Yes, as stated: it ends in 3, so the cube root must be a multiple of 3.
The "cube root" part contains the implicit information that we have a product ending in 3, of three identical factors.
I guess the missing assumption is that given the context of solving the cube root in ones head, it would more likely be a trick that only works with integers, and as such the correct answer might be expected to be one.
37*3 also ends in 3 but is not divisible by 3. Thus, were the range not capped at 30 it may have been illegitimately discarded by the criteria.
I think the 3 combined with the 5 digits and 20-30 range, made my guess a bit lucky. It's also contrived for that cube, so not useful for others, other than perhaps as a direction to go in (and likely arrive at the article).
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
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.
The "cube root" part contains the implicit information that we have a product ending in 3, of three identical factors.
I guess the missing assumption is that given the context of solving the cube root in ones head, it would more likely be a trick that only works with integers, and as such the correct answer might be expected to be one.