, 2020

After the elimination of the 64 digits the final number has $73 - 64 = 9$ digits.
If in the composition of $A$, every digit appears at most 8 times, then $A$ would have at most $9 \cdot 8 = 72$ digits, which is false.
Therefore, there is a digit different from 0, let that be $a$, that appears 9 times in $A$. Then we can erase 64 digits from $A$ such that the final number would have the form $\overline{aaaaaaaaa} = a \cdot 111111111 = a \cdot 1001001 \cdot 3 \cdot 37$, so the number is divisible by 37.