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by liferocks » Wed Jun 02, 2010 4:36 am
(x+4)! / x! = (x + 1)!
or (x+4)/(x+1)!=x!
or (x+2)(x+3)(x+4)=x!

one thing we can notice here is x! cannot have a prime factor which is greater than x.
Among the options only for 6 this is true for everything else its not correct.
IMO ans is option A
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by jube » Wed Jun 02, 2010 5:02 am
liferocks wrote: one thing we can notice here is x! cannot have a prime factor which is greater than x.
Awesome observation! Would've never struck me!
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by vinee » Thu Jun 03, 2010 7:24 am
The OA is 'A' but could you explain a little bit more how you got (x+4) (x+3) (x+2) = x! ?
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by liferocks » Thu Jun 03, 2010 7:31 am
vinee wrote:The OA is 'A' but could you explain a little bit more how you got (x+4) (x+3) (x+2) = x! ?
(x+4)! / x! = (x + 1)!
or (x+4)!/(x+1)! =x!
or[(x+1)!*(x+2)*(x+3)*(x+4)]/(x+1)! = x!
or (x+4) (x+3) (x+2) = x!

hope this is clear now.
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by odod » Thu Jun 03, 2010 1:17 pm
great thread..one question though..I don't understand this statement

"one thing we can notice here is x! cannot have a prime factor which is greater than x. "

how come X! cannot have a prime factor greater than X? I must be missing something here

thanks
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by uwhusky » Thu Jun 03, 2010 3:57 pm
Use 5! for example, which is 5 x 4 x 3 x 2 x 1, and therefore 7 cannot be a prime factor of 5!.
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by liferocks » Thu Jun 03, 2010 7:45 pm
odod wrote:great thread..one question though..I don't understand this statement

"one thing we can notice here is x! cannot have a prime factor which is greater than x. "

how come X! cannot have a prime factor greater than X? I must be missing something here

thanks
X! is product of all integers <= x..now any prime which is >x will not appear in that product.But for composites >x this is not true as they may be formed out of the available numbers <=x.
Ex take x=7...x! or 7! will never have 11 as a factor but 12 is a factor of it.
Hope this helps.
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