Now that's huge

Number Theory Level 3

Find the last two digits of

$\large 403^{(10^{10}+6)}$ .

81 27 03 29 43 09

1 solution

Ken Mercado
Nov 22, 2017

Use euler's theorem to reduce exponent to 6 because tot(100) = 40 and 10^10 + 6 mod 40 = 6. So the answer is just the last two digits of 3^6 which is 29

Now that's huge

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