#### notes

**(i) Reversing the digits – two digit number****Addition****Step-1 :** Choose any 2-digit number of the form 10 x + y.**Step-2 :** Reverse the digits to get a new number i.e., **Step-3:** Add the reversed number to the original number.

(10x + y) + (10y +x) = 11x + 11y = 11(x +y)**Step-4 :** Divide the answer by 11.

11(x + y) ÷ 11 = (x + y)**Result:** There is no remainder.**Remark:** The sum of a two-digit number and the number formed by reversing its digits is exactly divisible by 11 and the quotient obtained is the sum of the digits of the original 2-digit number. Adding both the number, we get 36 + 63 = 99, which is exactly divisible by 11

**Subtraction****Step-1 :** Choose a two digit number in the form 10x + y.**Step-2 :** Reverse the digits to get a new number in the form 10y + x.**Step-3 :** Subtract both the numbers.

(10y + x) – (10x + y) = 9y – 9x = (9 (y – x)**Step-4 :** Divide the answer by 9.

9(y – x) ÷ 9 = (y – x)**Result:** There is no remainder.**Remark:** The difference of a two digit number and its reversed number is exactly divisible by 9 and the quotient obtained is either the difference of the digits of the original 2-digit number or 0.

**(ii) Reversing the digits – three digit number.**

**Addition : ****Step-1 :** Choose a three digit number xyz in the form 100x +10y + z.**Step-2 :** From 2 more numbers in a way yzx = 100z + 10x + y **Step-3 :** Add all three numbers

(100x + 10y + z) + (100y + 10z + x) + (100z + 10x + y)**Step-4 :** Divide the answer by 111.

= 111 (x + y + z) ÷ 111 = (x + y + z).**Remark :** The sum of a 3-digit number and the number formed by arranging its digits in such a way that each digit occupies a place value only once, is exactly divisible by 111

**Substraction:****Step-1 :** Take any three-digit number xyz in the form 100x + 10y + z. **Step-2 :** Reverse the digits : zyx = 100z + 10y + x.**Step-3 :** Substract both the numbers.

(100x + 10y + z) – (100z + 10y + x) = 99x – 99z = 99(x – z)**Remark:** The difference of a 3-digit number and the number formed by reversing the digits is exactly divisible by 99.