MIT Mathematics Tournament (II)

Algebra Level 3

Find the largest positive integer $n$ such that $1+2+3+\dots+n^2$ is divisible by $1+2+3+\dots+n.$

The answer is 1.

1 solution

Brian Moehring
Aug 7, 2018

Note that $\frac{1+2+3+\cdots +n^2}{1+2+3+\cdots + n} = \frac{\frac{n^2(n^2+1)}{2}}{\frac{n(n+1)}{2}} = \frac{n(n^2+1)}{n+1} = n^2 - n + 2 - \frac{2}{n+1}$

It follows that $1+2+3+\cdots + n^2$ is divisible by $1+2+3+\cdots + n$ if and only if $2$ is divisible by $n+1$ . Under this condition, $n$ is maximized when $n+1 = 2 \iff n=\boxed{1}$

Brian, the expression you used for the sum of the first n squares is not correct. That expression is for the sum of the first n cubes. I, too, when I first looked at the problem, thought it was cubes, but a careful look showed that I was wrong. Ed Gray

Edwin Gray - 2 years, 9 months ago

The sum of the first $n$ squares never appears in this problem. I'm guessing you're talking about $1 + 2 + 3 + \cdots + n^2 = \sum_{k=1}^{n^2} k = \frac{n^2(n^2+1)}{2}$ which is the sum of all the positive integers less than or equal to $n^2$ , not only the squares.

Brian Moehring - 2 years, 9 months ago

Brian, you are absolutely correct. Thes old eyes have failed me. Best Regards, Ed

Edwin Gray - 2 years, 9 months ago

MIT Mathematics Tournament (II)

The answer is 1.

1 solution

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