Hey! I have a question. If I have to calculate a pow b modulo mod, with b>mod, is it the same with a pow (b % (mod-1)?
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Hey! I have a question. If I have to calculate a pow b modulo mod, with b>mod, is it the same with a pow (b % (mod-1)?
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By Euler's Theorem, if gcd(a, n) = 1,
. If m happens to be a prime, then phi(m) = m - 1.
If "gcd(a,n)=1", you mean gcd(a,m)=1? I can not see the n
Upps, sorry. You're right, that was meant to be an m :)
Thanks!
Even when gcd(a, n) > 1, the following holds: If b ≥ φ(m),
. When m is large and you cannot efficiently factor it to compute φ(m) and b is given in decimal form, you can still use fast exponentiation. Process digits of b one by one starting from the most significant one and use a10b + c = (ab)10 × ac.