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).

Thank to DJ's noting that the square roots are in the form 6...67, the proof can proceed from there forward to the squares.

First note that 10^n-1 is a number represented by n 9's, so (10^n-1)/9 is represented by n 1's, and ((10^n-1)/9)d is represented by n occurrences of the digit d.

So let's start with a number in the form DJ found:
(((10^n-1)/9)6 + 1), where the +1 changes the last 6 to a 7. The square of this number is given by

((10^(2n)-2(10^n)+1)/81)36 + ((10^n-1)/9)12 + 1

Then as 36/81=4/9, this becomes

((10^(2n)-2(10^n)+1)/9)4 + ((10^n-1)/9)12 + 1

Then as we wish to combine the terms involving 10^n together, we make this

((10^(2n)-2(10^n-1)-1)/9)4 + ((10^n-1)/9)12 + 1

Then, actually moving those over,

((10^(2n)-1)/9)4 + ((10^n-1)/9)(12-(2)(4)) + 1
or
((10^(2n)-1)/9)4 + ((10^n-1)/9)4 + 1

But ((10^(2n)-1)/9)4, as mentioned at the beginning, is represented in the decimal notation by 2n 4's, while ((10^n-1)/9)4 is represented by n 4's. These two terms then represent a 2n-digit number of all 4's plus an n-digit number of all 4's, therefore converting the second half of the string to all 8's. The unit term at the end changes the last 8 to a 9.

This accounts for any value of n (or k in the statement of the puzzle).

Posted by Charlie on 2003-04-05 13:42:37