Let number= 100x+10y+z x is the first digit, y is the second and z is the third.

Six times the middle digit of a three digit number is the sum of the other two.

6y=x+z

1st equation: x-6y+z=0

If the number is divided by the sum of its digit, the answer is 51 and the remainder is 11.

(100x+10y+z)/(x+y+z)=51+[11/(x+y+z)]

Multiplying both sides by x+y+z, we get100x+10y+z=51(x+y+z)+11

100x+10y+z=51x+51y+51z+11

2nd equation: 49x-41y-50z=11

If the digits are reversed, the number becomes 198 lesser.

100z+10y+x=100x+10y+z-198

-99x+99z=-198

99x-99z=198

99(x-z) =198

3rd equation: x-z=2

x=z+2

Substitute x with z+2 for first equation: z+2-6y+z=0

2z-6y= -2. Remember this.

Same for 2nd equation. 49(z+2)-41y-50z=11

49z+98-41y-50z=11

-z-41y= -87

Multiply both sides by 2

-2z-82y= -174

So 2z-6y= -2 and -2z-82y= -174

Adding these 2 equations together, we get (2z-6y)+(-2z-82y)= -2+(-174)

Simplifying, we get -88y=-176

y=2

From a previous statement, 2z-6y= -2.

So 2z-6(2)= -2.

2z=10

Z=5

Now since x=z+2, then x=5+2=7

x=7, y=2, z=5

Number=725

You can verifiy if the number satisfies the statements in the problem.