# 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.    Q4. 901

Let the 3-digit number chosen by Minakshi = 100a + 10b + c.

After reversing the order of the digits, number = 100c + 10b + a.

On subtraction:

• 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.

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