Check what the result would have been if Minakshi had chosen the numbers shown below. In each case keep a record of the quotient obtained at the end.
Let the 3-digit number chosen by Minakshi = 100a + 10b + c.
After reversing the order of the digits, number = 100c + 10b + a.
• If a > c, then the difference between the original number & reversed number is
(100a + 10b + c) – (100c + 10b + a) = 100a + 10b + c – 100c – 10b – a = 99a – 99c = 99(a – c).
• If c > a, then the difference between the numbers is (100c + 10b + a) – (100a + 10b + c) = 99c – 99a = 99(c – a).
• If a & c are equal, the difference is 0.
here a = 9, b = 0 and c = 1
901- 109= 792= 99*8 = multiple of 99
quotient in each case = c - a.