First round – City competition

2.5. Let $a_n$ be the number the grasshopper is located at after the $n^{th}$ jump, i.e. $$a_1=1,a_n=1+k+\cdots +k^{n-1},n\ge 2.$$ We are looking for all numbers $k$ such that $2015\nmid a_n$ for all $n=1,\dots ,2015$. Suppose that $M(k,2015)=d>1$. Then every $a_n$ divided by $d$ gives the remainder 1, and since 2015 is divisible by $d$ we have that $2015\nmid a_n$ for all $n$. Therefore, all positive integers which are not relatively prime to 2015 comply with the terms of the problem. If $M(k,2015)=1$, we observe the remainders of dividing $a_1,\dots ,a_{2015}$ by 2015. If one of them is divisible by 2015, such a $k$ is not good. Otherwise, since there are 2014 possible remainders, at least two numbers give the same remainder. Let these numbers be $a_l$ and $a_m$, $m>l$. In this case, their difference is divisible by 2015. On the other hand, we have that $$a_m-a_l=k^l+\cdots +k^{m-1}=k^l(1+\cdots +k^{m-l-1})=k^l\cdot a_{m-l}.$$ From $2015\mid k^l\cdot a_{m-l}$ and $M(k,2015)=1$, it follows that $2015\mid a_{m-l}$, which is in contradiction with the assumption that none of the numbers $a_1,\dots ,a_{2015}$ is divisible by 2015. Therefore, if $M(k,2015)=1$, the grasshopper will jump into a hole. To conclude, the only numbers which are suitable for the terms of the problem are those which are not relatively prime to 2015.