Remind Me of the Remainder 2

Find the remainder if ( 2001 ) 2012 (2001)^{2012} is divided by 106 106 .


The answer is 95.

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1 solution

The number is 1 mod 2 . 2001 is -13 mod 53. And 2012 is 36 mod 52. So by fermat's little theorem. The number is 13^36 mod 53. Which is 10^18 mod 53. Which is -6^9 mod 53. Which is -4^3 mod 53 . Which is -11 mod 53 Which is 42 mod 53. So the number which is both 1mod2 and 42 mod 53 is (53+42) = 95. So the number is 95 mod 106. Or the remainder is 95 when divided by 106. I used the concept of Chinese remainder theorem to split 106 into 2 and 53 (which are both coprime). And then simultaneoualy solve the two congruences

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