Ramanujan’s Solution – House Number Problem Explained

algebra-precalculuscontinued-fractionselementary-number-theorymath-historyproblem solving

The wikipedia entry on Ramanujan contains the following passage:

One of his remarkable capabilities was the rapid solution for
problems. He was sharing a room with P. C. Mahalanobis who had a
problem,

Imagine that you are on a street with houses marked $1$
through $n$. There is a house in between $(x)$ such that the sum of the
house numbers to left of it equals the sum of the house numbers to its
right. If $n$ is between $50$ and $500$, what are $n$ and $x$?

This is a
bivariate problem with multiple solutions. Ramanujan thought about it
and gave the answer with a twist: He gave a continued fraction. The
unusual part was that it was the solution to the whole class of
problems. Mahalanobis was astounded and asked how he did it. "It is
simple. The minute I heard the problem, I knew that the answer was a
continued fraction. Which continued fraction, I asked myself. Then the
answer came to my mind," Ramanujan replied.

What was the continued fraction, and how did it give all solutions to the problem? Most importantly, how could someone derive such a solution? Do similar problems also have continued fractions that describe all solutions?

This seems like an interesting and powerful method, and I would like to learn more about it.

Best Answer

An expansion of Andre's comment into a detailed exposition can be found at http://www.johnderbyshire.com/Opinions/Diaries/Puzzles/2009-06.html. It's also at http://www.angelfire.com/ak/ashoksandhya/winners2.html#PUZZANS4.

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