# Six times the middle digit of a three digit number is the sum of the other two. if the number is divided by the sum of its digit, the answer is 51 and the remainder is 11. If the digits are reversed, the number becomes 198 lesser. Find the number?

1
by emot

2014-01-20T01:43:15+08:00
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.