This problem begs to be solved by brute force, but I was just wondering what it would look like if we tried to solve it analytically? I gave it a start, not sure how much further I could go with it.
It's obvious to start that we need to find n such that n^2 contains 4 digits and n^3 contains 6 digits. The cube root of 100,000 is ~46.4 so n must be greater than 46 and less than 100.
The sum of all digits from 0 through 9 is 0 mod 3, so n cannot be 1 mod 3, since n^2 and n^3 would both be 1 mod 3 and then the sum of their digits would be 2 mod 3.
n cannot end in a 0, 1, 5 or 6 since n^2 and n^3 would each end in the same digit.
We've already ruled out a pretty large chunk of numbers. From there I started looking at the leftmost digits of the square roots of thousands (1000, 2000, etc) and the cube roots of hundred-thousands. We can rule out 89 (they'd both start with 7), 93-94 (they'd both start with 8) and 97-99 (they'd all start with 9).
Similarly we can rule out n = 68 since n^2 would start and end with a 4, and n = 74 since n^3 would start and end with a 4.
We can rule out 48, 57, and 92 because n^3 would end in the same digit that n^2 starts with. We can rule out 78 because n^2 would end in the same digit that n^3 starts with.
Now the list of potential candidates has been whittled down to the following 12:
At this point I don't know if there's a good way to continue pruning the list, though at least it's now at a length you could feasibly brute-force by hand as opposed to using software.
Posted by tomarken
on 2021-04-07 08:05:55