The answer is 2019.

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his is the definition of Riemann zeta function.

$\zeta (s)=\sum_{n=1}^{\infty}\frac{1}{n^s} \ (*)$

here $s=2$

$gcd(n,2019)$ can be either $1,3,673,2019$

$\sum_{n=1}^{\infty}\frac{gcd(n,2019)}{n^2}=\sum_{gcd(n,2019)=1}\frac{1}{n^2}+\sum_{gcd(n,2019)=3}\frac{3}{n^2}+\sum_{gcd(n,2019)=673}\frac{673}{n^2}+\sum_{gcd(n,2019)=2019}\frac{2019}{n^2}=$

$\sum_{gcd(n,2019)=1}\frac{1}{n^2}+\sum_{gcd(k,673)=1}\frac{3}{(3k)^2}+\sum_{gcd(n,3)=1}\frac{673}{(673k)^2}+\sum_{k}\frac{2019}{(2019k)^2}=$

$\zeta(2)+\sum_{gcd(k,673)=1}\frac{3-1}{(3k)^2}+\sum_{gcd(n,3)=1}\frac{673-1}{(673k)^2}+\sum_{k}\frac{2019+1}{(2019k)^2}=$

$\zeta(2)+\frac{3-1}{3^2}\zeta(2)+\frac{673-1}{673^2}\zeta(2)+(\sum_{k}\frac{2019+1}{(2019k)^2}-\sum_{k}\frac{3-1}{(2019k)^2}-\sum_{k}\frac{673-1}{(2019k)^2})=$

$\zeta(2)+\frac{3-1}{3^2}\zeta(2)+\frac{673-1}{673^2}\zeta(2)+\frac{2019-3-673+3}{2019^2}\zeta(2)$

after summing up, one realises that the fraction is not reducible and, therefore, the denominator is $2019^2$ .

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2 Interesting
0 Brilliant
0 Confused

Yes, that is roughly what I had in mind; thank you!

There might be a typo where you write $2019+1$ in the numerator; isn't it $2019-1$ ?

Otto Bretscher
- 2 years, 5 months ago

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I think it is correct. Sorry if I didn't explain the steps well. The +1 comes from the two summations on the left, because multiples of 2019 had been counted twice and one of them would be added to 2019. Please investigate further and advise me accordingly.

A Former Brilliant Member
- 2 years, 5 months ago

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It seems to me that you need a minus 1. You get one summand $\frac{1}{2019^2}$ from $\zeta(2)$ , which you need to subtract. Multiples of 2019 are not counted in the two other sums because of the $\gcd$ conditions.

I must confess, though, that I'm currently vacationing in the Caribbean, so, I'm not really in a "mathy" mode ;)

Otto Bretscher
- 2 years, 5 months ago

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@Otto Bretscher – Haha, I know it is such a dilemma (same for me). It is either a "nice vacation" or a "logical mind", not both. Hope you are having fun there.

A Former Brilliant Member
- 2 years, 5 months ago

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@A Former Brilliant Member – If you write the expression as $\sum_{k=1}^{\infty}\frac{1}{k^2}+\sum_{k=1}^{\infty}\frac{2}{(3k)^2}+\sum_{k=1}^{\infty}\frac{672}{(673k)^2}+\sum_{k=1}^{\infty}\frac{a}{(2019k)^2}$ and consider the denominator $2019^2$ , you see that $1+2+672+a=2019$ , so $a=2019-1-2-672$ . (I wrote up a brief solution.)

Otto Bretscher
- 2 years, 5 months ago

How do we find the value of reimann zeta function for 2 ?? Also, how did u write the symbol for that (it's not in my phone's keyboard)??

Mr. India
- 2 years, 4 months ago

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honestly, I do not know how to find the value of Riemann zeta function, but I think there is a table for that and the calculation is done numerically. About the symbol, all keyboards in Germany has the symbol. You do not have it? :)

A Former Brilliant Member
- 2 years, 4 months ago

The values of the Riemann zeta function for even integers is given by: zeta(2n) = (-1)^(n + 1)* (B
*
2n) *(2*pi)^(2n)/[2(2n)!], where the B
*
2n are the Bernoulli numbers. The first few are zeta(2) = (pi^2)/6, zeta(4) = (pi^4)/90, and zeta(6) = (pi^6)/945.

Edwin Gray
- 2 years, 3 months ago

Do you think my solution is correct? It seems there is a high probability that the nominator and the denominator of the fraction are coprimes. That is why many people (including me) got the answer right. I appreciate if you have a look at the solution, taking Otto's comments into account.

A Former Brilliant Member
- 2 years, 5 months ago

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Write $S=\sum_{k=1}^{\infty}\frac{1}{k^2}+\sum_{k=1}^{\infty}\frac{a}{(3k)^2}+\sum_{k=1}^{\infty}\frac{b}{(673k)^2}+\sum_{k=1}^{\infty}\frac{c}{(2019k)^2}$ , with $a,b,c$ to be determined.

If $\gcd(n,2019)=3$ then $1+a=3$ so $a=2$ . If $\gcd(n,2019)=673$ then $1+b=673$ so $b=672$ . If $\gcd(n,2019)=2019$ then $1+a+b+c=2019$ so $c=1344$ . Thus $S=\left(1+\frac{2}{3^2}+\frac{672}{673^2}+\frac{1344}{2019^2}\right)\zeta(2)=\left(1+\frac{2}{3^2}\right)\left(1+\frac{672}{673^2}\right)\zeta(2)=\frac{4989611}{2019^2}\zeta(2)$ . The answer is $\boxed{2019}$ .