We have:

12 = 3 x 4 

So T \(⋮12\) <=> T \(⋮3and4\)

T \(⋮3\Leftrightarrow\)The sum of all the number of T \(⋮3\)

\(\Leftrightarrow\left(1+1+1+...+1+0+...+0\right)⋮3\)

But T's sum must be different than 0 

=> The smallest sum must be 3 = 1 + 1 + 1 + 0 + 0 + ... + 0

T divided by 4 <=> Two-last digit must be divided by 4

=> Two last digits must be 00

=> T = 11100

It's too big :)