ϕ\phind the Remainder

ϕ(n)\phi(n) is the number of positive integers less than nn that are relatively prime to n.n.

Find the remainder when ϕ(22018+1)\phi\big(2^{2018} + 1\big) is divided by 4036.


Bonus: Generalize for the remainder when ϕ(2n+1)\phi(2^n+1) is divided by 2n.2n.

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