### vkditya0997's blog

By vkditya0997, history, 4 years ago, Hello Codeforces,

I was trying this 603B - Moodular Arithmetic, I couldn't get the idea and read the editorial.

I have a doubt.

for k >  = 2 case :

we choose m such that km = 1 mod p

and also it has been proved that f(ki * n mod p ) = ki * f(n) mod p

and now if we choose some f(n) , we get the value of f(n) , f(k * n mod p ) ,.... f(km - 1 * n mod p )

Understood until this point

I didn't understand from here..

Therefore, if we choose f(n) of p - 1 / m integers n, we can get value of all p-1 non-zero integers. Since f(n) can be chosen in p ways for each of the p - 1 / m integers, the answer is pp - 1 / m  Comments (1)
 » p prime implies there exists g (generator) such that {gi: 1 ≤ i ≤ p - 1} = {1, ..., p - 1}. g will have order p - 1. So gn = k where n = (p - 1) / m since gp - 1 = gnm = km = 1.Consider the sets Gi = {gi, kgi, ..., km - 1gi} for all 1 ≤ i ≤ n. Every element from 1 to p - 1 appears in exactly one of the Gi (cosets).Consider any Gi and . Assigning f(x) a value defines f on all other elements in Gi. So to fully define f we only need to assign one value from each Gi. So p choices for each of the n sets or p(p - 1) / m choices overall.