Bookmarks Reviews Important topics
Author Message
Intern
Joined: 30 May 2008
Posts: 4

Kudos [?]: 4 [0], given: 0

If x and y are positive integers, is x^16 - y^8 + 345y^2 [#permalink]

### Show Tags

13 Jul 2009, 02:23
00:00

Difficulty:

(N/A)

Question Stats:

0% (00:00) correct 0% (00:00) wrong based on 0 sessions

### HideShow timer Statistics

This topic is locked. If you want to discuss this question please re-post it in the respective forum.

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

Statement (1) by itself is not sufficient. For a number to be a multiple of 15, it has to be divisible by both 3 and 5. As both 25 and 20 are multiples of 5, we need to check if $$x^{16} - y^8$$ is divisible by 3. $$x^{16} - y^8$$ can be rewritten as $$(x^2 - y)(x^2 + y)(x^4 + y^2)(x^8 + y^4)$$ . We only need to check if expressions from first two parentheses are divisible by 3. We'll pick 50 and 60 for that purpose:
$$50^2 - 60 = 2500 - 60 = 2440$$
$$50^2 + 60 = 2500 + 60 = 2560$$

Neither of these numbers is divisible by 3. Therefore, $$x^{16} - y^8$$ isn't divisible by 3 (neither by 15) either.

Based on the given explaination, I've the following questions:

1) How do we actually get x^16 - y^8 to (x^2 - y)(x^2 + y)(x^4 + y^2)(x^8 + y^4)?
2) Why do we only need to check if the first two expressions in parentheses are divisible by 3?

Thanks!

Kudos [?]: 4 [0], given: 0

Director
Joined: 03 Jun 2009
Posts: 785

Kudos [?]: 884 [0], given: 56

Location: New Delhi
WE 1: 5.5 yrs in IT
Re: gmatClub test 2, Qu2. [#permalink]

### Show Tags

13 Jul 2009, 02:44
sudimba wrote:
1) How do we actually get x^16 - y^8 to (x^2 - y)(x^2 + y)(x^4 + y^2)(x^8 + y^4)?

Solving backwards:
$$(x^2 - y)(x^2 + y)(x^4 + y^2)(x^8 + y^4)$$
...merging the 1st two using the formula $$(a+b)(a-b) = (a^2 - b^2)$$
$$= (x^4 - y^2)(x^4 + y^2)(x^8 + y^4)$$
$$= (x^8 - y^4)(x^8 + y^4)$$
$$=(x^16 - y^8)$$

sudimba wrote:
2) Why do we only need to check if the first two expressions in parentheses are divisible

Not sure
_________________

Kudos [?]: 884 [0], given: 56

Current Student
Joined: 20 Jun 2009
Posts: 168

Kudos [?]: 38 [0], given: 0

Schools: Yale SOM Class of 2013
Re: gmatClub test 2, Qu2. [#permalink]

### Show Tags

13 Jul 2009, 07:18
Curious to know the answer to this question. As I generally shy away from factoring complex expressions such as x^16 - y^8, my approach (which may very well be wrong) is more rudimentary:

For x^16 - y^8 + 345y^2to be divisible by 15, each of the terms in the expression must be divisible by 15.

Statement 1 is insufficient. Both x and y could be multiples of 15 (if, for example, x=75 and y=60), but we don't have enough information to know whether this is the case.

Statement 2 is sufficient. If y=x^2, we can simplify the expression as follows:

x^16 - (x^2)^8 + 345y^2
= x^16 - x^16 + 345y^2
= 345y^2

We know that 345y^2 is always divisible by 15 because 345 is divisible by both 3 and 5 and multiplying any number by a multiple of 15 yields a result that is divisible by 15. Thus, x^16 - y^8 + 345y^2 is divisible by 15 where y=x^2. The answer is B.

Last edited by carriedinterest on 13 Jul 2009, 08:02, edited 2 times in total.

Kudos [?]: 38 [0], given: 0

Director
Joined: 03 Jun 2009
Posts: 785

Kudos [?]: 884 [0], given: 56

Location: New Delhi
WE 1: 5.5 yrs in IT
Re: gmatClub test 2, Qu2. [#permalink]

### Show Tags

13 Jul 2009, 22:24
carriedinterest wrote:
For x^16 - y^8 + 345y^2to be divisible by 15, each of the terms in the expression must be divisible by 15.

Sorry to contradict you. But I don't agree on this.

Since 345y^2 is already divisible by 15, I would put it this way that for x^16 - y^8 + 345y^2 to be divisible by 15, we will need to prove (x^16 - y^8) is also divisible by 15.

Note: each of these terms may not be divisible by 15 separately, but may be divisible jointly.

e.g x = 25 and y = 20

Now from one of the previous calculations (posted in previous post)
(x^2 + y) is factor of (x^16 - y^8)

(x^2 + y) = 25^2 + 20 = 625 + 20 = 645, which is divisible by 15

This is just one case. If we can prove that for some other case, its not divisible, we can say option A is insufficient.
_________________

Kudos [?]: 884 [0], given: 56

Re: gmatClub test 2, Qu2.   [#permalink] 13 Jul 2009, 22:24
Display posts from previous: Sort by

# If x and y are positive integers, is x^16 - y^8 + 345y^2

 post reply Question banks Downloads My Bookmarks Reviews Important topics

 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®.