if the sum of the digits in the number is $2011$ then it isn't divisible by $3$

Marwanshtine wrote: if the sum of the digits in the number is $2011$ then it isn't divisible by $3$ Exactly. I think the question is misstated

No, you guys just misread it. The product of the digits is a power of $6$... Here's my solution (which might not be entirely correct): We want the number to have the fewest possible digits, i.e. the largest possible digits. This can be done by using as many $9$s and $8$s as possible. Additionally, we must use $3$ $9$s per $2$ $8$s, or $43$ at a time. Since $2011=33+46\times 43$, we want to minimize the number of digits (with minimum size) to achieve a sum of $33$. Clearly, $4$ digits is impossible (it would have to be $8889$ or $9996$, neither of which work), so we must have $5$ digits. The only possibility for this is $34899$, so for the final number to be minimized, it must be the following: \[\boxed{3488888\dots8889999\dots9999}\]This number has precisely $46\times 2+1=93$ $8$s and $46\times 3+2=140$ $9$s.

benstein wrote: No, you guys just misread it. The product of the digits is a power of $6$... Here's my solution (which might not be entirely correct): We want the number to have the fewest possible digits, i.e. the largest possible digits. This can be done by using as many $9$s and $8$s as possible. Additionally, we must use $3$ $9$s per $2$ $8$s, or $43$ at a time. Since $2011=33+46\times 43$, we want to minimize the number of digits (with minimum size) to achieve a sum of $33$. Clearly, $4$ digits is impossible (it would have to be $8889$ or $9996$, neither of which work), so we must have $5$ digits. The only possibility for this is $34899$, so for the final number to be minimized, it must be the following: \[\boxed{3488888\dots8889999\dots9999}\]This number has precisely $46\times 2+1=93$ $8$s and $46\times 3+2=140$ $9$s. Yes I misread it but came up with the same solution.