Last two digits of a power

What are the last two digits of 21002^{100}? (Give the two-digit number, e.g. an answer of …07\ldots07 would be 77.)

Show hints (2)+
  1. Euler needs gcd⁡(a,m)=1\gcd(a,m)=1; here gcd⁡(2,100)=2\gcd(2,100)=2, so split 100=4×25100=4\times 25.
  2. Mod 44 it's 00; mod 2525 use ϕ(25)=20\phi(25)=20; then recombine with CRT.

Answer

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76

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