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 numbedr & reversed numbers 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, then the difference is 0.
here a,b and c are 4, 6 & 9 respectively.
964 - 469 = 495 = 99*5 = multiple of 99