From the list $\frac{1}{3},\frac{1}{5},\frac{1}{7},\frac{1}{9},\frac{1}{11}$..... is it possible to chose a limited number of terms that sum to one? This can be done with even fractions: $\frac{1}{2},\frac{1}{4},\frac{1}{8},\frac{1}{12},\frac{1}{24}$
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1$\begingroup$ This would be true if an odd perfect number existed :) $\endgroup$– idokCommented Feb 3, 2018 at 19:08
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4$\begingroup$ Yes, that's a known problem. Look here $\endgroup$– user436658Commented Feb 3, 2018 at 19:12
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1$\begingroup$ @ProfessorVector Where exactly is the solution to the question? $\endgroup$– idokCommented Feb 3, 2018 at 19:14
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2$\begingroup$ Typically: $1/3+1/5+1/7+1/9+1/11+1/15+1/35+1/45+1/231 = 1$. $\endgroup$– WatsonCommented Feb 3, 2018 at 19:14
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2$\begingroup$ @Professor Vector: Please consider converting your comment into an answer, so that this question gets removed from the unanswered tab. $\endgroup$– Robert HowardCommented Feb 4, 2018 at 1:51
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1 Answer
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Such a representation of a fraction as the sum of fractions with numerator 1 and different denominators is called Egyption fraction, because that was the way fractions were written in ancient Egypt. It's clear that for 1, we must have an odd number of summands, because otherwise the numerator of the sum would be even and the denominator odd. As it turns out, the minimal number is 9, and there are the following 5 solutions: \begin{align} 1&=\frac13+\frac1{ 5}+\frac1{ 7}+\frac1{ 9}+\frac1{ 11}+\frac1{ 15}+\frac1{ 35}+\frac1{ 45}+\frac1{ 231}\\ 1&=\frac13+\frac1{ 5}+\frac1{ 7}+\frac1{ 9}+\frac1{ 11}+\frac1{ 15}+\frac1{ 21}+\frac1{ 231}+\frac1{ 315}\\ 1&=\frac13+\frac1{ 5}+\frac1{ 7}+\frac1{ 9}+\frac1{ 11}+\frac1{ 15}+\frac1{ 33}+\frac1{ 45}+\frac1{ 385}\\ 1&=\frac13+\frac1{ 5}+\frac1{ 7}+\frac1{ 9}+\frac1{ 11}+\frac1{ 15}+\frac1{ 21}+\frac1{ 165}+\frac1{ 693}\\ 1&=\frac13+\frac1{ 5}+\frac1{ 7}+\frac1{ 9}+\frac1{ 11}+\frac1{ 15}+\frac1{ 21}+\frac1{ 135}+\frac1{ 10395} \end{align} There are also solutions of length 11, 13, 15,..., and it can be shown that every odd length $\ge9$ is possible. This information (and further references) can be found in this article.