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New Scientist Enigma 501 – A Reciprocal Arrangement

by Christophe Maslanke

From Issue #653, 25th February 1989

“As you insist on disturbing my peace of mind with puzzles.” remarked Potter to Kugelbaum as they sat down to drinks at the Maths Club. “It is only fair that you submit to the same fate.” Kugelbaum agreed. “The Egyptians expressed fractions as sums of reciprocals.” continued Potter. “For example, they wrote 3/8 = 1/8 + 1/4. That and the notion of getting one over you inspires this puzzle.”

“The smallest integer, U such that 1/U may be expressed as the sum of exactly two reciprocals in exactly and only two distinct ways is 2: 1/2 = 1/4 + 1/4 and 1/2 = 1/3 + 1/6 and there are no other ways of doing it. The second smallest is 3, since 1/3 may be expressed as the sum of exactly two reciprocals in exactly and only two distinct ways: 1/3 = 1/6 + 1/6 and 1/3 = 1/4 + 1/12. Again there are no other ways. But the third smallest value of U is not 4, since 1/4 may be expressed as the sum of two reciprocals in three distinct ways.”

“Yes.” said Kugeibaum in reply. “namely 1/8 + 1/8 + 1/12, 1/6 + 1/12 and 1/5 + 1/20. What is your question”

Potter drew himself up: “What is the eighth smallest value of U. such that 1/U is expressible as the sum of exactly and only two reciprocals in exactly and only eight distinct ways?”

Kugelbaum’s eyes glazed over and the cogs began to whir. In fact, Potter didn’t even know if the question had an answer and so when Kugelbaum gave the answer, he had to take it on trust. Given that Kugelbaum is never wrong, what was his answer?

2 Comments Leave one →
  1. Brian Gladman permalink

  2. Frits permalink

    Based on above program. I noticed recursion is a bit of overkill for l=2.

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