Fermat's little theorem is key to this problem
It tells us that, as 101 is prime, 123^ 100 = 1 mod 101
Then, (123^1001) mod 101 =
123^(100*10+1) mod 101 =
123 mod 101 =
22
Og N.
asked 12/05/20Find 123^1001 mod 101 using the Modular Exponentiation algorithm?
Fermat's little theorem is key to this problem
It tells us that, as 101 is prime, 123^ 100 = 1 mod 101
Then, (123^1001) mod 101 =
123^(100*10+1) mod 101 =
123 mod 101 =
22
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