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I know that $2 + 2 = 2 \cdot 2$ and $1 + 2 + 3 = 1 \cdot 2 \cdot 3.$

My question: Are there other positive integers with sum equal to the product? (The number 1 can not appear more than once among them.)

I am therefore excluding the infinity of solutions obtained by repeating 1, such as:

$$1 + 1 + 2 + 4 = 1 \cdot 1 \cdot 2 \cdot 4$$

$$1 + 1 + 1 + 2 + 5 = 1 \cdot 1 \cdot 1 \cdot 2 \cdot 5$$

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First we shall look at 2 numbers.

Consider $a,b \in \mathbb{N}$. We have $ab=a+b$ $$\implies ab-a=b$$ $$\implies a(b-1)=b$$ $$\therefore b-1|b$$ $$\implies b=2$$ $$\implies a=2$$ And this is the only solution for two numbers.

Now let us look at $n$ numbers, $a_1,a_2,a_3, \dots a_n \in \mathbb{N}$. WlOG $a_1 \leq a_2 \leq \cdots \leq a_n$

We have $a_1a_2a_3 \cdots a_{n-1}a_n=a_1+a_2+a_3+ \cdots + a_n$. Clearly the way to optimize the situation (i.e minimizing the L.H.S while maximizing the R.H.S) is to put $a_1=1$ and $a_2,a_3, \cdots a_n=2$

Then $2^{n-1}=2n-1$ $$\implies2^{n-2}<n-1$$ $$\implies n\leq3$$

So now we look at $n=3$ (we have already done $n=2$).

If none of the terms is equal to one the most likely case is 2,2,2 which doesn't work so none can work. So $a_1=1$ then we have $$a_2a_3=1+a_2+a_3$$ $$\implies a_2a_3-a_2=a_3+1$$ $$\implies a_2(a_3-1)=a_3+1$$ $$\implies a_3-1 | a_3+1$$ $$\implies a_3=2,3$$ Which you can sub in and you find the only solution to be $a_1=1,a_2=2,a_3=3$.

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  • $\begingroup$ When you talk about "optimising", I think you meant to say $2^{n-1}=2n-1$, not $2(n-1)$. $\endgroup$
    – Glen O
    Commented May 24, 2013 at 1:16
  • $\begingroup$ Yes that's true edit made. Thanks for that. $\endgroup$
    – John Marty
    Commented May 24, 2013 at 3:07
  • $\begingroup$ It is not clear to me why $a_1 = 1$ and $a_2, \cdots ,a_n = 2$ "optimizes the situation". The LHS is minimized in this case surely, but the RHS is maximized by setting $a_i$ arbitrarily high. So perhaps you mean the difference between the two sides is optimized? $\endgroup$ Commented May 24, 2013 at 21:00
  • $\begingroup$ Is there a slight mistake here or have I missed my Monday morning coffee? $2^{n-1}=2n-1 \implies n-1 < 2^{n-2} = n - \frac12 < n$- dividing both sides by $2$. Obviously the argument will still go through, but I think we should actually have $n-1 < n$. $\endgroup$ Commented Jun 27, 2016 at 8:29
  • $\begingroup$ @JohnMarty: Is it possible to find a general formula for $a_1, ..., a_n$ through some parameter $k$? $\endgroup$
    – volond
    Commented Apr 5, 2020 at 18:47

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