Question 745477
Ok, I didn't expect the number to be whole integer, my reason is the remainder 11.
Anyway, let,
h - hundreds place
t - tens place
u - unit place
The number is: 100h + 10t + u

Writing the equations:
Six times the middle of a three digit number is the sum of the other two:
the middle digit is t so,{{{6t = h + u}}}

1){{{t=(h+u)/6}}}

If the number is divided by the sum of the digits, the answer is 51 and the remainder is 11: 
2) {{{(100h+10t+u)/(h+t+u)=51(11/51)}}}

If the digits are reversed, the number becomes smaller by 198:
3) {{{(100u+10t+h)=(100h+10t+u)-198}}}
Note: see the interchanging of h and u in the equation (reversing the number)

simplifying 2): (4){{{(2488/52)h-(2102/51)t-(2561/51)u=0}}}
simplifying 3): (5){{{99u-99h=-198}}}
substitute 1) to (4): (6){{{(6413/153)h-(8734/153)u=0}}}

Solve (5) and (6);
*[invoke linear "h", "u", -99, 99, -198, 6413/153, -8734/153, 0]

substitute value of h and u to (1)
therefore, h = 7.5261, t = 2.175 and u = 5.5261

Finally the number is: 100h + 10t + u = 779.886

Checking:
If the number is divided by the sum of the digits, the answer is 51 and the remainder is 11: 
{{{51.2166=51(11/51)}}}

If the digits are reversed, the number becomes smaller by 198:
lets just take the whole number: 977 - 779 = 198