THE 68th ROMANIAN MATHEMATICAL OLYMPIAD

a) Since $7n=7^2+7^3+7^4+\dots +7^{2017}+7^{2018}$, we have $6n=7n-n=7^{2018}-7$. It follows that $7^{2018}=6n+7=6(n+1)+1$, and the remainder of the division by $6$ of the number $7^{2018}$ is $1$.

Also, $n=7+7^2(1+7)+7^4(1+7)+\dots +7^{2016}(1+7)=7+8\cdot (7^2+7^4+\dots +7^{2016})=7+8p$, where $p=7^2+7^4+\dots +7^{2016}$. Hence, $7^{2018}=6(8p+7)+7=48p+42+7=48p+49=48(p+1)+1$, wherefrom we conclude that the remainder of the division of $7^{2018}$ by $48$ is $1$.

b) We have $6n=7^{2018}-7$. Denoting by $u_2(x)$ the number built with the last two digits of the number $x$, we have $u_2(7^{4k})=01$, $u_2(7^{4k+1})=07$, $u_2(7^{4k+2})=49$, $u_2(7^{4k+3})=43$.

Then $u_2(7^{2018})=u_2(7^{4\cdot 504+2})=49$. Hence $u_2(6n)=u_2(49-7)=42$.