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$

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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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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 $\ge 33$.