A problem I came up with on Chinese New Year

x x is the biggest integer such that x 2018 216 m o d x 216 x^{2018}\equiv216\mod x-216

Enter the last two digits of the value of the following expression: x 216 n = 0 2016 ( 21 6 n ) \frac{x-216}{\sum_{n=0}^{2016}(216^n)}


The answer is 40.

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1 solution

Leonel Castillo
Feb 18, 2018

We have to solve the congruence x 2018 216 m o d x 216 x^{2018} \equiv 216 \mod x - 216 . Notice that x 2018 21 6 2018 x^{2018} - 216^{2018} is divisible by x 216 x - 216 by the difference of powers formula so we can add it or subtract it without changing anything. If we subtract it we get the congruence 21 6 2018 216 m o d x 216 216^{2018} \equiv 216 \mod x - 216 and equivalently, 21 6 2018 216 0 m o d x 216 216^{2018} - 216 \equiv 0 \mod x - 216 . It is obvious that x = 21 6 2018 x = 216^{2018} works and no bigger x x can work because if x > 21 6 2018 x > 216^{2018} then x 216 > 21 6 2018 216 x - 216 > 216^{2018} - 216 , making divisibility impossible. Thus we have deduced that x = 21 6 2018 x = 216^{2018} .

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