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

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Director
Joined: 05 Jan 2008
Posts: 687
2^X)*(3^Y)*(5^Z) is the greatest positive divisor of (8!)^4. [#permalink]

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26 May 2008, 02:04
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(2^X)*(3^Y)*(5^Z) is the greatest positive divisor of (8!)^4. the value of x+y+z=?
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CEO
Joined: 17 Nov 2007
Posts: 3584
Concentration: Entrepreneurship, Other
Schools: Chicago (Booth) - Class of 2011
GMAT 1: 750 Q50 V40

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26 May 2008, 02:11
$$(8!)^4=(8*7*6*5*4*3*2)^4=(2^3*7*2*3*5*2^2*3*2)^4=(2^7*7*3^2*5)^4=2^{28}*3^8*5^4*7$$

x+y+z=28+8+4=40
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Director
Joined: 05 Jan 2008
Posts: 687

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26 May 2008, 02:18
yup it is
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Persistence+Patience+Persistence+Patience=G...O...A...L

SVP
Joined: 28 Dec 2005
Posts: 1559

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26 May 2008, 04:39
with walker ... 40 for me as well.

Start off with 8! , which we know is 8x7x6x5x4x3x2x1 ... we can write 8 as 2x2x2, 6 as 2x3, 4 as 2x2 ... you end up with 2^7*7*3^2*5 ... and now apply the power of 4 to get 2*28*7^4*3^8*5^4 ... add up the exponents to get 28+8+4=40
VP
Joined: 18 May 2008
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26 May 2008, 06:15
Nice explanation!
Re: PS: X+Y+Z   [#permalink] 26 May 2008, 06:15
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