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PS - counting

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Director
Director
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Joined: 19 Mar 2007
Posts: 524
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Kudos [?]: 6 [0], given: 0

PS - counting [#permalink] New post 17 Jul 2007, 23:53
00:00
A
B
C
D
E

Difficulty:

(N/A)

Question Stats:

0% (00:00) correct 0% (00:00) wrong based on 0 sessions
The sum = ?

Last edited by nick_sun on 28 Jun 2008, 02:21, edited 1 time in total.
Senior Manager
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Posts: 349
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Re: PS - counting (Gset 27 - 17) [#permalink] New post 18 Jul 2007, 00:01
nick_sun wrote:
If x is the product of the positive integers from 1 to 8, inclusive, and if i, k, m, and p are positive integers such that x = 2^i * 3^k * 5^m * 7^p, then i + k + m + p =

A. 4
B. 7
C. 8
D. 11
E. 12


8! = 8*7*6*5*4*3*2*1 = 2^i * 3^k * 5^m * 7^p

i=7, k=2, m=1, p=1

Answer = 11 (D)
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 [#permalink] New post 18 Jul 2007, 01:43
Quote:
If x is the product of the positive integers from 1 to 8, inclusive


x=8.7.6.5.4.3.2.1=8!

Quote:
if i, k, m, and p are positive integers such that x = 2^i * 3^k * 5^m * 7^p


we can simplify x as 2^7.3^2.5^1.7^1
Equating the values of x
we have i=7,k=2,m=1,p=1
Quote:
then i + k + m + p =

11. Answer.
  [#permalink] 18 Jul 2007, 01:43
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