For every $n>32$, prove that $n$ can be written as sum of a number of natural numbers such that sum of inverses of these numbers will be 1. Suppose the statement is correct for $33, 34, ..., 73$.
I want to prove this by strong induction. if $n$ is even, we know $n=2+2k$. Also $2k=k+k$ and we know $k=\sum{a_i}$ such that $\sum1/a_i$ is 1. Therefore $2k=\sum(2a_i)$ and for inverses we have: $\sum1/(2a_i) = 1/2*\sum1/a_i = 0.5$
Now since $n = 2 + 2k$, $n = 2 + \sum 2a_i$ and $(1/2 + \sum 1/(2a_i) = 0.5 + 0.5 = 1)$
I'm not sure how can i prove the statement for odd numbers though. Any ideas?
Thanks in advance.
Same trick, let $N=2k+3+6\ge 75$ be odd. Then $k\ge 33$ is by strong induction "in the good list" of numbers having the given property, we have a representation $k=a+b+\dots+z$ with $1=\frac 1a+\frac 1b+\dots +\frac 1z$, and from here we get $$ \begin{aligned} N &=(2a)+(2b)+\dots+(2z)+3+6\ ,\qquad\text{with}\\ 1 &=\frac 12+\frac 13+\frac 16 \\ &= \underbrace{\frac 1{2a}+\frac 1{2b}+\dots+\frac 1{2z}}_{=1/2}+\frac 13+\frac 16\ . \end{aligned} $$