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

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

It appears that you are browsing the GMAT Club forum unregistered!

Signing up is free, quick, and confidential.
Join other 350,000 members and get the full benefits of GMAT Club

Registration gives you:

Tests

Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.

Applicant Stats

View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more

Books/Downloads

Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

Re: Hard divisibility DS [#permalink]
27 Sep 2006, 15:29

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, 18:29, edited 1 time in total.

Re: Hard divisibility DS [#permalink]
29 Sep 2006, 00:07

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.

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.

This week went in reviewing all the topics that I have covered in my previous study session. I reviewed all the notes that I have made and started reviewing the Quant...

I was checking my phone all day. I wasn’t sure when I would receive the admission decision from Tepper. I received an acceptance from Goizueta in the early morning...

I started running as a cross country team member since highshcool and what’s really awesome about running is that...you never get bored of it! I participated in...