How to calculate 2 to the power of a large number modulo another large number?

Question

M = 115792089237316195423570985008687907853269984665640564039457584007908834671663

2^{96514807760119017459957299373576180339312098253841362800539826362414936958669} % M = ?

Is it possible to calculate this in Python? Or are there other methods?

Answer 1

To calculate the result, the three-argument pow does this efficiently, as mentioned by @MarkDickinson in the comments.

A simplified explanation of how this works:

• to calculate 2**N mod M, first find K = 2**(N//2) mod M
• if N was even, 2**N mod M = K * K mod M
• if N was odd, 2**N mod M = K * K * 2 mod M That way, there is no need to calculate huge numbers. In reality, pow uses more tricks, is more general and doesn't need recursion.

Here is some demonstration code:

def pow_mod(B, E, M):
    if E == 0:
        return 1
    elif E == 1:
        return B % M
    else:
        root = pow_mod(B, E // 2, M)
        if E % 2 == 0:
            return (root * root) % M
        else:
            return (root * root * B) % M

M = 115792089237316195423570985008687907853269984665640564039457584007908834671663
E = 96514807760119017459957299373576180339312098253841362800539826362414936958669

print(pow_mod(2, E, M))
print(pow(2, E, M))