## 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.

If you have comments about the problems, kindly type them below.

If you want to learn more about word problems, visit our Word Problem page.

If you want to learn more about mathematics, please go to the Math page.

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 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.