## The Solving Digit Problems in Algebra Tutorial Series

The Solving Digit Problems in Algebra Tutorial Series is a series of tutorials on solving digit problems. Digit Problems is one of the types of Word Problems in Algebra. Below is the list of posts.

• How to Solve Digit Problems Part I  explains the basics of digit problems including the concepts behind the strategy in solving. The particular concepts discussed are place values and the decimal number system.  One problem involving 2 digit numbers was also discussed including its detailed solution. The solution to the problem involves one variable.
• How to Solve Digit Problems Part II is the continuation of the first part. This is a review on the number system and one more sample problem involving 2 digit numbers is discussed. The solution to this problem also involves one variable
• How to Solve Digit Problems Part III introduces the use of two variables in solving digit problems with numbers containing 2 digits.
• How to Solve Digit Problems Part IV introduces the basic of solving digit problems involving 3-digit numbers. Three variables are used to solve the problem in this part.
• How to Solve Digit Problems Part V presents another detailed example on how to solve digit problems involving 3-digit numbers. Systems of Equations in 3 variables was used to solve the problem in this part.

To view the complete list of posts, you can visit the Post List page.

## How to Solve Digit Problems in Algebra Part V

This is the last part of the series on How to Solve Digit Problems in Algebra. In this post, we are going to solve another digit problem involving 3-digit numbers just like in the previous post.

Problem

The sum of the digits of a 3-digit number is 10. The hundreds digit is 3 more than the tens digits. If 198 is subtracted from the number, then the digits are reversed. What is the number?

Solution and Explanation

Since this is already the fifth part of the series, we won’t be as detailed as the previous parts.

Let

$h$ = hundreds digit of the number
$t$ = tens digit of the number
$u$ = units (or ones) digit of the number.

The first sentence says the sum of the digits of the number is 10. So,
$h + t + u = 10$ –> (Equation 1).

The hundreds digit is 3 more than the tens digit.
$h = t + 3$ –> (Equation 2).

If $198$ is subtracted from the number, the digits are reversed.

$100h + 10t + u - 198 = 100u + 10t + h$.

Simplifying the preceding equation, we have

$(100h - h) + (10t - 10t) + (u - 100u) = 198$
$99h - 99u = 198$.

Dividing both sides by $99$, we have
$h - u = 2$ –> (Equation 3).

We can eliminate $u$ by adding Equation 1 and Equation 3.
$(h + t + u) + (h - u) = 10 + 2$
$2h + t = 12$ (*).

We substitute $t + 3$ from  Equation 2 to $h$ in (*)
$2(t + 3) + t = 12$
$2t + 6 + t = 12$
$3t + 6 = 12$
$3t = 6$
$t = 2$.

Substituting the value of $t$ in Equation 2,
$h = t + 3$
$h = 2 + 3$
$h = 5$.

Substituting the values of $h$ and $t$ in Equation 1,
$h + t + u = 10$
$5 + 2 + u = 10$
$7 + u = 10$
$u = 10 - 7$
$u = 3$.

So, the hundreds digit is 5, tens digit is 2, and ones digit is 3. Therefore, the number is 523.

If we check with the conditions above, we have
(1) The sum of the digits is 10. That is, 5 + 2 + 3 = 10
(2) The hundreds digit is 3 more than the tens digit. The hundred digit is 5 is 3 more than 2
(3) If 198 is subtracted from the number, the digits are reversed. That is, 523 – 198 = 325.

## 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 $72 - 27 = 45$.