Go Back   TestMagic Forums > Test preparation > GMAT > GMAT Math > GMAT Data Sufficiency
Register Forum Rules FAQ Members List Calendar Search Today's Posts Mark Forums Read

Reply
 
LinkBack Thread Tools Search this Thread Display Modes
Old 2007 November 28th, 11:57 PM   #1 (permalink)
o_O
 
Join Date: Jul 2007
Posts: 165
jcmsolis just joined TestMagic.
Divisibility problem from Manhattan

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


(A) Statement (1) alone is sufficient, but statement (2) alone is not sufficient.
(B) Statement (2) alone is sufficient, but statement (1) alone is not sufficient.
(C) BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.
(D) Each statement ALONE is sufficient.
(E) Statements (1) and (2) TOGETHER are NOT sufficient.
jcmsolis is offline  
Digg this Post!Add Post to del.icio.usBookmark Post in TechnoratiFurl this Post!Google Bookmark this Post!Reddit!
Reply With Quote
Old 2007 November 29th, 04:36 AM   #2 (permalink)
TestMagic Guru-in-Training
 
DWarrior's Avatar
 
Join Date: Oct 2007
Location: New Jersey
Posts: 642
DWarrior 's dreams are becoming reality.
Statement 1:
Factors into k*m*(m^2-1)=k*m(m-1)(m+1). The factors will always be (m-1), m, and (m+1), meaning one of them will be divisible by 3. I don't see any way to determine if it's divisible by 10.

I can show that Statement 2 will always be divisible by 10:
We want the last digit to be divisible by 10. The last digit of n^5 will always be its first digit, then when you subtract n from that, the result's last digit will always be 0, thus it will be divisible by 10:
last digit of n=1: 1, 1, 1, 1, 1
last digit of n=2: 2, 4, 8, 6, 2
last digit of n=3: 3, 9, 7, 1, 3
last digit of n=4: 4, 6, 4, 6, 4
last digit of n=5: 5, 5, 5, 5, 5
last digit of n=6: 6, 6, 6, 6, 6
last digit of n=7: 7, 9, 3, 1, 7
last digit of n=8: 8, 4, 2, 6, 8
last digit of n=9: 9, 1, 9, 1, 9

We can also factor the right side of the equation: n(n^4-1), which can be further broken into: n(n^2-1)(n^2+1)=n(n-1)(n+1)(n^2+1). Once again, it's divisible by 3 because n, (n-1), and (n+1) are all factors, one of which must be divisible by 3. Maybe this can somehow be used to prove divisibility by 10, but I don't see it.

Thus, 2 will be sufficient.

B

PS, are Manhattan problems supposed to be doable in 2 minutes? I doubt if there are more than 10 people in the world who can solve this one in 2 minutes.
_ _ _ _ SIG _ _ _ _
If you have questions about my solutions, PM me.

Some gre/gmat stuff (mostly Math):
My del.icio.us bookmarks
My Clipmarks bookmarks
DWarrior is offline  
Digg this Post!Add Post to del.icio.usBookmark Post in TechnoratiFurl this Post!Google Bookmark this Post!Reddit!
Reply With Quote
Old 2007 November 29th, 12:37 PM   #3 (permalink)
TestMagic Guru-in-Training
 
Join Date: Aug 2007
Posts: 674
krusta80 's dreams are becoming reality.
Quote:
Originally Posted by jcmsolis View Post
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

(A) Statement (1) alone is sufficient, but statement (2) alone is not sufficient.
(B) Statement (2) alone is sufficient, but statement (1) alone is not sufficient.
(C) BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.
(D) Each statement ALONE is sufficient.
(E) Statements (1) and (2) TOGETHER are NOT sufficient.
x mod 30 = 0?

(1) x = k(m^3 - m), where m and k are both integers > 9

k*m*(m - 1)*(m+1) mod 30 = 0?

k doesn't really help us here. It could be anything from a multiple of 30 to a prime number like 11.

30 = 6*5 = 2*3*5

From m*(m-1)*(m+1), we know that we have a multiple of 2 and a multiple of 3. But there is nothing to indicate divisibility by 5. INSUFF

(2) x = n^5 - n, where n is an integer > 9

n^5 - n mod 30 = 0?

n*(n^2 - 1)*(n^2 + 1) mod 30 = 0?

n*(n+1)*(n-1)*(n^2+1) mod 30 = 0?

Again, we know that 2 and 3 are factors...let's take a look at n^2+1:

n^2 + 1 mod 5 = (n^2 mod 5 + 1) mod 5 = [(n mod 5)^2 mod 5 + 1] mod 5

Now we'll quickly look at all possibilities for n mod 5:

n mod 5 = 0 --> done
n mod 5 = 1 --> (n-1) mod 5 = 0 --> done
n mod 5 = 2 --> n^2+1 mod 5 = 0 --> done
n mod 5 = 3 --> n^2+1 mod 5 = 0 --> done
n mod 5 = 4 = -1 --> (n+1) mod 5 = 0 --> done

SUFFICIENT!

B
krusta80 is offline  
Digg this Post!Add Post to del.icio.usBookmark Post in TechnoratiFurl this Post!Google Bookmark this Post!Reddit!
Reply With Quote
Old 2007 November 29th, 04:17 PM   #4 (permalink)
TestMagic Guru-in-Training
 
DWarrior's Avatar
 
Join Date: Oct 2007
Location: New Jersey
Posts: 642
DWarrior 's dreams are becoming reality.
Your solution implies that numbers multiplied together get their remainders multiplied.

Is there any guide to basic modular arithmetic? I've never had it in school, and someone told me it's a topic of abstract algebra, so I didn't even bother.
_ _ _ _ SIG _ _ _ _
If you have questions about my solutions, PM me.

Some gre/gmat stuff (mostly Math):
My del.icio.us bookmarks
My Clipmarks bookmarks
DWarrior is offline  
Digg this Post!Add Post to del.icio.usBookmark Post in TechnoratiFurl this Post!Google Bookmark this Post!Reddit!
Reply With Quote
Old 2007 November 29th, 09:18 PM   #5 (permalink)
TestMagic Guru-in-Training
 
Join Date: Aug 2007
Posts: 674
krusta80 's dreams are becoming reality.
Quote:
Originally Posted by DWarrior View Post
Your solution implies that numbers multiplied together get their remainders multiplied.

Is there any guide to basic modular arithmetic? I've never had it in school, and someone told me it's a topic of abstract algebra, so I didn't even bother.
I came up with these myself (with "proofs" no less). I will follow up with some details soon, but here are the rules for mods:

(a+b) mod c = [(a mod c) + (b mod c)] mod c
(a*b) mod c = [(a mod c) * (b mod c)] mod c
(a^b) mod c = [(a mod c)^b] mod c

Cool, huh?
krusta80 is offline  
Digg this Post!Add Post to del.icio.usBookmark Post in TechnoratiFurl this Post!Google Bookmark this Post!Reddit!
Reply With Quote
Old 2007 November 29th, 10:52 PM   #6 (permalink)
TestMagic Guru-in-Training
 
DWarrior's Avatar
 
Join Date: Oct 2007
Location: New Jersey
Posts: 642
DWarrior 's dreams are becoming reality.
Damn, that's sick. I figured out the "remainders add" (your first part) by doing some problems here, but kudos on the other 2.
_ _ _ _ SIG _ _ _ _
If you have questions about my solutions, PM me.

Some gre/gmat stuff (mostly Math):
My del.icio.us bookmarks
My Clipmarks bookmarks
DWarrior is offline  
Digg this Post!Add Post to del.icio.usBookmark Post in TechnoratiFurl this Post!Google Bookmark this Post!Reddit!
Reply With Quote
Old 2007 November 30th, 03:00 PM   #7 (permalink)
Within my grasp!
 
Join Date: Sep 2005
Posts: 364
kevinspain is on the way!
Quote:
Originally Posted by jcmsolis View Post
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


(A) Statement (1) alone is sufficient, but statement (2) alone is not sufficient.
(B) Statement (2) alone is sufficient, but statement (1) alone is not sufficient.
(C) BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.
(D) Each statement ALONE is sufficient.
(E) Statements (1) and (2) TOGETHER are NOT sufficient.
Note that x=(n-1)n(n+1)(n^2+1), a multiple of 2 and 3 (as n - 1, n and n+1 are three consecutive integers). x is not a multiple of 30 if and only if x is not a multiple of 5. If none of the three consecutive integers n-1,n,n+1 is a multple of 5, n= 5m + 2 or n= 5m + 3, (i.e. either 2 or 3 greater than a multiple of 5). In either case, n^2 + 1 is a multiple of 5.
_ _ _ _ SIG _ _ _ _
Kevin Armstrong
GMAT Instructor
Manhattan Review
kevinspain is offline  
Digg this Post!Add Post to del.icio.usBookmark Post in TechnoratiFurl this Post!Google Bookmark this Post!Reddit!
Reply With Quote
Old 2009 January 3rd, 07:03 AM   #8 (permalink)
Within my grasp!
 
Join Date: Aug 2008
Posts: 475
hina_0611 just joined TestMagic.
let me try another way to prove n^5 - n Is devisible by 30:

n^5 -n = n(n-1)(n+1)( n^2 +1)= n(n-1)(n+1)(n^2 - 4 +5)
= (n-2)(n-1)n(n+1)(n+2) + 5(n-1)n(n+1)
product of 5 sucessive intergers must be divisible by 5,3,2 => divisible by 30
Product of 3 sucessive intergers must divisible by 3 and 2 => divisible by 6

=> n^5 -n divisible by 30
hina_0611 is offline  
Digg this Post!Add Post to del.icio.usBookmark Post in TechnoratiFurl this Post!Google Bookmark this Post!Reddit!
Reply With Quote
Old 2009 October 11th, 03:38 PM   #9 (permalink)
560 looser
 
mr.sanjogm's Avatar
 
Join Date: Oct 2009
Posts: 124
mr.sanjogm just joined TestMagic.
I picked numbers and got B . I took one even and odd number greater than 9 to test the choices.....but kevin's approach is bow wow yippee yay
mr.sanjogm is offline  
Digg this Post!Add Post to del.icio.usBookmark Post in TechnoratiFurl this Post!Google Bookmark this Post!Reddit!
Reply With Quote
Reply


Thread Tools Search this Thread
Search this Thread:

Advanced Search
Display Modes

What you can do
You cannot post new threads
You cannot post replies
You cannot post attachments
You cannot edit your posts

vB code is On
Smilies are On
[IMG] code is On
HTML code is Off
Trackbacks are On
Pingbacks are On
Refbacks are On


All times are GMT. The time now is 03:25 PM.

Contact TestMagic   TestMagic Forums      Archive   Privacy Statement

TestMagic Locations   Legal   Privacy


SEO by vBSEO 3.2.0
Copyright © 2010 TestMagic
Ad Management by RedTyger

Scroll Up