Manhattan Challenge Problem

[This topic has 21 member replies]

Free $100 Amazon.com Gift Card - Buy a GMAT course using a Beat The GMAT discount code between Mar 8-22 and get a $100 Amazon.com Gift Card. Learn more!

Goto page Previous 1, 2

« View previous topic

View next topic »

vinayakdl
Really wants to Beat The GMAT!

Joined: 06 May 2009
Posts: 131

Thanks given: 2
Thanked 5 times in 4 posts
Location: Chicago

Sat Jul 04, 2009 5:53 pm

sanjay_dce wrote: another apporach. total no of possible factors= 144. total number that has 6 as common factor= 81 so total number of factor that doesn't have 6= 144-81 = 63 Can you explain how you got these numbers? Vinayak

chipbmk
Rising GMAT Star

Joined: 05 Oct 2009
Posts: 70

Thanks given: 8
Thanked 0 times in 0 posts

Tue Nov 17, 2009 8:13 am

zuleron wrote: So the primes of 264600 are 2, 2, 2, 3, 3, 3, 5, 5, 7, 7 combinations that don't make 6 are (a) the 2s 5s 7s (b) 3s 5s 7s (a) 2^3 * 5^2 * 7^2 = 36 combinations (b) 3^3 * 5^2 * 7^2 = 36 combinations Can someone please explain how you get 36 combinations from 2^3 * 5^2 * 7^2? I do not get it .... thanks.

Reply from Top Beat The GMAT Member

See All Top Members

papgust
Moderator

Most Thanked Member Most Responsive Member

Joined: 10 Aug 2009
Posts: 874

Thanks given: 19
Thanked 81 times in 78 posts

Target GMAT Score: 700+
GMAT Score: 500

Tue Nov 17, 2009 9:52 pm

chipbmk wrote: zuleron wrote: So the primes of 264600 are 2, 2, 2, 3, 3, 3, 5, 5, 7, 7 combinations that don't make 6 are (a) the 2s 5s 7s (b) 3s 5s 7s (a) 2^3 * 5^2 * 7^2 = 36 combinations (b) 3^3 * 5^2 * 7^2 = 36 combinations Can someone please explain how you get 36 combinations from 2^3 * 5^2 * 7^2? I do not get it .... thanks. There is a shortcut formula basically to calculate the number of integral factors for a given number, if a^p * b^q * c^r ... so on ==> then number of integral factors is (p+1)(q+1)(r+1)... so on Here its, 2^3 * 5^2 * 7^2 take the powers alone and add 1, (3+1) * (2+1) * (2+1) = 36 (This can also be named as combinations) Same applies for (b) as well. Hope this helps..!

Thanked by: chipbmk

chipbmk
Rising GMAT Star

Joined: 05 Oct 2009
Posts: 70

Thanks given: 8
Thanked 0 times in 0 posts

Tue Nov 17, 2009 10:15 pm

papgust wrote: chipbmk wrote: zuleron wrote: So the primes of 264600 are 2, 2, 2, 3, 3, 3, 5, 5, 7, 7 combinations that don't make 6 are (a) the 2s 5s 7s (b) 3s 5s 7s (a) 2^3 * 5^2 * 7^2 = 36 combinations (b) 3^3 * 5^2 * 7^2 = 36 combinations Can someone please explain how you get 36 combinations from 2^3 * 5^2 * 7^2? I do not get it .... thanks. There is a shortcut formula basically to calculate the number of integral factors for a given number, if a^p * b^q * c^r ... so on ==> then number of integral factors is (p+1)(q+1)(r+1)... so on Here its, 2^3 * 5^2 * 7^2 take the powers alone and add 1, (3+1) * (2+1) * (2+1) = 36 (This can also be named as combinations) Same applies for (b) as well. Hope this helps..! That is exactly what I was looking for. Thanks for the explanation!

Nigogo
Just gettin' started!

Joined: 11 Jul 2009
Posts: 13

Thanks given: 1
Thanked 0 times in 0 posts

Wed Nov 18, 2009 8:16 pm

sanjay_dce wrote: another apporach. total no of possible factors= 144. total number that has 6 as common factor= 81 so total number of factor that doesn't have 6= 144-81 = 63 Your approach is very interesting and what I like very laconic, unfortunately i didnt understand how you get 81...

adam15
Rising GMAT Star

Joined: 12 May 2008
Posts: 69

Thanks given: 7
Thanked 3 times in 3 posts

Wed Nov 18, 2009 11:15 pm

the answer is 63 the first set : 5,7,2,2,2 this yields the following set of factors: 5c5+4c5+3c5+2c5+1c5= 6+ 25 the second set without 3 yields, 5,7,3,3,3, again this gives the following factors: 5c5+4c5+3c5+2c5+1c5=2+25 the total is 62 but don't forget 1 which makes 63 in total

sureshbala
GMAT Destroyer!

Joined: 04 Feb 2009
Posts: 318

Thanks given: 1
Thanked 65 times in 54 posts
Location: Delhi

Thu Nov 19, 2009 1:33 am

Number of factors of 264600 = 144 Number of factors of 264600 that are divisible by 6 = 81 Hence number of factors of 264600 which are not divisible by 6 = 144 - 81 = 63 Folks, here is the detailed explanation. Hope this helps...... [/img] _________________ Averages Accelerated: Complete guide to solve Averages Quickly Data Sufficiency for 780+ Aspirants Problem Solving for 780+ Aspirants.