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Perfect Squares (Posted on 2003-04-05) Difficulty: 5 of 5
Show that the numbers of the form:


[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).

See The Solution Submitted by Ravi Raja    
Rating: 4.0000 (3 votes)

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Solution solution | Comment 2 of 10 |
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
((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
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