#
Problem
of the Week

Problem
E and Solution

All
Square

## Problem

The positive multiples of \(3\) from
\(3\) to \(2400\), inclusive, are each multiplied by
the same positive integer, \(n\). All
of the products are then added together and the resulting sum is a
perfect square.

Determine the value of the smallest positive integer \(n\) that makes this true.

## Solution

What does the prime factorization of a perfect square look like?
Let’s look at a few examples: \(9=3^2\), \(36=6^2=2^23^2\), and \(129\,600=360^2=2^63^45^2\). Notice that the
exponent on each of the prime factors in the prime factorization in each
of the three examples is an even number. In fact, a positive integer is
a perfect square exactly when the exponent on each prime in its prime
factorization is even. Can you convince yourself that this is true?

The positive integer \(n\) is the
smallest positive integer such that \[3n+6n+9n+\cdots + 2394n + 2397n+2400n
\tag{1}\]

is a perfect square.

Factoring expression \((1)\), we
obtain \[3n(1+2+3+\cdots+798+799+800)\] Then, using
the formula \(1 + 2 + 3 + \cdots + n =
\dfrac{n(n+1)}{2}\), with \(n=800\), we see that this expression is
equal to \[3n\left(\frac{800\times
801}{2}\right)=3n(400)(801)\] Factoring \(3\times 400\times 801\) into the product of
primes, we have that expression \((1)\)
is equal to \[3n[(2)(2)(2)(2)(5)(5)][(3)(3)(89)]=n(2^4)(5^2)(3^3)(89)\tag{2}\]

We need to determine what additional factors are required to make the
quantity in expression \((2)\) a
perfect square such that \(n\) is as
small as possible. In order for the exponent on each prime in the prime
factorization to be even, we need \(n\)
to be \(3\times 89=267\). Then the
quantity in expression \((2)\) is the
perfect square \[n(2^4)(5^2)(3^3)(89)=(3)(89)(2^4)(5^2)(3^3)(89)=(2^4)(5^2)(3^4)(89^2)=\left[(2^2)(5)(3^2)(89)\right]^2\]

Therefore, the smallest positive integer is \(267\) and the perfect square is \[267\times 3\times 400\times
801=256\,640\,400=(16\,020)^2\]

Note: When solving this
problem, we could have instead noticed that \(3n + 6n + 9n+ \cdots + 2400n\) is an
arithmetic series with \(t_1 = 3n\) and
\(t_{800}=2400n\).

Substituting these values for \(t_1\) and \(t_{800}\) into the formula for the sum of
the terms in an arithmetic series, we get \[S_{800} = \frac{800}{2}\left( 3n + 2400n\right) =
400(2403n)\] When we factor \(400(2403n)\) into the product of primes, we
get the same expression as \((2)\), and
then we can continue from there to get the solution of \(267\).