# What was Ramanujan's solution?

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.

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The problem reduces to a Pell equation whose solutions can be obtained from the continued fraction expansion of $\sqrt{2}$. –  André Nicolas May 19 '12 at 20:32

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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See also math.auckland.ac.nz/~butcher/miniature/miniature2.pdf an article by John Butcher, where he guesses that Ramanujan was thinking of a (more general type of) continued fraction namely $$3 - \cfrac{1}{6 - \cfrac{1}{6 - \cfrac{1}{6 - \dots}}}$$. –  ShreevatsaR Dec 13 '12 at 10:00