Inspired wrote:Is y!/x! an integer?

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

y!/x! is an integer only when y>= x

1) says (x + y)(x - y) = 5! + 1 =>
x^2-y^2 = 121

now assuming that x and y are positive (Factorial operation is possible only on +ve integers and zero), x is greater than y which helps answering the question whether y!/x! is an integer (of course it is not)

2) is not sufficient

1) alone is sufficient

excellent, thank you very much

ajith wrote:

Inspired wrote:Is y!/x! an integer?

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

y!/x! is an integer only when y>= x

1) says (x + y)(x - y) = 5! + 1 =>
x^2-y^2 = 121

now assuming that x and y are positive (Factorial operation is possible only on +ve integers and zero), x is greater than y which helps answering the question whether y!/x! is an integer (of course it is not)

2) is not sufficient

1) alone is sufficient

____________
Hi ajith,
I was wondering ,in the statement 1,whether we do not have to consider the solutions of x^2-y^2 = 121 if we are getting a unique solution or not.
In the case that no unique solution is found,the option choice may change.

harsh.champ wrote:
Hi ajith,
I was wondering ,in the statement 1,whether we do not have to consider the solutions of x^2-y^2 = 121 if we are getting a unique solution or not.
In the case that no unique solution is found,the option choice may change.

x=11 y=0
x =61 and y=60

Are the integer solutions I can think about.

Now, it is not unique. y!/x! is not an integer in both cases

ajith wrote:

harsh.champ wrote:
Hi ajith,
I was wondering ,in the statement 1,whether we do not have to consider the solutions of x^2-y^2 = 121 if we are getting a unique solution or not.
In the case that no unique solution is found,the option choice may change.

x=11 y=0
x =61 and y=60

Are the integer solutions I can think about.

Now, it is not unique. y!/x! is not an integer in both cases

__________________________
Hey ajith,
That was what I was talking about.
As we are having 2 solutions,so we cannot arrive at a particular answer.
(though in both the cases here y!/x! is not an integer)
Although not in this question,but suppose in another question if without calculating the answers we go forward and if y>x comes out in one of the answers,then it cannot be determined if y!/x! is an integer or not .In one case it is and in another it is not.Guess,it is okay to invest some more time to actually solve the questions.
One of those trick questions where wrong answer may come if we proceed too quickly.

harsh.champ wrote: Hey ajith,
That was what I was talking about.
As we are having 2 solutions,so we cannot arrive at a particular answer.
(though in both the cases here y!/x! is not an integer)
Although not in this question,but suppose in another question if without calculating the answers we go forward and if y>x comes out in one of the answers,then it cannot be determined if y!/x! is an integer or not .In one case it is and in another it is not.Guess,it is okay to invest some more time to actually solve the questions.
One of those trick questions where wrong answer may come if we proceed too quickly.

Agree - it can happen that two answers will contradict and we may have to select E

But in this particular case, it doesnt hence the answer is C

very good analysis by ajith