Find all School-related info fast with the new School-Specific MBA Forum

It is currently 23 Jul 2014, 08:00

Close

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
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.

Events & Promotions

Events & Promotions in June
Open Detailed Calendar

2^X)*(3^Y)*(5^Z) is the greatest positive divisor of (8!)^4.

  Question banks Downloads My Bookmarks Reviews Important topics  
Author Message
TAGS:
SVP
SVP
User avatar
Joined: 05 Jul 2006
Posts: 1542
Followers: 5

Kudos [?]: 65 [0], given: 39

2^X)*(3^Y)*(5^Z) is the greatest positive divisor of (8!)^4. [#permalink] New post 12 Oct 2006, 15:49
2^X)*(3^Y)*(5^Z) is the greatest positive divisor of (8!)^4. Find the product XYZ
Manager
Manager
avatar
Joined: 28 Aug 2006
Posts: 160
Followers: 1

Kudos [?]: 10 [0], given: 0

GMAT Tests User
 [#permalink] New post 12 Oct 2006, 16:09
8! ^4 is (8.7.6.5.4.3.2.1)^4
=(2^7.3^2.5.7)^4=>2^x.3^y.5^z
=xyz=7.2.1.4.=56

Answer is 56
Senior Manager
Senior Manager
User avatar
Joined: 31 Jul 2006
Posts: 440
Followers: 3

Kudos [?]: 6 [0], given: 0

GMAT Tests User
 [#permalink] New post 12 Oct 2006, 16:35
vijay2001 wrote:
8! ^4 is (8.7.6.5.4.3.2.1)^4
=(2^7.3^2.5.7)^4=>2^x.3^y.5^z
=xyz=7.2.1.4.=56

Answer is 56


So I see that

2^7 => 2^x
3^2 => 3^y
5^1 => 5^z

but how did you get 7^1 => 4?

Thx.
VP
VP
User avatar
Joined: 25 Jun 2006
Posts: 1176
Followers: 2

Kudos [?]: 31 [0], given: 0

GMAT Tests User
 [#permalink] New post 12 Oct 2006, 17:48
No.

I got 896.

(8!)^4 = 2^28*3^8*5^4*7^4

so xyz = 28*8*4
Senior Manager
Senior Manager
User avatar
Joined: 31 Jul 2006
Posts: 440
Followers: 3

Kudos [?]: 6 [0], given: 0

GMAT Tests User
 [#permalink] New post 12 Oct 2006, 17:56
tennis_ball wrote:
No.

I got 896.

(8!)^4 = 2^28*3^8*5^4*7^4

so xyz = 28*8*4


I understand now. he multiplied 7^1*4 but forgot to multiply powers of other factors by 4
Director
Director
avatar
Joined: 23 Jun 2005
Posts: 847
GMAT 1: 740 Q48 V42
Followers: 3

Kudos [?]: 21 [0], given: 1

GMAT Tests User
 [#permalink] New post 12 Oct 2006, 18:04
Ans 896.

8! = (2^3)*7*(2*3)*5*(2^2)*3*2 = 7*5*(3^2)*(2*7)

So, (8!)^4= (2^28)*(3^8)*(5^4)*(7^4)

xyz = 28*8*4 = 896
Senior Manager
Senior Manager
User avatar
Joined: 28 Aug 2006
Posts: 302
Followers: 10

Kudos [?]: 72 [0], given: 0

GMAT Tests User
 [#permalink] New post 13 Oct 2006, 13:37
Folks, i have already explained u earlier how to calculate the highest power of any prime number in a factorial .Please use that and save time in the exam.............

Anyway, i am repeating it again........
Let's calculate highest power of 11 in 274!

274/11 = 24 (don't worry about the remainder)
24/11 = 2
Add up now 24+2 = 26 will be the highest power of 11 in 274!

Let's look at one more example.
Highest power of 5 in 350!
350/5 = 70
70/5 = 14
14/5 = 2
So 70+14+2 = 86 is the highes power of 5 in 350!.
I guess the explanation is lucid

Given problem is 8!^4

Highest power of 2 in 8!
8/2 = 4
4/2 =2
2/2 =1
So highes power of 2 in 8! is 7

Similarly highest power of 3 is 8!
8/3 = 2
So highest power of 3 in 8! is 2
Similarly highes powr of 5 in 8! is1

So 8!^4 = (2^7)^4 x (3^2)^4 x(5^1)^4 x K
ie 8!^4 = 2^28 x 3^8 x 5^4 x K

So xyz = 28x8x4 = 896

This would be faster even if the given factorial is big........

ie

_________________

Averages Accelerated:Guide to solve Averages Quickly(with 10 practice problems)

Director
Director
avatar
Joined: 23 Jun 2005
Posts: 847
GMAT 1: 740 Q48 V42
Followers: 3

Kudos [?]: 21 [0], given: 1

GMAT Tests User
 [#permalink] New post 13 Oct 2006, 18:24
cicerone wrote:
Let's look at one more example.
Highest power of 5 in 350!
350/5 = 70
70/5 = 14
14/5 = 2
So 70+14+2 = 86 is the highes power of 5 in 350!.
I guess the explanation is lucid

Thanks for the neat trick, Cicerone.
  [#permalink] 13 Oct 2006, 18:24
    Similar topics Author Replies Last post
Similar
Topics:
Experts publish their posts in the topic 2^X)*(3^Y)*(5^Z) is the greatest positive divisor of (8!)^4. prasannar 4 26 May 2008, 01:04
f(n) = (2^x)*(3^y)*(5^z), where x, y, and z are hundreds', nfa1rhp 6 29 Jul 2007, 10:35
meaning of greatest positive divisor cloaked_vessel 2 06 Jan 2005, 13:25
What is the greatest common divisor of positive integers m twixt 4 22 Nov 2004, 06:50
What is the Greatest Common Divisor of positive integers m Guest 2 02 Sep 2004, 23:19
Display posts from previous: Sort by

2^X)*(3^Y)*(5^Z) is the greatest positive divisor of (8!)^4.

  Question banks Downloads My Bookmarks Reviews Important topics  


GMAT Club MBA Forum Home| About| Privacy Policy| Terms and Conditions| GMAT Club Rules| Contact| Sitemap

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®.