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

Question

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

(2) b + a = 11

Bunuel · Accepted Answer

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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gmatprav · Answer

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.

unceldolan · Answer

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.

mvictor · Answer

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.

jhshin2 · Answer

Bad question bc Statements 1 and 2 are mutually exclusive. Something that will never happen in the real GMAT.

CAMANISHPARMAR · Answer

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.

QuantMadeEasy · Answer

To find whether a is greater than b, when a and b are integers.

(1) b^2 - a^2 = 7! + 1
7!+1 is an odd positive number, therefore b is greater than a. sufficient

(2) b+a =11
b can be greater or smaller than a. insufficient

A is correct.

bumpbot · Answer

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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?

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