Show that the numbers of the form:

444444....4444888888....8889

[Where there are 'k' Fours, '(k-1)' Eights and 'Exactly One' 9],

are always perfect squares.

(For example the sequence of numbers: 49, 4489, 444889, ....etc. and so on are always perfect squares).

I'm not sure exactly how to prove it, but those numbers are squares of 7, 67, 667, 6667, etc.

The 9 comes from the square of 7, then there will as many 8s as there are 6s in the root, and the remaining digits will all be 3s (from 6²=36) with a carryover digit. So, there are as many 4s in the answer as the rest of the digits; ie, the same number as of 8s plus one for the 9.

That's not much of a proof, just a bunch of observations. I have to collect my thoughts.

I just woke up.

Posted by DJ on 2003-04-05 05:56:26