Given n>5 , when (n!+n+1) is divided by (n+1) , what is the remainder ?
(1) (n+2) is a prime number.
(2) (n−2) is a prime number.
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iven n>5 , when (n!+n+1) is divided by (n+1)
This topic has expert replies
if we simplify (n!+n+1)/(n+1)
we need to find remainder of this term n!/(n+1) coz remainder of (n+1)/(n+1) is 0.
as n>5,
a. n+2 is prime --> n is definitely odd
n!/n+1 will have remainder 0. (n=9,11,15..)
Sufficient
b. n-2 is prime --> n is again definitely odd
n!/n+1 will have remainder 0. (n=7,9,13..)
Sufficient
hence D
we need to find remainder of this term n!/(n+1) coz remainder of (n+1)/(n+1) is 0.
as n>5,
a. n+2 is prime --> n is definitely odd
n!/n+1 will have remainder 0. (n=9,11,15..)
Sufficient
b. n-2 is prime --> n is again definitely odd
n!/n+1 will have remainder 0. (n=7,9,13..)
Sufficient
hence D
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Hi guerrero,
This question appeared not too long ago in this Forum. I posted an explanation here:
https://www.beatthegmat.com/please-help- ... 73744.html
GMAT assassins aren't born, they're made,
Rich
This question appeared not too long ago in this Forum. I posted an explanation here:
https://www.beatthegmat.com/please-help- ... 73744.html
GMAT assassins aren't born, they're made,
Rich
