If x and y are positive integers, is y!/x! an integer?

Question

(1) (x + y)(x - y) = 5! + 1
(2) x + y = 11^2

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gmatophobia · Answer

Question   Is  \frac{y!}{x!}  an integer ?

Inference :    \frac{y!}{x!}  will be an integer when y > x. Also y & x are non negative, as we are referring x and y in context of factorials

Statement 1

(x + y)(x - y) = 5! + 1

(x + y)(x - y) = 121

Possible cases :

(x+y) = 11 & (x-y) = 11
x = 11 & y = 0

(x+y) = 121 & (x-y) = 1 
x = 61 & y = 60

(x+y) = 1 & (x-y) = 121
x = 61 & y = -60

We can discard the first case (due to constraint) and third case (as its an invalid one).

In the second case, x > y. Hence we have sufficient information to answer.

Statement 2

(2) x + y =  11^2

x + y = 121

Possible cases :

x = 1 & y = 120

x = 120 & y = 1

As we have two contradictory answers one in which x > y and other in which x < y, the statement is not sufficient.

Option A

Archit3110 · Answer

gmatophobia

question stem says " if x and y are  positive integers "

for #1 
 (x+y) = 11 & (x-y) = 11 
x = 11 & y = 0 ; would not be valid..

gmatophobia · Answer

Agree. We can discard the scenario in this case (can be valid in general). However the exclusion / inclusion wouldn't change the outcome.

If x and y are positive integers, is y!/x! an integer?

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