Right. The correct argument is that 19683 is a *multiple* of 3, because its digit sum 27 is a multiple of 3. So if it's a perfect cube, then the cube root mus

Right. The correct argument is that 19683 is a *multiple* of 3, because its digit sum 27 is a multiple of 3. So if it's a perfect cube, then the cube root must be a multiple of 3. https://news.ycombinator.com/item?id=34652053 https://news.ycombinator.com/item?id=34652053 Also, if it's a perfect cube, since it ends in 3, the cube root must end in 7 (cubing integers is 1:1 on the units digit, and 7->3). So it must be 27, 57, 87, ... Since it's too small for the cube root to be 57, 27 is the answer.