It is currently 25 Jun 2017, 23:07

### 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

# Is x divisible by 30? (1) X = k*(m^3 - m), where m and k are

 post reply Question banks Downloads My Bookmarks Reviews Important topics
Author Message
SVP
Joined: 05 Jul 2006
Posts: 1747
Is x divisible by 30? (1) X = k*(m^3 - m), where m and k are [#permalink]

### Show Tags

27 Sep 2006, 15:14
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.

Is x divisible by 30?

(1) X = k*(m^3 - m), where m and k are both integers > 9
(2) X = n^5 - n, where n is an integer > 9
Intern
Joined: 25 Sep 2006
Posts: 23
Re: Hard divisibility DS [#permalink]

### Show Tags

27 Sep 2006, 16:29
yezz wrote:
Is x divisible by 30?

(1) X = k*(m^3 - m), where m and k are both integers > 9
(2) X = n^5 - n, where n is an integer > 9

B
factors of 30 = 2*3*5

1
k*(m^3 - m) = k*m*(m^2-1) = k*m*(m+1)*(m-1)
any 3 consecutive integers > 9 will allow for divisibility by 2 and 3
but they may or maynot allow for for divisibility by 5
for eg,
m=21, k= 13 do not allow for divisibility by 5
m=23, k= 24 allows for divisibility by 5

insuffi

2
x = n^5 - n = n*(n^4-1) = n*(n^2+1)*(n+1)*(n-1)
any 3 consecutive integers > 9 will allow for divisibility by 2 and 3
but they may or maynot allow for for divisibility by 5,

the new expression here is (n^2+1)
for n= 11 , (n^2+1) = 122 is not divisible by 5, but 10=n-1 is
for n= 12 , (n^2+1) =145 is divisible by 5

suffi

Last edited by muddypaws on 27 Sep 2006, 19:29, edited 1 time in total.
Manager
Joined: 15 Aug 2006
Posts: 54

### Show Tags

27 Sep 2006, 16:45
One more for E....good explanation muddypaws.
Manager
Joined: 28 Aug 2006
Posts: 243
Location: Albuquerque, NM

### Show Tags

27 Sep 2006, 18:57
Answer is B, the second condition is always satisfied for n > 9

and that is because of term n^2 + 1

since if the n leaves a remainder of 2 or 3 when divided by 5, then n-1 or n+1 is not divisible by 5 otherwise they are

If n leaves a remainder of 1 then n-1 is divisible by 5 and if n leaves remainder of 4 then n +1 is divisible by 5

for cases n leaves remainder of 2 and 3 the unit digit could be 2, 3, 7, 8

and n^2 + 1 will have last digit of (2X2) +1 , (3X3) +1, (7X7) +1, (8X8) +1

all of them divisible by 5

thus n^5 - n is divisible by 30
Current Student
Joined: 29 Jan 2005
Posts: 5218

### Show Tags

27 Sep 2006, 19:30
jainan24 wrote:
Answer is B, the second condition is always satisfied for n > 9

and that is because of term n^2 + 1

since if the n leaves a remainder of 2 or 3 when divided by 5, then n-1 or n+1 is not divisible by 5 otherwise they are

If n leaves a remainder of 1 then n-1 is divisible by 5 and if n leaves remainder of 4 then n +1 is divisible by 5

for cases n leaves remainder of 2 and 3 the unit digit could be 2, 3, 7, 8

and n^2 + 1 will have last digit of (2X2) +1 , (3X3) +1, (7X7) +1, (8X8) +1

all of them divisible by 5

thus n^5 - n is divisible by 30

Good explanation. Indeed the answer should be (B)
Manager
Joined: 15 Aug 2006
Posts: 54

### Show Tags

28 Sep 2006, 10:13
Opps sorry...perfect example for careless error.
Senior Manager
Joined: 28 Aug 2006
Posts: 304
Re: Hard divisibility DS [#permalink]

### Show Tags

29 Sep 2006, 01:07
yezz wrote:
Is x divisible by 30?

(1) X = k*(m^3 - m), where m and k are both integers > 9
(2) X = n^5 - n, where n is an integer > 9

statement 1: X= K*(m^3-m)

i.e X= K*(m-1)(m)(m+1).

Product of any n consecutive positive numbers is always divisible by n!.

So clearly (m-1)(m)(m+1) is divisible by 3! =6
Hence X is divisible by 6.
We are not sure whether x is divisible by 5 or not.
So 1 is not sufficient.

Statement 2: X= n^5-n
This will always end with 0.
So X is clearly divisible by 5.
X=n^5-n = n(n^4-1) = n[((n^2)^2) - 1^2]
ie X= n(n^2+1)(n^2-1)
X= n(n^2+1)(n+1)(n-1)
X= (n-1)(n)(n+1)(n^2+1)
Clearly X is divisible by 6.

So X must be divisibly by 30.

Hence B
_________________

Last edited by cicerone on 25 Sep 2008, 01:46, edited 1 time in total.
Director
Joined: 02 Mar 2006
Posts: 575
Location: France

### Show Tags

23 Oct 2006, 07:11
GMATT73 wrote:
jainan24 wrote:
Answer is B, the second condition is always satisfied for n > 9

and that is because of term n^2 + 1

since if the n leaves a remainder of 2 or 3 when divided by 5, then n-1 or n+1 is not divisible by 5 otherwise they are

If n leaves a remainder of 1 then n-1 is divisible by 5 and if n leaves remainder of 4 then n +1 is divisible by 5

for cases n leaves remainder of 2 and 3 the unit digit could be 2, 3, 7, 8

and n^2 + 1 will have last digit of (2X2) +1 , (3X3) +1, (7X7) +1, (8X8) +1

all of them divisible by 5

thus n^5 - n is divisible by 30

Good explanation. Indeed the answer should be (B)

Can please someone explain! I don't uderstand at all!
Director
Joined: 06 Sep 2006
Posts: 736
Re: Hard divisibility DS [#permalink]

### Show Tags

23 Oct 2006, 08:34
cicerone wrote:
yezz wrote:
Is x divisible by 30?

X= n(n^2+1)(n+1)(n-1)
X= (n-1)(n)(n+1)(n^2+1)
Clearly X is divisible by 6.

So X must be divisibly by 30.

Hence B

X= (n-1)(n)(n+1)(n^2+1)
Clearly X is divisible by 6.

So X must be divisibly by 30
.

Dude, You wana expand on 'X must be divisible by 30? I understand 3 consecutive numbers are always divisble by 6.
SVP
Joined: 03 Jan 2005
Posts: 2233

### Show Tags

23 Oct 2006, 09:21
N^5-n=n(n^2+1)(n^2-1)=n(n^2+1)(n+1)(n-1)
We already know n(n+1)(n-1) is divisible by 6, just need to know if the whole thing is divisible by 5.

if n=5k then it is divisible by 5.
if n=5k+1, n^2-1=25k^2+10k, divisible by 5.
if n=5k-1, n^2-1=25k^2-10k, divisible by 5.
if n=5k+2, n^2+1=25k^2+20k+5, divisible by 5.
if n=5k-2, n^2+1=25k^2-20k+5, divisible by 5.

I went the long way to find out that it is also divisible by 5. Cicerone, can you explain why n^5-n is always multiples of 10? Did you have to go through this long derivation or if there's a more straight forward way that we can see the result immediately?
_________________

Keep on asking, and it will be given you;
keep on seeking, and you will find;
keep on knocking, and it will be opened to you.

23 Oct 2006, 09:21
Display posts from previous: Sort by

# Is x divisible by 30? (1) X = k*(m^3 - m), where m and k are

 post reply Question banks Downloads My Bookmarks Reviews Important topics

 Powered by phpBB © phpBB Group and phpBB SEO 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®.