anonymous 4 years ago 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 $$10^3$$.

1. Mr.Math

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

2. anonymous

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

3. Mr.Math

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.

4. anonymous

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