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Square Rods

Problem

Using rods of fixed length 1, 2, 3, ..., n, the smallest value of n for which a square can be formed is n = 7.


Find the values of n for the next two solutions.


Solution

The next solution occurs when n = 8: the perimeter of the square will be, 1 + 2 + 3 + ... + 8 = 36. Therefore one side will be 36/4=9 units long.

It can be seen that 1 must be paired with 8, then 2 must be paired with 7, and so on. Hence we must use the following pairings: (1,8), (2,7), (3,6), (4,5).


In general, the perimeter, P = 1 + 2 + 3 + ... + n, must be divisible by 4.

We note that P, the sum of the first n positive integers, is given by n(n+1)/2. And as the product of two consecutive numbers, n and n+1, will contain one odd factor and one even factor, the even factor must be divisible by 8 in order to be divisible by 4 after being halfed. This explains why n=8 is a solution, and n=7 is a solution because n+1=8. In other words, it is necessary for n or n+1 to be divisible by 8.

Hence the next solution should occur when n=15, as n+1=16; that is, 1 + 2 + ... + 15 = 15times16/2 = 15times8 = 120.

As the side of the square will be 120/4 = 30, we can form a square using the following sets: (15,14,1), (13,12,5), (11,10,9), and (8,7,6,4,3,2).

If n or n+1 is divisible by 8, will you always be able to form a square?

Problem ID: 188 (28 Nov 2004)     Difficulty: 2 Star

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