Find a six-digit multiple of 64 that consists only of the digits 3 and 2.

Answer 1

The prime factorization of 64 is $2^6$. Therefore, a multiple of 64 must have at least six factors of 2. We see that $2^6=64$ and $2^7=128$. To begin, we put the number $2$ in the ten-thousands place (the left-most digit) since we want a six-digit number of the form $2yyyyy$. Now we have five spaces left to fill with two's and three's with the constraint that there must be at least five of them:

$$2\textbf{\_}\textbf{\_}\textbf{\_}\textbf{\_}\textbf{\_}$$Since there can't be six two's (because there is only one digit $2$ remaining), there must be five two's. So we have:

$$22\textbf{\_}\textbf{\_}\textbf{\_}\textbf{\_}$$The number can't end in a two, so it must end in three.

$$223\textbf{\_}\textbf{\_}\textbf{\_}$$Since a multiple of 64 must have at least two factors of 8, the last three digits must be a multiple of 8. The only such digit remaining is 3, so we have:

$$22383$$Therefore, $\boxed{22,\!383}$ is a six-digit multiple of 64 using only the digits 3 and 2.