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drmurat · 2020-09-15, 16:06

Quote:

Originally Posted by paulunderwood

Code:

Mod(10,107)^100000000000000000000000000000000000000000
Mod(34, 107)

##
  ***   last result computed in 0 ms

.

It actually works by left-right binary exponentiation modulo n

And since 107 is prime we can use Fermat's little theorem:

Code:

Mod(10,107)^(100000000000000000000000000000000000000000%106)
Mod(34, 107)

Or simething even "bigger":

Code:

Mod(10,107)^(100000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000%106)
Mod(4, 107)

Thanks