Saudi Arabia Mathematical Competitions

For $j=1,2,\dots ,p-1$, we have $$\frac{j^{p-1}-1}{p}=a_{j}p+r_j$$ for some integer $a_j$. It follows $$\frac{j^p-j}{p}=j{a}_{j}p+j{r}_j,$$ hence $$\frac{j^p-j+(p-j{)}^p-(p-j)}{p}=j{a}_{j}p+j{r}_j+(p-j)a_{p-j}p+(p-j)r_{p-j}.$$ We obtain $$\frac{j^p+(p-j{)}^p}{p}=j{a}_{j}p+j{r}_j+(p-j)a_{p-j}p+(p-j)r_{p-j}+1.$$ Because $$j^p+(p-j{)}^p=(\genfrac{}{}{0ex}{}{p}{0})p^p-(\genfrac{}{}{0ex}{}{p}{1})p^{p-1}j+\dots +(\genfrac{}{}{0ex}{}{p}{p-1})p{j}^{p-1}$$ we obtain that $p^2\mid j^p+(p-j{)}^p$ and we get for all $j=1,2,\dots ,p-1$, $$j{r}_j+(p-j)r_{p-j}+1\equiv 0(\mathrm{mod}p)$$ Adding up all these relations it follows $$2\left(r_1+2r_2+\dots +(p-1)r_{p-1}\right)\equiv -(p-1)(\mathrm{mod}p)$$ hence $$r_1+2r_2+\dots +(p-1)r_{p-1}\equiv \frac{p+1}{2}(\mathrm{mod}p)$$