HMMT November

It is not difficult to see that for all of the logarithms to be non-integers, they must lie strictly between $n$ and $n+1$ for some integer $n$. Therefore, we require $b^{n+1}-b^{n}>2018$, and so $m_{b}=b^{n}+1$ where $n$ is the smallest integer that satisfies the inequality. In particular, this means that $b^{n}-b^{n-1} \leq 2018$. Note that $m_{2}=2^{11}+1=2049\left(\right.$ since $\left.2^{12}-2^{11}=2048>2018\right)$ and $m_{3}=3^{7}+1=2188$ (since $3^{8}-3^{7}=4374>2018$ ). we now show that 2188 is the maximum possible value for $m_{b}$. If $n=0$, then $m_{b}=1+1=2$. If $n=1$, then $b-1 \leq 2018$ and thus $m_{b}=b+1 \leq 2020$. If $n=2$, then $b^{2}-b \leq 2018$, which gives $b \leq 45$, and thus $m_{b}=b^{2}+1 \leq 2018+b+1 \leq 2065$. If $n=3$, then $b^{3}-b^{2} \leq 2018$, which gives $b \leq 12$, and thus $m_{b}=b^{3}+1 \leq 12^{3}+1=1729$. If $n=4$, then $b^{4}-b^{3} \leq 2018$, which gives $b \leq 6$, and thus $m_{b}=b^{4}+1 \leq 6^{4}+1=1297$. It then remains to check the value of $m_{4}$ and $m_{5}$. Indeed, $m_{4}=4^{5}+1=1025$ and $m_{5}=5^{4}+1=626$, so no values of $m_{b}$ exceeds 2188.