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Ps: divisor

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VP
Joined: 29 Dec 2005
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Ps: divisor [#permalink]

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13 Jun 2006, 23:04
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what is XYZ if (2^X)*(3^Y)*(5^Z) is the greatest positive divisor of (8!)^4?

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SVP
Joined: 30 Mar 2006
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Re: Ps: divisor [#permalink]

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13 Jun 2006, 23:53
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

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Current Student
Joined: 29 Jan 2005
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14 Jun 2006, 08:13
Why isn`t the answer 7*2*1

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GMAT Club Legend
Joined: 07 Jul 2004
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Location: Singapore

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14 Jun 2006, 08:45
(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

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VP
Joined: 25 Nov 2004
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14 Jun 2006, 09:08
GMATT73 wrote:
Why isn`t the answer 7*2*1

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

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VP
Joined: 02 Jun 2006
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14 Jun 2006, 10:39
(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??

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VP
Joined: 29 Dec 2005
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14 Jun 2006, 13:15
thanx everybody, you all hit the ball into the post.

goal = 28 x 8 x 4

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Current Student
Joined: 29 Jan 2005
Posts: 5210

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15 Jun 2006, 07:02
MA wrote:
GMATT73 wrote:
Why isn`t the answer 7*2*1

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.

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15 Jun 2006, 07:02
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