59th Ukrainian National Mathematical Olympiad

Suppose Arsenii's claim is correct and by changing the last digit of number $N$ he obtained $13^k$, where $k$ is a positive integer. Clearly, he had to change the last digit, because it is $5$. Since $2^{100}\equiv 1(\mathrm{mod}3)$, $3^{100}\equiv 0(\mathrm{mod}3)$ and $5^{100}\equiv 1(\mathrm{mod}3)$, the sums of their digits have the same remainders modulo $3$. The change of the last digit $5$ to digit $l$ means that we subtracted $2$ (or added $1$) and added $l$ modulo $3$. Thus, the new number (denoted by $M$) equals $l\equiv 0(\mathrm{mod}3)$. Since $13^k\equiv 1(\mathrm{mod}3)$, we could add one of the following digits: $1$, $4$ or $7$. Since the last digit of $13^k$ can be $1$, $3$, $7$ or $9$, we need to consider two cases. Case 1. $13^k$ ends with $1$, it is possible in the case $k\equiv 0(\mathrm{mod}4)$, then the following is true modulo $8$: $M=13^k=13^{4j}=169^{2j}\equiv 1(\mathrm{mod}8)$. Note that $M=N-4$. $N\equiv 5^{100}=25^{50}\equiv 1(\mathrm{mod}8)\Rightarrow M=N-4\equiv 5(\mathrm{mod}8)$, so we get a contradiction. Case 2. $13^k$ ends with $7$, it is possible if $k\equiv 3(\mathrm{mod}4)$, then the following is true modulo $8$: $M=13^k=13^{4j+3}=169^{2j}\cdot 13\equiv 5(\mathrm{mod}8)$. Note that $M=N+2\Rightarrow M\equiv 3(\mathrm{mod}8)$ – contradiction. These contradictions complete the proof.