In the **previous post**, we proved a theorem that the sum of a 2-digit number with reversed digit is divisible by 11. In this post, we will learn another theorem about numbers with reversed digits. This time, we will explore the difference between two 3-digit numbers with reversed digits.

Let’s have the following examples.

231 – 132 = 99

512 – 215 = 297

741 – 147 = 594

543 – 345 = 198

Before continuing reading, try other numbers and see if you can find a pattern.

You have probably observed that all the differences of a 3-digit number and a number formed by reversing its digit is a multiple by 99. Now, is this true for all possible cases? You may want to prove this on your own before continuing below. The proof on the 2-digit number with reversed digits is would be a good hint.

**Theorem:** The difference of a 3-digit number and the number formed by reversing the order of its digits is divisible by 99.

**Explanation and Proof**

Any 3-digit number can be written in the form 100x + 10y + z, where x, y, and z are the hundreds, tens, and ones digits, respectively. For example, the number 325 can be written as

100(3) + 10(2) + 5.

Now, the number formed by reversing the digits will be 100z + 10y + x. For example, in the case of 325, the number with the reversed digits becomes

100(5) + 10(2) + 3.

Now, subtracting we have

100x + 10y + z – (100z + 10y + x) = 100x + 10y + z –100z – 10y – x = 99x – 99z.

Factoring, we have 99(x – z). Since x and z are 1-digit numbers, x – z is an integer. Since 99 is a factor of 99(x – z), 99(x – z) is divisible by 99. Therefore, the difference of a 3-digit number and the number formed by reversing the order of its digits is divisible by 99