2. Reversing the two digit numbers

Playing with numbers is always an enjoyable task. We tend to figure out a few fantastic facts and these facts keep us massively entertained in the process.

 

Let us now observe a number game played by Alka and Yash.

 

Alka and Yash started to play a game. Alka asked Yash to think of any two-digit number. Once Yash has thought of one, she asked him to reverse the number. She then bet that the sum of both the two-digit numbers is a multiple of \(11\). Yash was shocked to find it right.

 

 

Step \(1\): Yash thought of \(29\) at first.

 

Step \(2\): On Alka's instructions, he then reversed the number to \(92\).

 

Step \(3\): To check Alka's statement, he added both the numbers.

 

\(29 + 92 = 121\)

 

Step \(4\): He was then impressed to know that the total obtained is a multiple of \(11\).

 

How do you think Alka could strongly bet on her statement?

 

It's simple logic.

 

There are two standard logics concerning two-digit numbers and their reverse.

The general form of the two-digit number '\(ab\)' is \((10 \times a) + b\).

 

We know that to reverse a number, we should swap the numbers in TENS and ONES position.

 

Therefore, the general form of its reverse '\(ba\)' is \((10 \times b) + a\).

 

Let us try to find the sum of both these numbers.

 

\(ab + ba = [(10 \times a) + b] + [(10 \times b) + a]\)

 

\(= 10a + b + 10b + a\)

 

\(= 11a + 11b\)

 

\(= 11(a + b)\)

 

Therefore, the sum of two-digit numbers and their reverse is always a multiple of \(11\).

Let us now find the difference between the two numbers.

 

\(ab - ba = [(10 \times a) + b] - [(10 \times b) + a]\)

 

\(= 10a + b - 10b - a\)

 

\(= 9a - 9b\)

 

\(= 9(a - b)\)

 

Thus, it is understood that the difference between the two numbers is a multiple of \(9\).