Let the digits be a, b, and c, so our number is abc.

Since it's a three digit number, we can represent the original number as (a*100) + (b*10) + c

(Why? Take the number "two hundred thirty seven" as an example. This equals 200 + 30 + 7, or 2 * 100 + 3 * 10 + 7. We apply the same concept to our original number)

So if we multiply the sum of the digits by 31, we get the original number. Therefore,

31 * (a + b + c) = (a*100) + (b*10) + c.

If we reverse the digits (**cba**) and add 99, we get the original number, so

(c*100) + (b*10) + a + 99 = (a*100) + (b*10) + c

Five times the middle digit [b] is 7 times the sum of the outside digits [a and c], so

5b = 7(a + c)

Simplify each equation by distributing, etc as necessary and combining like terms, then solve by either elimination or by substitution. This should be enough to get you started.