# How to Solve Digit Problems Part II

In the previous post, we have discussed the **basics of digit problems**. We have learned the decimal number system or the number system that we use everyday. In this system, each digit is multiplied by powers of 10. For instance, 871 means

.

Recall that .

In this post, we continue this series by providing another detailed example.

**Problem**

The sum of the digits of a 2-digit number is . If the digits are reversed, the new number is more than the original number. What are the numbers?

**Solution and Discussion**

If the tens digit of the number is , then the ones digit is (can you see why?).

Since the tens digit is multiplied by , the original number can be represented as

.

Simplifying the previous expression, we have **10x – x + 9 = 9x + 9.**

Now, if we reverse the number, then becomes the tens digit and the ones digit becomes . So, multiplying the tens digit by 10, we have

.

Simplifying the expression we have **10 – 10x + x = ****90 – 9x**.

As shown in the problem, the new number (the reversed number) is more than the original number. Therefore,

**reversed number** – **original number** = 45.

Substituting the expressions above, we have

**90 – 9x** – (**9x + 9**) =** 45**.

Simplifying, we have

.

Therefore, the tens digit of the original number is 2 and the ones digit is .

So, the original number is and the reversed number is .

Now, the problem says that the new number is more than the original number. And this is correct since .