Prime is a divisor

Number Theory Level 2

Let k > 2 k >2 be a positive integer and p < k p < k be a prime. Is it possible that p k p^k is a divisor of k ! k! ?

Yes, it's possible. No, it's not possible.

Áron Bán-Szabó by Áron Bán-Szabó , 17, Hungary

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

1 solution

Ben A
Jul 17, 2017

According to Fermat's Little Theorem, if k is a prime and p is not a multiple of k, then p^k is congruent to 1, mod k. p cannot be a multiple of k, since it is less than k, and if a number is not congruent to k, then it cannot be congruent to k! since k is a factor of k!.

1 Helpful 0 Interesting 0 Brilliant 0 Confused

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...