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If a and b are positive integers, is a!/b! an integer?

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If a and b are positive integers, is a!/b! an integer? [#permalink] New post 20 Sep 2013, 13:01
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If a and b are positive integers, is a!/b! an integer?

(1) (b - a)(b + a) = 7! + 1

(2) b + a = 11
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Re: If a and b are positive integers, is a!/b! an integer? [#permalink] New post 20 Sep 2013, 13:08
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If a and b are positive integers, is a!/b! an integer?

In order a!/b! to be an integer, a must be more than or equal to b. Thus the question basically asks whether a\geq{b}.

(1) (b - a)(b + a) = 7! + 1 --> b^2-a^2=7! + 1=positive. Since a and b are positive integers, then b>a. So, the answer to the question is NO. Sufficient.

(2) b + a = 11. We cannot determine whether a\geq{b}. Not sufficient.

Answer: A.

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Re: If a and b are positive integers, is a!/b! an integer? [#permalink] New post 08 Feb 2014, 02:15
Given that a & b are both positive integers. For \frac{a!}{b!} to be integer a must be greater than or equal to b. otherwise a!/b! is going to be a fraction. We can then re-phrase the question to

Is a >= b?


Statement 1: (b-a)(b+a) = 7!+1. 7!+1 is +ve. b+a is +ve. b-a also has to be positive. note that b != a. then b>a. so a!/b! can not be integer. Sufficient.

Statement 2: a+b=11. a=1 b=10 ++> a!/b! is not integer. reverse the values, that is b=1, a=10 then yes. Since we have a yes and no. Insufficient.

A is answer.
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Re: If a and b are positive integers, is a!/b! an integer? [#permalink] New post 05 Aug 2014, 03:11
From question: in order to a!/b! = integer, a has to be greater or equal to b. Hence the question becomes is a>=b?

(1) (b-a)(b+a) = 7! + 1 --> b²-a² = 7!+1. Since 7! + 1 is >0, b is > a (since both are positive integers) and the answer is NO. Suff.
(2) b+a = 11, if b = 5 and a = 6 then answer is YES, if b = 6 and a = 5 answer is NO. IS.

A.
Re: If a and b are positive integers, is a!/b! an integer?   [#permalink] 05 Aug 2014, 03:11
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