This number's third power is the only cube in which three distinct digits each occur three times.
a. Jeopardy style: you provide the question.
b. What else can be said about this special number?
(In reply to
What is... (solution) by Jer)
Actually there are only 181440 ninedigit numbers have three occurrences of three distinct digits.
There are C(10,3) = 120 ways of choosing which 3 digits are used.
The number of ways of arranging the three digits in groups of 3 is 9!/(3!)^3 = 1680.
Allowing all arrangements, including those beginning with zero, gives 120*1680 = 201600.
We need to subtract those beginning with zero (otherwise we'd be choosing from a universe of 1,000,000,000 instead of 900,000,000):
There are C(9,2) = 36 choices of the other two numbers.
There are 8!/(((3!)^2)*2) = 560 ways of arranging the 8 digits past the initial zero for each of the 36 choices.
So we need to subtract 36*560 = 20160 from the initial 201600, leaving 181440, making the quoted expected number of cubes even smaller: .107856.

Posted by Charlie
on 20150303 11:01:33 