Problem 343

We note that $20=2^{2} \cdot 5$ and $16=2^{4}$ and $2016=16 \cdot 126=2^{5} \cdot 3^{2} \cdot 7$. For an integer to be divisible by each of $2^{2} \cdot 5$, $2^{4}$, and $2^{5} \cdot 3^{2} \cdot 7$, it must include at least 5 factors of 2, at least 2 factors of 3, at least 1 factor of 5, and at least 1 factor of 7. The smallest such positive integer is $2^{5} \cdot 3^{2} \cdot 5^{1} \cdot 7^{1}=10080$. The tens digit of this integer is 8.