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).

This also goes from the square roots up, but can be worked the reverse way as well.

Since (a+1)² =a² +2a+1, 7² =6² +2(6)+1, 67² = 66² +2(66)+1, 667² = 666² + 2(666) + 1,

Since (x 6s)² =( x 4s then x 5s) - x 9s (Or, (x-1) 4s, then a 3, then (x-1) 5s, then a 6), adding 2(x 6s ) will just change the 3 to a 4 (adding 1), the 5s to 8s (adding 3) and the 6s to 8s (adding 2). You get (x 4s then x 8s) as a result, and adding 1 to the end gives 9 at the end instead of an 8. So you get (x 4s, then (x-1) 8s, then a 9) which is what the problem asks for.

I haven't found how to prove the 36, 4356, 443556, 44435556 sequence, which is the next problem with this.

Posted by Gamer on 2003-04-06 05:03:37