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# Is y!/x! an integer?

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Manager
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13 Jul 2012, 02:01
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Is y!/x! an integer?

(1) (x + y)(x-y) = 5! + 1
(2) x + y = 112
[Reveal] Spoiler: OA

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Last edited by Bunuel on 13 Jul 2012, 04:17, edited 1 time in total.
Edited the question.

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Re: Is y!/x! an integer? [#permalink]

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13 Jul 2012, 04:47
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This is a flawed question.

Is y!/x! an integer?

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 $$y\geq{x}$$.

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.

Hope it's clear.
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14 Oct 2012, 12:47
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lrajiv74 wrote:
Is y!/x! an integer?

(1) (x + y)(x – y) = 5! + 1
(2) x + y = 112

y!/x! will be integer if y>=x

from 1. x^2-y^2 = positive value

Thus |x| > |y|

Since negative values cant have factorials. we have x>y . Hence it is NEVER an INTEGER. HENCE SUFFICIENT. Therefore A

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14 Oct 2012, 16:09
With (1) we know that x^2 - y^2 = 5! +1

Is y!/x! an integer?

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.

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13 Jun 2017, 19:34
1. Is (y!/x!) an integer?
(1) (x + y)(x − y) = 5! + 1
(2) x + y = 11^2

Last edited by chetan2u on 13 Jun 2017, 19:38, edited 1 time in total.
formatted the Q

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Re: Is (y!/x!) an integer? [#permalink]

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13 Jun 2017, 19:43
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.

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Re: Is (y!/x!) an integer? [#permalink]

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13 Jun 2017, 19:48
SajjitaKundu wrote:
1. Is (y!/x!) an integer?
(1) (x + y)(x − y) = 5! + 1
(2) x + y = 11^2

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
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Re: Is (y!/x!) an integer? [#permalink]

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13 Jun 2017, 19:53
[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,
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?

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Re: Is (y!/x!) an integer? [#permalink]

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13 Jun 2017, 20:17
chetan2u wrote:
SajjitaKundu wrote:
1. Is (y!/x!) an integer?
(1) (x + y)(x − y) = 5! + 1
(2) x + y = 11^2

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,
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?

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Re: Is (y!/x!) an integer? [#permalink]

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13 Jun 2017, 20:30
SajjitaKundu wrote:
ashudhall wrote:
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,
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.

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Re: Is (y!/x!) an integer? [#permalink]

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13 Jun 2017, 20:57
ashudhall wrote:
SajjitaKundu wrote:
ashudhall wrote:
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,
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.

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13 Jun 2017, 22:34
SajjitaKundu wrote:
1. Is (y!/x!) an integer?
(1) (x + y)(x − y) = 5! + 1
(2) x + y = 11^2

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.

Hope this helps.
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Re: Is y!/x! an integer? [#permalink]

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13 Jun 2017, 22:44
SajjitaKundu wrote:
1. Is (y!/x!) an integer?
(1) (x + y)(x − y) = 5! + 1
(2) x + y = 11^2

Merging topics.

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Re: Is y!/x! an integer?   [#permalink] 13 Jun 2017, 22:44
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