The House in the Middle

On a long street the houses are numbered $1, 2, 3, \dots$ in order. A man living there noticed that the sum of all the house numbers below his own was exactly equal to the sum of all the numbers above it. His house number was even and lay between $100$ and $999$. What was it, and how many houses were on the street?

Solution

Let his house be number $n$ and the last house on the street be $N$. The condition is $$1 + 2 + \dots + (n-1) = (n+1) + (n+2) + \dots + N.$$ Add $1 + 2 + \dots + n$ to both sides and the right becomes the whole sum $1 + \dots + N$, while the left becomes twice $1 + \dots + (n-1)$ plus $n$. Tidying up, the condition is exactly $$n^2 = \frac{N(N+1)}{2},$$ so his house number squared must be a triangular number. A number that is at once a perfect square and triangular is rare; the pairs $(n, N)$ run $$(1,1), \ (6,8), \ (35,49), \ (204,288), \ (1189,1681), \ \dots$$ The only even $n$ between $100$ and $999$ is $204$, paired with $N = 288$. So he lived at house $204$ on a street of $288$ houses. As a check, $1 + \dots + 203 = 20706$, and $205 + \dots + 288 = 41616 - 20910 = 20706$ as well.