Rotating digits of a three digit number.

As mentioned in the question, let $\overline{abc}$ be a three digit number.

A very elementary approach to the problem is:

$\overline{abc} = 100\times a + 10\times b + c$

Implies,

$\overline{abc} + 9 = 100\times a + 10\times b + c + 9 = \overline{acb} = 100\times a + 10\times c + b$

From this we get a relation between $b$ and $c$ , which is

$c = b + 1$ .........................(1)

$\overline{acb} + 90 = 100\times a + 10\times c + b = \overline{cab} = 100\times c + 10\times a + b$

From this we get a relation between $a$ and $c$ , which is $c = a + 1$ .........................(2)

$a = b = c - 1$

Since we have a three digit number, $a$ cannot be 0, and since $a + 1 = c$ , $a$ cannot be 9.

Implies, $a$ and $b$ can take values from 1 to 8, 1 and 8 inclusive.

Hence, $c$ correspondingly , can take values from 2 to 9, 2 and 9 inclusive.

The smallest three digit number formed by such conditions is $\boxed{112}$ and the largest is $\boxed{889}$ .

$112 + 889 = \boxed{1001}$