2860.11 – Integer Sum of Unknown Length!


Find the integer nn such that 1+2+3++n3n=36\dfrac{1 + 2 + 3 + \ldots + n}{3n} = 36.


Solution

You're supposed to know that 1+2+3++n2=n(n+1)2\dfrac{1 + 2 + 3 + \ldots + n}{2} = \dfrac{n(n+1)}{2}.

So all we need to do is set n(n+1)6n=36\dfrac{n(n+1)}{6n} = 36, and cross-multiply to get n2+n=216nn^2 + n = 216n, a quadratic equation with roots 00 and 215215. The latter root is the answer.