Let us observe the sum of a 2-digit number and a number formed by reversing its digits.

23 + 32 = 55

63 + 36 = 99

43 + 34 = 77

21 + 12 = 33

26 + 62 = 88.

Notice that all the numbers are divisible by 11. Now, are all such numbers divisible by 11?

**Theorem:** The sum of 2-digit numbers whose digits are reversed is divisible by 11.

**Explanation and Proof**

All 2-digit numbers with digits a and b can be written as 10a + b. For example,

34 = 10(2) + 4 (a = 3, b = 4)

89 = 10(8) + 9 (a = 8, b = 9)

57 = 10(5) + 7 (a = 5, b = 7).

In effect, when the digits are reversed, the number become 10b + a.

43 = 10(4) + 3

98 = 10(9) + 8

75 = 10(7) + 5.

So, in general, we have

10a + b: number

10b + a: number with reversed digits.

So, adding the two numbers we have

10a + b + 10b + a = 11a + 11b = 11(a + b).

As we can see, 11 is a factor of 11(a + b). Therefore, 11(a + b) is divisible by 11.

So, the sum of a 2-digit numbers and the same number with its digit reversed is divisible by 11.

Sir for not only two digit it possible for all