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  • If and <img alt="\mathrm{n}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%

If \mathrm{n>1} and \mathrm{n} divides \mathrm{(n-1) !+1} , then

Option: 1

n must be prime


Option: 2

n must be divisible by exactly two primes


Option: 3

n  must be a composite number


Option: 4

none of these.


Answers (1)

best_answer

If \mathrm{n} is not prime, then there exists \mathrm{r \in \mathbf{N}} such that \mathrm{2 \leq r \leq n-1} and \mathrm{r \mid n}.

As \mathrm{r \ln\, and \, n ![(n-1) !+1]}. we get

\mathrm{r ![(n-1) !+1]}
As  \mathrm{2 \leq r \leq n-1, r \mid(n-1)!} therefore \mathrm{r l1}. A contradiction.
 

Posted by

Irshad Anwar

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