Find all three distinct natural numbers such that sum of their reciprocals is 1.
$1/2+1/3+1/6=1$ is one such solution.
2026-05-05 15:47:53.1777996073
Find all three distinct natural numbers such that sum of their reciprocals is 1.
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That's all there is.
Note that the smallest can't be 1 (no room for the other two reciprocals), and it must therefore be 2 (otherwise the sum would be at most $\frac13+\frac14+\frac15<1$), so you want two distinct natural numbers $a,b>2$ such that $\frac1a+\frac1b=\frac12$, i.e., $ab = 2(a+b)$. But this is $(a-2)(b-2)=4$ and the only way we can only factor $4$ as the product of two distinct natural numbers is $1\times 4$.