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4 solutions
Let ≡ 1 0 1 denote congruence m o d 1 0 1 . Then, by Fermat's Little Theorem ,
2 1 0 1 0 ≡ 1 0 1 2 ( 1 0 1 0 m o d 1 0 0 ) ≡ 1 0 1 2 ( 1 0 0 5 m o d 1 0 0 ) ≡ 1 0 1 2 0 ≡ 1 0 1 1
Ummm............a direct application of Fermat's Little Theorem would suffice.........
Compute consecutive powers of 2 until its modulus 1 0 1 is 1 or the problem is solved by brute force.
N.B., when the power of 2 becomes larger than 1 0 1 only the modulus needs to be retained.
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Rewrite 2 1 0 1 0 as 2 1 0 0 5 .
The problem can now be rewritten as ( 1 5 m o d 1 0 1 ) .
The answer is 1
Touché! I learned something, thanks to both of you.
2 1 0 1 0 = ( 2 1 0 0 ) 1 0 8 ≡ 1 1 0 8 = 1 ( m o d 1 0 1 ) by Fermat's little theorem.