# 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 $$267$$.