Good number $n=a_1+a_2+a_3+\cdots+a_k$ with ${1\over {a_1}} + {1\over {a_2}} + {1\over {a_3}} + \cdots+{1\over{a_k}}=1$

An integer n will be called good if we can write $n=a_1+a_2+a_3+\cdots+a_k$, where $a_1,a_2,a_3 \ldots a_k$ are positive integers (not necessarily distinct) satisfying: $${1\over {a_1}} + {1\over {a_2}} + {1\over {a_3}} + \cdots +{1\over{a_k}}=1$$ Given the information that the integers $33$ through $73$ are good, prove that every integer $\ge$ 33 is good.

We can use recurrence in this case, and suppose that $n$ is good and prove that $n+1$ is good too. then $n+1=a_1+a_2+a_3+\cdots+a_k +1$.

Can you help me carry on? Thanks in advance.

• I don't see why $n$ good implies $n+1$ good. It's easy to show that if $n$ is good then $2n+2$ is good. (Not that that gives a complete solution, just an example of the sort of thing that might come up in a solution) May 18 '16 at 15:58
• using recurrence i suppose that n is good and prove that n+1 is good too.I have to prove that every n>33 is good not only 2n+2 May 18 '16 at 16:01
• Is $k$ a fixed integer? May 18 '16 at 16:01
• If $n$ is good, $n+1$ isn't necessary good ($1$ is good, $2$ is not). May 18 '16 at 16:10
• I understand how induction works. If there were an easy way to show that $n$ good implies $n+1$ good then there'd be no need to know that all the numbers between $33$ and $73$ are good - knowing just that $1$ is good would imply that every $n$ is good. So it must be a little more complicated... May 18 '16 at 16:10

If we have $\sum a_i=n$, then take $b_i=2a_i$ and an additional term $b=2$. That gives $2n+2$. Similarly taking two additional terms $3,6$ gives $2n+9$.
Clearly repeating gives us all numbers $\ge 33$.
• And I was so close. $\frac13+\frac16=\frac12$; lemme write that down... May 18 '16 at 16:26
• @DavidC.Ullrich Oddly, your comment did not help me. I did not understand it and ignored it. It was only after I had found $2n+2$ that I realised you were way ahead of me! May 18 '16 at 16:29