is the biggest integer such that
Enter the last two digits of the value of the following expression:
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We have to solve the congruence x 2 0 1 8 ≡ 2 1 6 m o d x − 2 1 6 . Notice that x 2 0 1 8 − 2 1 6 2 0 1 8 is divisible by x − 2 1 6 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 2 1 6 2 0 1 8 ≡ 2 1 6 m o d x − 2 1 6 and equivalently, 2 1 6 2 0 1 8 − 2 1 6 ≡ 0 m o d x − 2 1 6 . It is obvious that x = 2 1 6 2 0 1 8 works and no bigger x can work because if x > 2 1 6 2 0 1 8 then x − 2 1 6 > 2 1 6 2 0 1 8 − 2 1 6 , making divisibility impossible. Thus we have deduced that x = 2 1 6 2 0 1 8 .