Question

the sum of the digits of a three digit number is 12 if the digits are reversed then the new number obtained diminished by 198 if the middle term digit is half the sum of the digit in the extreme

Answer 1

let the number be xyz and can also be expressed as 100x+10y+z. it is given that the middle digit is half the sum of the extreme digits => y=x+z/2 => 2y=x+z
When reversed it becomes zyx=> 100z+10y+x. 100z+10y+x+198=100x+10y+z=> 99z-99x=198=> x-z=
2. So x+y+z=12=> 3y=12=> y=4. So x+z=8 and x-z=2 => 2x=10=> x=5 and z=5-2=3. So the number is 5(100)+4(10)+3=543.