Question

find thousand natural numbers such that their sum equals to their product

Answer 1

worked on this question for lesser cases :

2×2=2+22×3×1=2+3+13×3×1×1×1=3+3+1+1+17×7×1×1×⋯×1 (35 times) =7+7+1+1.... (35 times)

2×2=2+22×3×1=2+3+13×3×1×1×1=3+3+1+1+17×7×1×1×⋯×1 (35 times) =7+7+1+1.... (35 times)

Using this logic, I seemed to have reduced the problem in the following way.

a×b×1×1×1×⋯×1=a+b+1+1+...a×b×1×1×1×⋯×1=a+b+1+1+...

This equality is satisfied whenever ab=a+b+(1000−n)ab=a+b+(1000−n) Or abc⋯n=a+b+⋯+n+...+(1000−n)abc⋯n=a+b+⋯+n+...+(1000−n) In other words, I need to search for n numbers such that their product is greater by 1000−n1000−n than their sum. This allows the remaining spots to be filled by 11's. I feel like I'm close to the answer.

Note : I have got the answer thanks to Henning's help. It's 112×10×1×1×1×...112×10×1×1×1×... (998998 times)=10+112+1+1+1+...=10+112+1+1+1+... (998998 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)