chetan2u wrote:
SajjitaKundu wrote:
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!
(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?
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