Can you count?

For convenience, we can have the numbers 0, 1, 2, ..., 9 instead be expressed as 0,0,1,1,...,4,4 in mod 5. Then we can bash super hard to get the following cases: 01234, 00113, 00221, 00334, 00442, 11224, 11332, 11440, 22330, 22443, 33441. The first case has 5!x32 ways, because there are 5! ways to order it, and also we must multiply by 2^5 because either 0 or 5 can be 0. either 1 or 6 can be 1, etc etc. The next 10 cases each have 5!x2 ways. So the answer is 5!x52=6240

There are 52 ways to choose a set $A\subset\{0,\ldots,9\}$ in order that $|A|=5$ and $\sum A\equiv 0\pmod{5}$ , since, once set $\xi=\exp\left(\frac{2\pi i}{5}\right)$ : $\frac{1}{5}\sum_{k=0}^{4}[y^5]\prod_{j=0}^{9}(1+ y (x \xi^k)^j) = 52,$ so the answer is just $52\cdot 5! = 6240$ .

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An alternative easy solution :D

#include<iostream>
#include<stdio.h>
using namespace std;
int main()
{int conta=0;
for(int a=0;a<10;a++)
   for(int b=0;b<10;b++)
      for(int c=0;c<10;c++)
         for(int d=0;d<10;d++)
            for(int e=0;e<10;e++)
               if((a+b+c+d+e)%5==0 && a!=b && a!=c && a!=d && a!=e && b!=c && b!=d && b!=e && c!=d && c!=e && d!=e) conta++;
               cout<<conta;
 system("pause");
return 0;
} 

conta=6240 :D