Bonkers Tetration

Computer Science Level 5

Find the sum of all positive integers n n less than 100 satisfying the congruence

2 3 4 ⋅ ⋅ ⋅ n m o d ( n + 1 ) = 1 . \Large 2^{3^{4^{\cdot^{\cdot^{\cdot^ n} }}}} \bmod {(n+1)} = 1 \; .


The answer is 1473.

Pi Han Goh by Pi Han Goh , 30, Malaysia

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1 solution

We want the sum of all odd integers n n in the interval [ 2 , 100 ] [2,100] such that ϕ ( n ) \phi(n) is divisible by 3. 

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totient :: Int -> Int
totient 1 = 1
totient a = length $ [ b | b <- [1..a-1], gcd a b == 1]

answer :: Int
answer = sum $ [n | n <- [2..100], odd n, (totient n) `mod` 3 == 0]

main = print answer

1 Helpful 0 Interesting 0 Brilliant 0 Confused

n n should be even.

Abdelhamid Saadi - 4 years, 4 months ago

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