Euler's Totient Function Game

ϕ ( n ) \phi(n) ( read phi(n) ) denotes how many positive integers which less or equal to n n that relatively prime with n n . Let m m and x x be the positive integers which satisfy ϕ ( 201 5 2015 ) = m \phi(2015^{2015}) = m and x = n ( A ) x =n(A) which A = { r ϕ ( r ) = 201 5 2015 , r N } A = \{r | \phi(r) = 2015^{2015} , r \in N\} Find the value of m x mx


The answer is 0.

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1 solution

Rimba Erlangga
Jan 13, 2015

Hint: ϕ ( n ) \phi(n) is never gonna be an odd integer

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