jmc

If $n$ is divisible by both 2 and 5, then we can write $n$ in the form $10^a\cdot 2^b$ or $10^a\cdot 5^b$, where $a$ and $b$ are positive integers. Since $10^a$ simply contributes trailing zeros, we can continue dividing by 10 until $n$ is a power of two or a power of 5. We generate a list of powers of 2. $$\begin{array}{rl}{2}^1& =2\\ 2^2& =4\\ 2^3& =8\\ 2^4& =16\\ 2^5& =32\\ 2^6& =64\\ 2^7& =128\\ 2^8& =256\\ 2^9& =512\\ 2^{10}& =1024\\ 2^{11}& =2048\\ 2^{12}& =4096\end{array}$$Therefore, we can conclude that $n\le 4096$. Looking at the powers of 5, we note that the first five powers of five do not contain the digit 9, and since $5^6=15625$, the smallest integer that works is $n=\boxed{4096}$.