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First of all, factorial is defined only for non-negative integers, so realistic GMAT question would mention that \(x\) and \(y\) are non-negative integers.

Next, \(\frac{y!}{x!}=integer\) will hold true if \(y\geq{x}\). So, the question basically asks whether \(y\geq{x}\).

(1) (x + y)(x-y) = 5! + 1 --> \(x^2-y^2=121\). As discussed, since \(x\) and \(y\) must be non-negative integers, then \(x>y\) and the asnwer to the question is NO. Sufficient.

(2) x + y = 112. Not sufficient to answer whether \(y\geq{x}\).

Answer: A.

Now, even though formal answer to the question is A, this is not a realistic GMAT question, as: on the GMAT, two data sufficiency statements always provide TRUE information and these statements never contradict each other. But from (1) the only non-negative integer solutions for \(x\) and \(y\) are: (11, 0) and (61, 60), so \(x+y\) cannot equal to 112 as the second statement says, which means that the statements clearly contradict each other.

The question is flawed. You won't see such a question on the GMAT.

Well, that's a factorial divided by a factorial. We know that 0! = 1. And since x^2 - y^2 = some positive integer, we know that x and y are going to be integers as well (no decimals, a 7.13! would be weird!)

So...x and y are integers. y! / x! ...okay we well can get into dangerous territory if the denominator is a large number --- that introduces fractions. If y>x, then numerator will always be larger and we won't necessarily get that fraction problem. But in this case, we know that x^2 - y^2 = some positive integer...

... so x>y. That means we know DEFINITIVELY the denominator is going to be bigger than the numerator and we know DEFINITIVELY that we get a fraction less than one.

In that case, is y! / x! an integer? Well we know definitively that it's a fraction less than 1, thus not an integer. Thus, we have enough information to say whether YES integer or NO, not an integer. The answer would be NO not an integer. But don't confuse that with the actual question it's asking...it's not do we have an integer yes or no? THe question is asking...do we have SUFFICIENT INFORMATION to determine whether the answer to that question is yes or no. And the answer is we DO have SUFFICIENT INFORMATION to make a definitive yes or no answer.

1) (x+y)(x-y)=5!+1 121*1=121(no other combination is possible. 121=11*11 or 121*1 Since sum and difference cannot be same.hence we are left with only 121*1) This means x=61, y=60 i.e y!/x! Is not an integer. Hence sufficient. 2) x+y = 121 This can have multiple values for x & y. Hence not sufficient.

1. Is (y!/x!) an integer? (1) (x + y)(x − y) = 5! + 1 (2) x + y = 11^2 Please detail with explanation. Thanks in advance!

Hi,

please post the Q along with topic name and OA. y and x are integers

Nw for the Q.. when can \(\frac{y!}{x!}\) be an integer.. when either y>x or y=x..

so lets check the statements (1) (x + y)(x − y) = 5! + 1 as we are looking for y!/x!, we can take that y and x are positive integers as negative / fractions do not have factorials.. RHS is positive and LHS has x-y, so x-y must be POSITIVE and hence x>y.. so our answer will be NO always. y!/x! will be a fraction. suff

(2) x + y = 11^2 so x+y =121.. x and y can take various values. with y as 61 and x as 60 ans is YES with y as 1 and x as 120 ans is NO insuff

[quote="ashudhall"]1) (x+y)(x-y)=5!+1 121*1=121(no other combination is possible. 121=11*11 or 121*1 Since sum and difference cannot be same.hence we are left with only 121*1) This means x=61, y=60 i.e y!/x! Is not an integer. Hence sufficient. 2) x+y = 121 This can have multiple values for x & y. Hence not sufficient.

Hi, Thanks for the reply! But for the 1st statement, if x+y =11 and x-y =11, it leaves us with x=11 and y=0. Thus making the of (y!/x!) as 0. And the other case as you said, not an integer. So, shouldn't the answer be C?

1. Is (y!/x!) an integer? (1) (x + y)(x − y) = 5! + 1 (2) x + y = 11^2 Please detail with explanation. Thanks in advance!

Hi,

please post the Q along with topic name and OA. y and x are integers

Nw for the Q.. when can \(\frac{y!}{x!}\) be an integer.. when either y>x or y=x..

so lets check the statements (1) (x + y)(x − y) = 5! + 1 as we are looking for y!/x!, we can take that y and x are positive integers as negative / fractions do not have factorials.. RHS is positive and LHS has x-y, so x-y must be POSITIVE and hence x>y.. so our answer will be NO always. y!/x! will be a fraction. suff

(2) x + y = 11^2 so x+y =121.. x and y can take various values. with y as 61 and x as 60 ans is YES with y as 1 and x as 120 ans is NO insuff

A

Hi, Thanks for the reply. My query is for statement (1), there is a possibility that x=11 and y=0, or x=121 and y=1. (Statement 2 is insufficient alone.) So this is where statement 2 helps us. So, shouldn't the answer be C?

1) (x+y)(x-y)=5!+1 121*1=121(no other combination is possible. 121=11*11 or 121*1 Since sum and difference cannot be same.hence we are left with only 121*1) This means x=61, y=60 i.e y!/x! Is not an integer. Hence sufficient. 2) x+y = 121 This can have multiple values for x & y. Hence not sufficient.

Hi, Thanks for the reply! But for the 1st statement, if x+y =11 and x-y =11, it leaves us with x=11 and y=0. Thus making the of (y!/x!) as 0. And the other case as you said, not an integer. So, shouldn't the answer be C?

0! is 1 not 0, hence in that case as well 1/11! Would not be an integer. Hence the answer should be A only.

1) (x+y)(x-y)=5!+1 121*1=121(no other combination is possible. 121=11*11 or 121*1 Since sum and difference cannot be same.hence we are left with only 121*1) This means x=61, y=60 i.e y!/x! Is not an integer. Hence sufficient. 2) x+y = 121 This can have multiple values for x & y. Hence not sufficient.

Hi, Thanks for the reply! But for the 1st statement, if x+y =11 and x-y =11, it leaves us with x=11 and y=0. Thus making the of (y!/x!) as 0. And the other case as you said, not an integer. So, shouldn't the answer be C?

0! is 1 not 0, hence in that case as well 1/11! Would not be an integer. Hence the answer should be A only.

1. Is (y!/x!) an integer? (1) (x + y)(x − y) = 5! + 1 (2) x + y = 11^2

Please detail with explanation. Thanks in advance!

This question is basically testing whether x! Completely divides into y! Or not. In other words, what you have to check is whether X! < Y! or X! > Y! as Factorial is basically a product of all numbers from n to 1.

For e.g. 5! = 5*4*3*2*1

S1 -> (X+Y)(X-Y) =5! + 1 (X+Y)(X-Y) = 121 X=61 , Y=60 => Y!/X! Is not an integer. X = 60, Y = 61 violates the equation as the left hand side becomes negative => -121 is not equal to 121. Sufficient.

S2 -> X+Y = 121 Values can interchange I.e. X =60 and Y=61 and vice versa. Insufficient.

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