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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
by detrites 4y ago
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.
- dmurray 4y ago> Yes, as stated: it ends in 3, so the cube root must be a multiple of 3. I don't get it. 7 cubed is 343. The article explains better what numbers might have cubes ending in 3.
- detrites 4y agoYes, the article is great (now I've read it). 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).
- nigamanth 4y agoNo, he meant that the cube root must be a multiple of 3. 27 is a multiple of 3.
- jstanley 4y agoBut 7 isn't a multiple of 3.
- mtlmtlmtlmtl 4y agoThe difference is that 343 is not a multiple of 3, which excludes its cube root from being a multiple of 3.
- less_less 4y agoRight. 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.