Euler all around!

Let ϕ ( n ) \phi(n) denote the number of positive integers which are less than or equal to n n and relatively prime to n n .Let d ( n ) d(n) denote the number of positive integral divisors of n n .Find the sum of all odd integers n 5000 n\le 5000 such that n ϕ ( n ) d ( n ) n| \phi(n)d(n) .


The answer is 2903.

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1 solution

Bob Kadylo
May 17, 2018

The set of solutions for n n is {1, 9, 243, 625, 2025} and the sum is 2903 \boxed {2903}

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