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For which of the following values of X is the equation

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Manager
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For which of the following values of X is the equation [#permalink] New post 29 Dec 2007, 01:17
For which of the following values of X is the equation
(X+4)!/X! = (X+1)! true ?

a) 6

b)7

c) 8

d) 9

e) 10

Is there any easier way to solve the problem apart from backsolving ?
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Re: Probability problem [#permalink] New post 29 Dec 2007, 01:27
dynamo wrote:
For which of the following values of X is the equation
(X+4)!/X! = (X+1)! true ?

a) 6

b)7

c) 8

d) 9

e) 10

Is there any easier way to solve the problem apart from backsolving ?


all i can think of, besides backsolving, is watching the intersection of numbers -- since X! in the denominator removes the "edge" the left side has and forces intersection with the right. i'd love to hear someone else's take. what i've said here amounts to backsolving.
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 [#permalink] New post 29 Dec 2007, 03:32
Expert's post
(X+4)!/X! = (X+1)!

(X+4)*(X+3)*(X+2)=X!

X+4,X+3,X+2 - should not be prime numbers.

6: 6,5,4
7: 7,6,5
8: 8,7,6
9: 9,8,7
10: 10,9,8 - BINGO
11: 11,10,9
12: 12,11,10
13: 13,12,11
14: 14,13,12

added: x+4=10 ==> x=6. A
Maybe for clarity I should write "2: 6,5,4" instead of "6: 6,5,4"

Last edited by walker on 29 Dec 2007, 09:20, edited 3 times in total.
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 [#permalink] New post 29 Dec 2007, 04:00
I think it is A...x=6
What is OA?
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 [#permalink] New post 29 Dec 2007, 04:04
A. for me as well if i am not missing any crucial point
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 [#permalink] New post 29 Dec 2007, 07:28
Sorry mate. Could only come up with back solving to give A.
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 [#permalink] New post 29 Dec 2007, 07:53
Walker u are right . the Answer is 10.
Thanks everyone.

As per the rule

n^Cr = n^Cn-r , 6,8 and 10 are correct. but not sure how to distinguish between them.

Walker, can you please explain as to why the nos shd not be prome.
Also :- x = 6 so 3!
but how do find the the factorial value for x = 7,8,9 and 10.
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 [#permalink] New post 29 Dec 2007, 09:00
Expert's post
dynamo wrote:
Walker u are right . the Answer is 10.

x+4=10 ==> x=6 :)

dynamo wrote:
but how do find the the factorial value for x = 7,8,9 and 10.


(X+4)*(X+3)*(X+2)=X!

all prime factors of X!=X*(X-1)....1 less then X
therefore, if, for example, (X+2) is prime, X! can not be divided by (X+2)
  [#permalink] 29 Dec 2007, 09:00
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