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

 It is currently 04 May 2015, 04:11

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

Ps: divisor

Author Message
TAGS:
VP
Joined: 29 Dec 2005
Posts: 1349
Followers: 7

Kudos [?]: 30 [0], given: 0

Ps: divisor [#permalink]  13 Jun 2006, 22:04
00:00

Difficulty:

(N/A)

Question Stats:

0% (00:00) correct 0% (00:00) wrong based on 0 sessions
what is XYZ if (2^X)*(3^Y)*(5^Z) is the greatest positive divisor of (8!)^4?
SVP
Joined: 30 Mar 2006
Posts: 1739
Followers: 1

Kudos [?]: 46 [0], given: 0

Re: Ps: divisor [#permalink]  13 Jun 2006, 22:53
gmat_crack wrote:
Professor wrote:
what is XYZ if (2^X)*(3^Y)*(5^Z) is the greatest positive divisor of (8!)^4?

28* 4*8

8! can be written as 2^7*3^2*5*7

Hence GCD
28*8*4
Current Student
Joined: 29 Jan 2005
Posts: 5244
Followers: 23

Kudos [?]: 179 [0], given: 0

Why isnt the answer 7*2*1
GMAT Club Legend
Joined: 07 Jul 2004
Posts: 5078
Location: Singapore
Followers: 22

Kudos [?]: 184 [0], given: 0

(8!)^4

= (8*7*6*5*4*3*2*1)^4
= (2^3 * 7 * 2*3 * 5 * 2^2 * 3 * 2 * 1)^4
= (2^7 * 7 * 5 * 3^2 * 1)^4
= 2^28 * 7^4 * 5*4 * 3*8

TO divide (8!)^4 evenly, x = 28, y = 8, z = 4
VP
Joined: 25 Nov 2004
Posts: 1495
Followers: 6

Kudos [?]: 38 [0], given: 0

GMATT73 wrote:

you missed "greatest +ve divisor".

= (8!)^4
= (8x7x6x5x4x3x2x1)^4
= (2^3 x 7 x 2 x 3 x 5 x 2 x 2 x 3 x 2 x 1)^4
= (2^7 x 7 x 3^2 x 5)^4
= 2^28 x 7^4 x 3^8 x 5^4
so x = 28, y = 8 and z = 4
xyz = 28 x 8 x 4
VP
Joined: 02 Jun 2006
Posts: 1267
Followers: 2

Kudos [?]: 42 [0], given: 0

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

As (2^x)x(3^y)x(5^z) is GCD of (8!)^4,

(8!)^4/(2^x)x(3^y)x(5^z) = k

or (2^28)x(3^8)x(5^4)x(7^4) = k x(2^28)x(3^8)x(5^4)x(7^4)

=> x = 28, y = 8, z= 4

xyz = 28 x 8 x 4 = 28 x 32

Am I even going the right way with this??
VP
Joined: 29 Dec 2005
Posts: 1349
Followers: 7

Kudos [?]: 30 [0], given: 0

thanx everybody, you all hit the ball into the post.

goal = 28 x 8 x 4
Current Student
Joined: 29 Jan 2005
Posts: 5244
Followers: 23

Kudos [?]: 179 [0], given: 0

MA wrote:
GMATT73 wrote:

you missed "greatest +ve divisor".

= (8!)^4
= (8x7x6x5x4x3x2x1)^4
= (2^3 x 7 x 2 x 3 x 5 x 2 x 2 x 3 x 2 x 1)^4
= (2^7 x 7 x 3^2 x 5)^4
= 2^28 x 7^4 x 3^8 x 5^4
so x = 28, y = 8 and z = 4
xyz = 28 x 8 x 4

Crystal clear explanation! Thanks MA.
Similar topics Replies Last post
Similar
Topics:
Divisors 2 31 Mar 2010, 05:35
Divisor 1 13 Aug 2009, 06:36
divisors 3 28 May 2009, 08:52
PS: Divisors 6 25 Feb 2009, 15:19
Display posts from previous: Sort by