The 3-digit perfect cubes are:
8-character perfect cubes range from
216^3 = 10077696
464^3 = 99897344
so from "TEN positive perfect cubes not more than THOUSAND", T could at most be 4, but no possibility for TEN (from being 2 away from a perfect cube) can allow T=4, so at most T could be 3.
8-digit perfect cubes beginning with a 3 range from
311^3 = 30080231
341^3 = 39651821
T can't be a 1, as there would be more than 198 cubes under the smallest 8-digit number.
The possibilities for TEN, are thus 214, 218, 341 and 345.
But 214^3 begins with a 9 and 218^3 begins with a 1, so both are eliminated.
And 345^3 begins with a 4. This leaves 341^3 = 39651821.
Now, THOUSAND need not be a perfect cube, and indeed 39651821 does not have the appropriate tens-position digit to match the 1 in 341 for TEN. So THOUSAND can be a larger number so long as the first digit remains a 3 (which changes before getting to the next perfect cube), so the second digit (the H in THOUSAND) is still a 9.
So THOUSAND is above 39651821, has no repeated digits, and is less than 40000000. There are thus allowable numbers where T=3, H=9, N=1, with E remaining 4.
So THEN = 3941.
Posted by Charlie
on 2008-11-12 13:45:13