Fool's problem of the day,

If $ n $ and $ m $ are natural numbers such that $ 1 + 2 + 3 + … + n = m^2 $, Find the sum of digits of the greatest such $ n $ , smaller than $ 10^3 $.

Question

Fool's problem of the day,

If $ n $ and  $ m $  are natural numbers such that  $ 1 + 2 + 3 + … + n = m^2 $, Find the  sum of digits of the greatest such  $ n $ , smaller than $ 10^3 $.

mr.math · Answer

We're looking for a number $m^2$ that's both triangular and perfect square. $n<10^3 \implies m^2=\frac{n(n+1)}{2}<500500.$ The largest such $m$ is $41616$, that gives $n(n+1)=2(41616) \implies n=288$. I cheated from here http://en.wikipedia.org/wiki/Square_triangular_number

anonymous · Answer

That's interesting MR.Math, however the challenge was a develop a solution that could be implemented without electronic aid ;-)

mr.math · Answer

I know, but I'm so lazy. Plus I know that a list of such numbers exist and there are not so many of them that are less than 500500. Steps to finding these numbers explicitly can be found in the link.

anonymous · Answer

lol, Mr.Math this is a quantitative aptitude problem, think elementary ;-)