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,
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?
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.
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