a^p/p 'then the remainder is a when p is a prime?

Answer 1

this is fermat's little theorerm please refer to

http://mathworld.wolfram.com/FermatsLittleTheorem.html

Answer 2

Yes, and you can check it quite compactly, because you can take remainders anywhere through the calculation and it will still be the same remainder at the end as if you had done the full working.

p = 43, a = 5

a^2 = 25
a^4 = 625 = 23
a^5 = 115 = 29
a^10 = 841 = 24
a^20 = 576 = 17
a^21 = 85 = 42
a^42 = 1764 = 1
a^43 = 5

Nowadays, most books present this theorem as:
a^(p-1) leaves a remainder of 1 when divided by p.

Answer 3

I'm not sure exactly what you're asking, but it is true that if p is prime, then for any a, a^p ≡ a (mod p). This is Fermat's Little Theorem.

Answer 4

i go with the first guy who anwsered....

Answer 5

No
Please give me best answer thanks!