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.