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.
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.
Multiplying both sides by x+y+z, we get100x+10y+z=51(x+y+z)+11
2nd equation: 49x-41y-50z=11
If the digits are reversed, the number becomes 198 lesser.
3rd equation: 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
Multiply both sides by 2
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
From a previous statement, 2z-6y= -2.
So 2z-6(2)= -2.
Now since x=z+2, then x=5+2=7
x=7, y=2, z=5
You can verifiy if the number satisfies the statements in the problem.