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

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

(2) b + a = 11

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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Practice makes Perfect.

I rephrased the question to:
is a greater or equal to b? otherwise, if b>, then a!/b! will not be an integer.

1 - b^2 - a^2 = 7!+1
so we know that both are positive integers, and b^2 is greater than a^2. it means that b is greater than a, and we know for sure that a!/b! is not an integer.
B, C, and E - out.

2 - b+a=11.
if a=6, b=5, then it is an integer
if a=5, b=6, then not.
B is out, so the answer is:

A.

According to statement 1: b > a hence we have a definite "No" to the question asked: Hence statement 1 alone is sufficient.

Statement 2: Multiple combinations of a & b are possible. hence insufficient.

Hence Option A is the correct answer.
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Manish

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