More unusual π \pi connections.

Number Theory Level 3

In honor of the coprime problems , consider the case where you pick four positive integers n 1 , n 2 , n 3 , n 4 n_1, n_2, n_3, n_4 at random. What is the probability that they don't share any common divisors? (1 doesn't count, as usual.)


The answer is 0.9239.

Max Yuen by Max Yuen , 42, USA

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1 solution

It's the reciprocal of Riemann Zeta Function for argument 4, which is equal to 90/(π^4)=0.9239

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Can you also show the steps in computing it? Thanks!

Max Yuen - 2 years, 1 month ago

I can't use Latex, and hence the symbols for zeta etc. It's the reciprocal of the sum of terms of the form (1/n^4), being the product of the terms of the form (1-(1/(p^4))), where p's are primes, 2,3,5,7,11 etc.

A Former Brilliant Member - 2 years, 1 month ago

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