CEMC Pascal, 2016

Among the list 10, 11, 12, 13, 14, 15, the integers 11 and 13 are prime.

Also, $10 = 2 \times 5$ and $12 = 2\times 2 \times 3$ and $14 = 2\times 7$ and $15 = 3 \times 5$.

For an integer $N$ to be divisible by each of these six integers, $N$ must include at least two factors of 2 and one factor each of 3, 5, 7, 11, 13.

Note that $2^2 \times 3 \times 5 \times 7 \times 11 \times 13 = 60\,060$.

(This is the least common multiple of 10, 11, 12, 13, 14, 15.)

To find the smallest six-digit positive integer that is divisible by each of 10, 11, 12, 13, 14, 15, we can find the smallest six-digit positive integer that is a multiple of $60\,060$.

Note that $1 \times 60\,060 = 60\,060$ and that $2 \times 60\,060 = 120\,120$.

Therefore, the smallest six-digit positive integer that is divisible by each of 10, 11, 12, 13, 14, 15 is $120\,120$.

The tens digit of this number is 2.