# 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

$(8 \times 10^2) + (7 \times 10^1) + (1 \times 10^0)$.

Recall that $10^0 = 1$.

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

Problem

The sum of the digits of a 2-digit number is $9$. If the digits are reversed, the new number is $45$ more than the original number. What are the numbers?

Solution and Discussion

If the tens digit of the number is $x$, then the ones digit is $9 - x$ (can you see why?).

Since the tens digit is multiplied by $10$, the original number can be represented as

$10x + (9 - x)$.

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

Now, if we reverse the number, then $9 - x$ becomes the tens digit and the ones digit becomes $x$. So, multiplying the tens digit by 10, we have

$10(9 - x) + x$.

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

As shown in the problem, the new number (the reversed number) is $45$ more than the original number. Therefore,

reversed numberoriginal number = 45.

Substituting the expressions above, we have

90 – 9x – (9x + 9) = 45.

Simplifying, we have

$90 - 9x - 9x - 9 = 45$
$81 - 18x = 45$
$18x = 81 - 45$
$18x = 36$
$x = 2$.

Therefore, the tens digit of the original number is 2 and the ones digit is $9 - 2 = 7$.

So, the original number is $27$ and the reversed number is $72$.

Now, the problem says that the new number is $45$ more than the original number. And this is correct since $72 - 27 = 45$.