TITLE: Find a thousand natural numbers such that their sum equals their product
QUESTION [23 upvotes]: The question is to find a thousand natural numbers such that their sum equals their product. Here's my approach :
I worked on this question for lesser cases :
\begin{align}
&2 \times 2 = 2 + 2\\
&2 \times 3 \times 1 = 2 + 3 + 1\\
&3 \times 3 \times 1 \times 1 \times 1 = 3 + 3 + 1 + 1 + 1\\
&7 \times 7 \times 1 \times 1 \times \dots\times 1 \text{ (35 times) } = 7 + 7 + 1 + 1 .... \text{ (35 times) }
\end{align}
Using this logic, I seemed to have reduced the problem in the following way.
$a \times b \times 1 \times 1 \times 1 \times\dots\times 1 = a + b + 1 + 1 +...$
This equality is satisfied whenever $ ab = a + b + (1000-n)$
Or $ abc\cdots n = a + b + \dots + n + ... + (1000 - n)$ 
In other words, I need to search for n numbers such that their product is greater by $1000-n$ than their sum. This allows the remaining spots to be filled by $1$'s. I feel like I'm close to the answer. 
Note : I have got the answer thanks to Henning's help. It's 
$112 \times 10 \times 1 \times 1 \times 1 \times ...$ ($998$ times)$ = 10 + 112 + 1 + 1 + 1 + ...$ ($998$ times)
This is for the two variable case. Have any of you found answers to more than two variables ?
$abc...n = a + b + c + ... + n + (1000 - n) $

REPLY [5 votes]: A solution with four numbers different from 1 is:
$$16 \times 4 \times 4 \times 4 \times 1^{996} = 16 + 4 + 4 + 4 + (996 \times 1) = 1024$$
How was this found?  $1024 = 2^{10}$ appeared to be a promising candidate for the sum and product because it's slightly larger than 1000 and has many factors. The problem then was to find a, b, c, d such that:
$$a+b+c+d=10$$
$$2^a+2^b+2^c+2^d=1024 -(1000-4)=28$$
None of a-d can be more than 4 since $2^5=32 > 28$. But on trying $a=4$, reducing the problem to finding b,c,d such that $b+c+d=6$ and $2^b+2^c+2^d = 12$, the solution was apparent.