The natural number ABGDE is greater than 1, and its product is equal to 2000.

Then these five numbers must be included in the above seven.

Maximum values: 2,2,2,5× 5× 5 and 2+2+ 125 = 133.

The minimum values are 2×2, 2×2, 5, 5, 5 and 4+4+5+5+5 = 23.

2:0,0,0,0,0,...,0

A ***2005 zeros, their sum is equal to their product.

4:(n+2)^2-n^2=4n+4=4*(n+ 1).

5. Solution: Let the original tens of thousands be A, thousands be B, hundreds be C, tens be D, and the unit be E,

Then the original number = a *10000+b *1000+c *100+d *10+e.

Numbers in reverse order = e *10000+d *1000+c *100+b *10+a.

Then reciprocal-original number = e *10000+d *1000+c *100+b *10+a-(a *10000+b * 650.

=9999e+990d-990b-9999a

= 99( 10 1e+ 10d- 10b- 10 1a)

So the result can be divisible by 99, and only the result of B is among the four results.

34056/99=344 is divisible by 99, so it is B.