 All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars  perplexus dot info  Phrase a question II (Posted on 2015-03-03) 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. Comments: ( Back to comment list | You must be logged in to post comments.) re: What is... (solution) Comment 2 of 2 | (In reply to What is... (solution) by Jer)

Actually there are only 181440 nine-digit 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 2015-03-03 11:01:33 Please log in:

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