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iven n>5 , when (n!+n+1) is divided by (n+1)

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by siddhu161 » Tue Feb 04, 2014 6:51 pm
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
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by [email protected] » Wed Feb 05, 2014 1:54 am
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

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