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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.

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  • $\begingroup$ 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) $\endgroup$ – David C. Ullrich May 18 '16 at 15:58
  • $\begingroup$ 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 $\endgroup$ – xAminex May 18 '16 at 16:01
  • $\begingroup$ Is $k$ a fixed integer? $\endgroup$ – almagest May 18 '16 at 16:01
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    $\begingroup$ If $n$ is good, $n+1$ isn't necessary good ($1$ is good, $2$ is not). $\endgroup$ – Abstraction May 18 '16 at 16:10
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    $\begingroup$ 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... $\endgroup$ – David C. Ullrich May 18 '16 at 16:10
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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 $2n+9$.

Using the first gives us all even numbers in the range 66 to 148. Using the second gives us all odd numbers in the range 75 to 155. So we now have all numbers in the range 33 to 149.

Clearly repeating gives us all numbers $\ge 33$.

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  • $\begingroup$ And I was so close. $\frac13+\frac16=\frac12$; lemme write that down... $\endgroup$ – David C. Ullrich May 18 '16 at 16:26
  • $\begingroup$ @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! $\endgroup$ – almagest May 18 '16 at 16:29
  • $\begingroup$ Story of my life - first out of the gate, finish last. heh-heh... $\endgroup$ – David C. Ullrich May 18 '16 at 16:39

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