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$11$.
Others: 4 = $2^2$, 6 = $2\cdot3$, 10 = $2\cdot5$, 12 = $2^2\cdot3$. So highest powers:
\mathrm{lcm} = 2^3 \cdot 3^2 \cdot 5 \cdot 7 \cdot 11 = 8 \cdot 9 \cdot 5 \cdot 7 \cdot 11.
$8 \cdot 9 = 72$,
$72 \cdot 5 = 360$,
$360 \cdot 7 = 2520$,
$2520 \cdot 11 = 27720$.
So $\mathrm{lcm} = 27720$.
Now find smallest $k$ such that $27720k > 1,\!000,\!000$:
k > \frac{1,\!000,\!000}{27720} \approx 36.06.