Find a thousand natural numbers such that their sum equals their product 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) $
 A: 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.
A: There's a sign error in your final equation; you want
$$ a+b+998=ab $$
which simplifies to
$$ (a-1)(b-1) = 999 $$
from which it should be easy to extract several integer solutions.
A: Do we have an alternative to fill up with ones?
Let $N=1000$, then
\begin{align}
\sum_{i=1}^N n_i &= \prod_{i=1}^N n_i \iff \\
N n_a &= n_g^N
\end{align}
where $n_a$ is the arithmetic mean and $n_g$ the geometric mean of the numbers $n_i$.
Those numbers on the left and right hand side of the equation 
drift apart very fast.
Already for $n_a = n_g = 2$ we would have $2000$ vs. $2^{1000} \approx  10^{301}$.
Assuming $n_a \approx n_g$ we estimate:
$$
N x = x^N \Rightarrow \\
N = x^{N-1} \Rightarrow \\
x = \sqrt[N-1]{N}
$$
For $N=1000$ this gives $x=\sqrt[999]{1000}=1.0069\dotso$. 
Such an estimated mean gives not much room for numbers $n_i > 1$.
$$
1.007 = \frac{1000+a}{1000} = 1 + a/1000 \Rightarrow \\
a = 7
$$
Spreading an excess of $7$ over a couple of numbers is too pessimistic.
$$
1.007 = \sqrt[1000]{b} \Rightarrow \\
b \approx 1070
$$
That looks better, the product has to be equal with a sum something above $1000$.

Have any of you found answers to more than two variables ?

A solution with three numbers different from $1$ is
$x=67$, $y=z=4$. This gives $67+4+4+997=1072$. 
Also $67\times 4 \times 4 \times 1^{997}=1072$.
