Computing $a ^ {^nC_r}$ % prime

Question

I have uploaded my problem as screenshot.

Answer 1

Fermat's little theorem says

x^p mod p = x mod p     or     x^(p-1) mod p = 1  (if p does not divide x)

We can use this to reduce the number of operations for computing x^m (m arbitrary) when p doesn't divide x:

1. Divide m by (p-1) and get: s = (m mod p-1); no need to compute the quotient
2. Note that m = (p-1)q + s and x^(p-1) = 1 mod p. So,
   • x^m = x^((p-1)q)x^s = (1^q)(x^s) mod p = x^(m mod p-1)

The problem that remains, however, is how to compute

choose(n,r) = n(n-1)...(n-r+1)/(r(r-1)...1) mod p-1

Addendum

Further thinking on the problem above leads us to consider the following. We have three integers:

c = choose(n,r)
m = n(n-1)...(n-r+1)
f = factorial(r)
q = p-1

which satisfy

c = m/f,                                 eq 1

the question we have to answer is whether it is valid to compute c mod q as

(c mod q) = (m mod q)/(f mod q)          eq 2

right? Because this would allow us to reduce the calculation of choose(n,r) to two series of modular multiplications plus a single division (an easy and efficient algorithm).

Now, eq 1 can be rewritten as

 c*f = m

which enables us to apply mod q to both sides:

 (c mod q)(f mod q) = (m mod q)          eq 3

because mod is known to commute with products (and sums), which is the very property we will use to compute the series of modular multiplications mentioned above.

And since the tree quantities involved in eq 3 are integers, we can divide both sides by (f mod q) arriving at eq 2. My answer is now complete.

Addendum 2

I said My answer is now complete. Not quite. The problem that still remains happens when f mod q = 0, in this case we cannot divide eq 3 as we did above. This case requires an special treatment and will result in a more complex algorithm.

One idea worth trying is to factor q, i.e., p-1 as a product of primes and consider these primes with their exponents one by one. Take one of them, say t^e. We know that t^e must divide the factorial f. Therefore, it must divide the right hand side of eq 3 too. So, we have to peek enough factors among n, n-1, n-2, ..., n-r+1 that are divisible by t until we exhaust e divisions by t. Then we need to replace those factors with their quotient by the corresponding power of t. After repeating this procedure for all the primes inside q=p-1, we will get f/q on the left, and the new list of factors on the right. This will be our new version of eq 3. Of course the new f (=f/q) may be again divisible by q. Because of this we have to repeat the same procedure until f mod q is no longer 0. At that point we will be able to divide and get the value of c mod q.

Answer 2

https://en.m.wikipedia.org/wiki/Fermat%27s_little_theorem implies that (x^y) mod p = x^(y mod (p-1)) mod p when p does not divide x.