## How to Solve Digit Problems Part III

This is the third part of the tutorial series on Solving Digit Problems. In Part 1 and Part 2, we used one variable to solve digit problems. In this post, we learn how to use two variables to solve digit problems. We still use the problem in Part 2.

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. Find the numbers.

Solution and Explanation

Let t = tens digit and u = units digit.

From the first sentence in the problem, we know that

t + u = 9 (1).

Also, as we have learned in the first two parts of this series, 2-digit numbers with tens digit t and ones digit u can be represented (or has value) 10t + u. For example, the number 25 with t = 2 and u = 5 has value 10(2) + 5.

So, the we can represent the original number as

10t + u.

If we reverse the digit, the specific example which is 25 becomes 52. This becomes 10(5) + 2. Hence, we can represent the reverse number as

10u + t.

Therefore, we our representation is as follows:

original number: 10t + u
new number (with digits reversed): 10u + t.

In the second sentence in the problem, it says when the digits are reversed, the new number is 45 more than the original number. That means that if we add 45 to the original number, they will be equal. That is,

original number + 45 = new number.

Substituting the representations above, we have

10t + u + 45 = 10u + t.

We can simplify the equation by putting the variables on the right.

45 = 10u + t – (10t + u)
45 = 10u + t – 10t – u
45 = 9u – 9t (2).

Thus, we have 2 systems of equations.

t + u = 9 (1)
9u – 9t = 45 (2).

Note: We just change reverse the position of the expressions in equation (2).

We can solve this using elimination or substitution. In this solution, we use substitution.

First, we find the value of t in (1)

t + u = 9.

Subtracting u from both sides, we have
t = 9 – u.

Next, we substitute 9 – u to the value of t in (2)

9u – 9t = 45
9u – 9(9-u) = 45
9u – 81 + 9u = 45
18u – 81 = 45
18u = 45 + 81
18u = 126.

Dividing both sides by 18,

u = 7

So, the units digit is 7.

To find t, we substitute in one of the equations in (1) and (2). We substitute in (1),

t + u = 9
t + 7 = 9
t = 2

So, our number has tens digit 2 and ones digit 7. Therefore, the number is 27.

If we check, if the number is reversed, it becomes 72. Let’s see if the number with reversed digit is 45 more than the original number.

72 – 27 = 45.

Therefore, we are correct.

Having two equations in two variables is an example of systems of equation. In the process above, we solved for the value of one of the variables in (1) and substituted it in (2). We will discuss systems of equations, particularly linear equations in two variables in details in the next posts.

## 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 $72 - 27 = 45$.

## How to Solve Digit Problems Part I

Digit Problems is one of the word problems in Algebra. To be able to solve this problem, you must understand how our number system works. Our number system is called the decimal number system because the numbers in each place value is multiplied by powers of 10 (deci means 10). For instance, the number 284 has digits 2, 8, and 4 but has a value of 200 + 80 + 4. That is,

$(100 times 2) + (10 times 8) + (4 times 1) = 284$.

As you can observe, when our number system is expanded, the hundreds digit is multiplied by 100, the tens digit is multiplied by 10, and the units digit (or the ones digit) is multiplied by 1. Then, all those numbers are added. The numbers 100, 10, and 1 are powers of 10: $10^2 = 100$, $10^1 = 10$, and $10^0 = 1$. So, numbers with $h$, $t$, and $u$ as hundreds, tens, units digits respectively has value

$100h + 10t + u$.

It is clear that this is also true for higher number of digits such as thousands, ten thousands, hundred thousands, and so on.

Many of the given numbers in this type of problem have their digits reversed. As we can see, if 10t + u is reversed, then it becomes $10u + t$. For instance, $32 = 10(3) + 1(2)$ when reversed is $23 = 10(2)+ 1(3)$. Now, that we have already learned the basics, we proceed to our sample problem.

Worked Example

The tens digit of a number is twice the units digit. If the digits are reversed, the new number is 18 less than the original. What are the numbers?

Solution and Explanation

The tens digit of a number is twice the unit digit. This means that if we let the units digit be $x$, then the tens digit is $2x$. As we have mentioned above, we multiply the tens digit with 10 and the units digit with 1. So, the number is

$(10)(2x) + x$.

Now, when the digits are reversed, then x becomes the tens digit and $2x$ becomes the ones digit. So, the value of the number is

$(10)(x) + 2x$.

From the problem above, the number with reversed digit is 18 less than the original number. That means, that if we subtract 18 from original number, it will equal the new number. That is,

$(10)(2x) + x - 18 = 10(x) + 2x$
$20x + x - 18 = 12x$
$21x - 18 = 12x$
$9x = 18$
$x = 2$
$2x = 4$

So, the number is 42 and the reversed number is 24.

Check: 42 – 24 = 18.