Solve this alphametic, where each of the capital letters in bold denotes a different decimal digit from 0 to 9. None of the numbers can contain any leading zero.
3√(HOW)+ 3√(AND) = 3√(WHEN)
(In reply to
re(2): two views by brianjn)
Not quite understanding what an implementation of infinite precision would be, but I think if indeed it were infinite, there'd be a match, as at no point would the lhs and rhs disagree, so a match would be found.
The more precision you have the less chance of false positives. But regardless of the precision there will always be an inaccuracy at the termination of that precision, and the error could go either way, so that the lhs might be a little too low and the rhs too high or vice versa, so long as an exact equality is demanded by the programming.
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Posted by Charlie
on 2009-07-30 22:24:08 |
re(3): two views
