There is a three digit number. The digit at the tenth place is 3. If the place of all the digits are replaced with each other symmetrically about the tens position,then the new number obtained is 396 more than the original three digit number. The sum of the digits at unit & hundredth place is 14. Find out the original number?
Correct Answer: 539
Explanation:
Ans.
Let digit = xyz
Given y = 3
According to the question = (100z+10y+x) - (100x+10y+z) = 396x99z-99x = 396
Z-x = 4
Also, Z+ x = 14
Solving (i) & (ii)
Z = 9 , x= 5
The original number is 539
