# What Is A Standard Algorithm? Explained for Kids, Parents & Teachers

**The standard algorithm is a set of steps taken to solve a math problem. In elementary school, students will learn an increasing number of standard algorithms to solve different problems such as addition and subtraction, division and multiplication problems. **

- What is a standard algorithm?
- When do students learn the standard algorithm?
- Why is the standard algorithm important?
- Standard algorithm in the school curriculum
- Problem solving using standard algorithm
- Standard algorithms and decimals
- Standard algorithm worked examples
- Standard algorithm practice questions
- Standard algorithm FAQs

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### What is a standard algorithm?

A standard algorithm is a set of steps to complete a process. In this context, the process is addition, subtraction, multiplication, or division of multi-digit numbers.

For example, addition using the standard algorithm looks like this:

Before learning the standard algorithm as shown in this example, students should already have the conceptual basis of addition and regrouping, as this is simply a quicker, neater process for performing the operation. This is extremely important for all operations and their algorithms.

### When do students learn the standard algorithm?

The word “algorithm” is first mentioned in the Common Core State Standards in grade 3. In 2nd grade, however, it is possible that students will be introduced to the standard algorithm of addition and subtraction with small numbers.

In 3rd grade, students are expected to fluently add and subtract within 1,000 using strategies and algorithms, and then add and subtract within 1,000,000 using the standard algorithm in 4th grade.

In 5th grade, students are introduced to the standard algorithm of multiplication, and will learn strategies similar to the standard algorithm of division, such as the partial quotients model. The CCSS introduces the standard algorithm of division, also known as long division, in 6th grade.

### Why is the standard algorithm important?

It is extremely important for students to have the conceptual basis of the operation they are performing before they begin to learn the standard algorithm for each operation.

Once students have a strong foundation set, each standard algorithm will prove to be an effective, organized way for students to add, subtract, multiply, or divide. The standard algorithms of each operation are likely what you know as the “old way” or the way you learned in school. These algorithms are the strategies that students will carry with them throughout their school careers and into adulthood.

### Standard algorithm in the school curriculum

Each school district likely follows their own curriculum maps with most of these being based on the Common Core State Standards. Some states, however, such as Texas, follow their own set of standards. Although other state standards may vary slightly from these, here you can see where the CCSS and Texas standards (TEKS) focus on the standard algorithms for each operation throughout the grade levels.

#### Common Core State Standards

3rd grade:

**3.NBT.2**– Fluently**add**and**subtract**within 1000 using strategies and**algorithms**based on place value, properties of operations, and/or the relationship between addition and subtraction

4th grade:

**4.NBT.4**– Fluently**add**and**subtract**multi-digit whole numbers using the**standard algorithm**

5th grade:

**5.NBT.5**– Fluently**multiply**multi-digit whole numbers using the**standard algorithm multiplication strategy.**

6th grade:

**6.NS.2**– Fluently**divide**multi-digit numbers using the**standard algorithm**.**6.NS.3**– Fluently**add**,**subtract**,**multiply**, and**divide**multi-digit**decimals**using the**standard algorithm**for each operation

It is important to note that for each of these standards, the previous grade level standard focuses on teaching strategies for the respective operation based on place value, properties of operations, and/or the relationship between the inverse operation.

This sets the foundation of the concept before students move on to the standard algorithm in the next grade.

**Note: Not all states follow the same math standards. **

### Problem solving using standard algorithm

The standard algorithm for each operation follows a specific set of steps. Read through those steps below alongside an example of each one in action. Note that for the addition and subtraction examples, regrouping is needed and shown.

#### Addition

The standard algorithm of addition goes as follows:

**Step one**: Line up the digits of each addend by place value. If one number has more digits than the other, that number should go on top. See how we set up the standard algorithm with the example 378 + 542.

**Step two**: Start by adding the two digits in the ones places. Once you add the two digits, place the sum underneath the line, also in the ones place. In this example, we add 8 + 2, which is 10. Since 10 has two digits (1 ten and 0 ones), we write the 0 in the ones place and we carry the 1 (which represents 1 ten) to the tens place in the next column.

**Step three**: Continue the process by adding the digits in the tens place next. In this case, we add 1 + 7 + 4, which equals 12. Again, we will need to regroup since we have a 2 digit number.

It is important to recognize that although we have added up 1 + 7 + 4 to get 12, what we have really added is 10 + 70 + 40 to get 120. We now write the 2 in the tens place and carry the 1 up to the hundreds place.

**Step four**: Again, we continue the same process by adding the digits in the next column, which is the hundreds place. So we add 1 + 3 + 5, which gives us 9. It is vital to understand the concept happening while we add these digits.

Although the algorithm allows us to easily add 1, 3, and 5, we are really adding 100 + 300 + 500 during this step, which is 900. We can place the 9 in the hundreds place in our answer to represent our 900.

Since there are no more columns with digits in either number, we are finished! The sum of 378 and 542 is 920.

#### Subtraction

The standard algorithm of subtraction follows a very similar pattern of steps as addition. However, the regrouping involved with subtraction may be a little more complicated for students.

Let’s look at the steps while we solve the equation 715 – 234. As with addition, we can also use the standard algorithm for subtraction without regrouping, which you will see an example of in the next section.

**Step one**: First, we set up our subtraction problem by lining up the digits in each place value. When subtracting, the number with the highest value must be placed on top.

**Step two**: After setting up our subtraction problem, we subtract the digits in the ones place first. In this case, 5 – 4 = 1.

**Step three**: Next, we move onto the tens place and subtract the digits there. Since we can’t subtract 3 from 1, we will need to borrow from the next place value, which is the hundreds place.

With the way we record this step in the algorithm, it will look like we’re borrowing 1, and then adding 10 to the digit in the tens place. However, conceptually, we are actually borrowing 100 from the 700 in 715, and adding the 100 to the 1 in the tens place (whose value is 10.) This gives us 6 in the hundreds place (actual value = 600) and 11 in the tens place (actual value = 110.)

This gives us enough tens to complete the subtraction in the tens place. We subtract the digits in the tens place as 11 – 3 = 8, although it really represents 110 – 30 = 80.

**Step four**: Now that we subtracted the digits in the ones place and the tens place, we move on to the hundreds place. Since the digit in the top number is larger than the digit in the bottom number, we can go ahead and subtract 6 – 2, which equals 4. Remember, this actually represents 600 – 200 = 400.

Since there are no more digits in either number, we are finished! Our answer is 481.

#### Multiplication

Students aren’t expected to use the standard algorithm to multiply until 5th grade according to the Common Core standards. There is a good reason that students are introduced to multiplication in 3rd grade but not the standard algorithm of multiplication until 5th grade.

**See also: **Long multiplication

It is imperative that students have a strong, solid foundation of the concept of multiplication and are able to use other strategies fluently to multiply multi-digit numbers, such as an area model, before they begin to learn and use the standard algorithm.

As you will see in the example, we are still multiplying our numbers in parts, which will give us partial products like an area model would, but we just record it differently and more efficiently. Here are the steps alongside the example equation 458 x 25:

**Step one**: As with both addition and subtraction, we start by setting up our problem with the numbers stacked on each other and the place values lined up. The number with the most digits should go on top.

**Step two**: Just as the previous standard algorithms, we start at the right in the ones place. We multiply the two digits in the ones place, which in this case are 8 and 5. Since 8 x 5 gives us 40, we will need to regroup. So, we place the 0 under the line in the ones place and we place the 4 (which represents 4 tens) above the top number in the tens place.

**Step three: **This is where our standard algorithm for multiplication differs from the addition and subtraction algorithms. Instead of moving from one column to the next completely, we actually multiply the ones digit in the bottom number by each digit in the top number before moving on to the next digit of our bottom number.

In step one, we multiplied 8 x 5, so the next two numbers we will multiply are 5 (from the ones place in 25) and 5 (from the tens place in 458). Be sure to recognize that we’re really multiplying 5 x 50 here.

Since we had to carry the 4 tens from our previous step, we also need to **add **those tens to our answer from this step. This may be a tricky step for students. So we multiply 5 x 5 and then we add 4 which gives us 29. (In reality, 290.) We can put the 9 in the tens place under the line and then carry the 2 above the digit in the hundreds place.

**Step four**: We continue multiplying the 5 by the last digit in the number 458, which is the 4 in the hundreds place. 5 x 4 = 20, and then we add the 2 that was carried, so we get 22. In reality, the value of this 22 is 2,200. Since there are no more digits in the top number, we can write the 22 under the line. One of the 2s would be in the hundreds place and the other in the thousands place in our answer.

Notice that what we have done so far in steps one through four is multiplied 458 by 5, which has given us a partial product of 2,290. The next set of steps will give us the other partial product by multiplying 458 by 20.

**Step five**: We’ll now follow the same process as we did with the 5 in 25, but now we will multiply each digit of 458 by the 2 in 25. Since the 2 actually represents 20, we will place a zero in the ones place underneath our first partial product.

Some teachers refer to this as a placeholder, but realize that it is part of our second partial product. It may be helpful at this point to cross off the 5 that we already multiplied by our top number, as well as the previously carried numbers so that you don’t confuse yourself.

**Step six**: Now we multiply the 2 in 25 by each digit in 458, starting with the ones place. 2 x 8 = 16, so we write the 6 in our answer and carry the 1 to the tens place. Since the value of this equation was actually 20 x 8, our answer is 160, but since we already put our zero as the placeholder, we can just move on with multiplying the digits as they are.

**Step seven**: Next, we multiply the 2 by the 5 in the top number. 2 x 5 = 10. We also add the 1 that we carried from the previous step, which gives us 11. We write 1 down below and carry the other 1 to the next place value. Hopefully, you see a pattern and get into the groove of the process at this point!

**Step eight: **The last two numbers we need to multiply are the 2 and 4. 2 x 4 = 8, and then we add the 1 that was carried, which gives us 9. We put the 9 down below in the last spot! Notice now that our second partial product, 9,160, comes from multiplying 458 by 20.

**Step nine: **Finally, we combine our two partial products. 2,290 + 9,160 = 11,450, which is our final product!

#### Division

Dividing using the standard algorithm is also referred to as “long division.” This is a long-standing algorithm that has been taught for many years and which all students will eventually need to master.

The standard algorithm for division does not appear in the 5th grade math standards for common core, although many teachers may still introduce students to it in this grade. Students will be expected to master the standard algorithm of division in 6th grade.

Note that students will first be introduced to long division with single-digit divisors, before moving on to 2-digit or 3-digit number divisors. Let’s look at the step-by-step process of division with a 4-digit dividend and 2-digit divisor using the example 4,832 ÷ 15.

Here are the steps we follow as we divide using the standard algorithm:

Divide, Multiply, Subtract, Bring down

**Step one**: To set up our division problem, we draw a “house” or a “bus stop.” The dividend goes under the “house” and the divisor goes outside to the left.

**Step two**: We start by looking at our divisor, which is 15. We work through the digits of the dividend and determine how many groups of the divisor we can make out of each digit.

Since our divisor is a 2-digit number, we would start by looking at the first two digits: 4 and 8. We can make 3 groups of 15 out of 48. We place the 3 above the 8. This completes the step “divide.”

**Step three**: We move onto the next step in the process: Multiply. 3 x 15 = 45. We write the 45 beneath the 48.

**Step four**: The next step is to subtract. 48 – 45 = 3.

**Step five**: Now, we bring down the next digit, which is 3. Draw an arrow to show that you have completed this step.

**Step six**: We repeat the “divide, multiply, subtract, bring down” process again, now looking at 33 as our new dividend. We can divide 15 into 33 two times. We place the 2 above the line next to the 3. Then we multiply 2 x 15 = 30. We subtract 33 – 30 = 3. Then we bring down the 2.

**Step seven**: We follow the four steps one last time. We can make 2 groups of 15 from 32. We place the 2 above the line in the remaining spot. We multiply 2 x 15 = 30. We subtract 32 – 30 = 2. Then, there is nothing left to bring down! We are left with a remainder of 2. 4,832 ÷ 15 = 322 r2.

*Note: As students progress, they will learn how to represent their remainder as a fraction and a decimal, as well.*

### Standard algorithms and decimals

#### Adding and subtracting decimals

Adding and subtracting decimals using the standard algorithm follow the same exact steps as adding and subtracting whole numbers. The only difference is that there are decimal points in the numbers.

Just as we line up the digits by place value when we add and subtract whole numbers, we also line up the decimal points. Let’s look at an addition example; 1.93 + 4.2, and a subtraction example, 9.8 – 0.76.

#### Multiplying decimals

Multiplying decimals using the standard algorithm follows the same process as multi-digit multiplication with whole numbers. The only difference is the decimals points! Let’s look at the example 4.28 x 3.6.

To start, we set up our decimals the same way we would set up whole numbers while multiplying – one on top of the other with the number with the most digits on top – but the difference here is that you will **not **line up the digits in the same place value.

It also differs from adding decimals because you also do **not **line up the decimal points. In fact, when we multiply two decimals using the standard algorithm, we can “ignore” the decimal points until the very end.

As you see in the example, we have followed all of the same steps as multiplication of whole numbers. At the very end, we look back to our original equation, which is 4.28 x 3.6. In total, there are 3 decimal places in the numbers; 4.28 has two digits that come after the decimal point and 3.6 has 1 number that comes after the decimal point, which gives us 3 total.

Now that we have added up our partial products, we place our decimal point in our product so that it has the same number of decimal places as there were in our original numbers together – again, there were 3.

Look at the numbers that were multiplied – 4.28 and 3.6 – and then look at your product – 15.608 – and ask yourself if the answer is reasonable. In this example, we are multiplying a number close to 4 and a number close to 3 ½. The answer is a number close to 15 ½, so that does in fact seem like a reasonable answer. This can help prevent students placing the decimal in the wrong place in the answer.

#### Dividing decimals

Dividing decimals using the standard algorithm, or long division, will follow the same steps as long division with whole numbers: Divide, Multiply, Subtract, Bring Down. However, there is one step we need to take before we can begin dividing, and it has to do with the decimal point.

We can determine where to place the decimal point above the line before we begin dividing. If the divisor is a whole number, we will simply place the decimal point in the same place it is in the dividend, but above the line.

If the divisor is a decimal, we will need to “move” the decimal point so that the divisor becomes a whole number. It is important to understand that what we are really doing here is multiplying the divisor by a power of 10.

If it has one decimal place, we will multiply by 10, if it has two, we multiply by 100, and so on. We multiply the dividend by the same number we multiply the divisor by. This is extremely important as it will keep the value of each number the same in relation to the other.

Once we determine how many decimal places the point will need to move to make the divisor a whole number, we “move” the decimal point that many places in both the divisor **and **the dividend. The decimal point in the dividend then goes straight up above the line (above the new location of the decimal point) to be placed in our quotient.

### Standard algorithm worked examples

#### Example 1: Addition

*Aligned to CCSS 3.NBT.3 and 4.NBT.4*

Question: Matthew has 615 cards in his card collection. His friend has 439 cards in his collection. How many cards do they have altogether?

**Step one**: Determine what operation is needed to solve the problem. Set up your numbers according to the standard algorithm for that operation.

**Step two**: Add the digits in the ones place; 5 + 9 = 14. Put the 4 in the ones place below the line. Carry the 1 (10) to the top above the tens place.

**Step three**: Add the digits in the tens place. Be sure to add the 1 that was carried from the previous step. 3 + 1 + 1 = 5. Place the 5 beneath the line in the tens place.

**Step four**: Add the digits in the hundreds place. 6 + 4 = 10. Since this is the final place value, we can put both digits into the answer.

**Step five**: Since this is a word problem, we write a sentence or label our answer.

#### Example 2: Subtraction

*Aligned to CCSS 4.NBT.4*

Carlos is playing his favorite arcade game. The high score on the game is 3,142 points. So far, Carlos has scored 2,598 points. How many more points does Carlos need to score to beat the high score?

**Step one**: Determine what operation is needed to solve the problem. Set up your numbers according to the standard algorithm for that operation.

**Step two**: Start in the ones place. Since we can’t subtract 8 from 2, we need to borrow from the tens place. The 4 in the tens place becomes a 3, and we add the 10 to the 2 in the ones place, which gives us 12. 12 – 8 = 4.

**Step three**: Subtract the digits in the tens place. Since we can’t subtract 9 from 3, we need to borrow from the hundreds place. The 1 in the hundreds place becomes a zero, and the 3 in the tens place becomes 13. (Note that it really represents 130 now!) 13 – 9 = 4.

**Step four**: Subtract the digits in the hundreds place. Since we can’t subtract 5 from zero, we need to borrow again, this time from the thousands place. The 3 in the thousands place becomes a 2 and the 0 in the hundreds place becomes a 10. (Note that the actual value is 100.) 10 – 5 = 5.

**Step f**ive: Subtract the digits in the thousands place. 2 – 2 = 0. Since this is the first place value of the number, we do NOT have to place the 0 at the beginning of our answer.

**Step **six: Since this is a word problem, students will need to reason about their answer. Some students may think that 544 is the answer since that is the answer to the computation. Advanced learners may be able to reason independently, but others may need help:

**Question:** How many more points does Carlos need to score to beat the high score?**Answer:** Carlos needs **545 **points to **beat **the high score. 544 points would tie the high score.

#### Example 3: Multiplication

*Aligned to CCSS 5.NBT.5*

**Part A**: The seats in a concert hall are organized into 48 rows. If there are 132 seats in each row, how many total seats are in the concert hall?

**Part B**: On Saturday night, the concert hall sells enough tickets to fill every single seat. If each ticket costs $36, how much money did the concert hall make from the tickets?

**Part A:Step one**: This is a two-part word problem. Starting with Part A, determine what operation is needed to solve the problem. Set up your numbers according to the standard algorithm for that operation.

**Step two**: Start in the ones place. 2 x 8 = 16. We place the 6 in the ones place below the line, and we carry the 1 (actual value = 10) above the 3 in the tens place.

**Step three**: We continue multiplying the 8 in 48 by the remaining digits in 132. 8 x 3 = 24. We add the 1 that we carried from the previous step. 24 + 1 = 25. We place the 5 down in the tens place and carry the 2 above the hundreds place.

**Step four**: Next, we multiply the 8 x 1, which equals 8. We add the 2 that we carried from the previous step. 8 + 2 = 10. Since this is the last place value, we can place both digits of the 10 in our answer.

**Step five**: We have now completed part of the multiplication equation, 132 x 8, and we have our first partial product of 1,056. Now we will move on to the second part of the standard algorithm, where we will multiply 132 x 40. First, we add a 0 in the ones place under the 6. We can also cross out the 8 and the numbers we carried from the previous steps, if it is helpful.

**Step six**: Multiply 4 x 2, which equals 8. Place the 8 in the tens place next to the zero. Here you can see why we put the placeholder zero. We actually multiplied 40 x 2, which equals 80.

**Step seven**: Now we multiply 4 x 3 = 12. We place the 2 next to the 8 and then carry the 1 over to the top of the next place value, which is the hundreds place.

**Step eight**: Finally, we multiply 4 x 1, then add the carried 1 from the previous step. 4 x 1 + 1 = 5. We place the 5 below the 1 in our answer, which gives us our second partial product of 5,280.

**Step nine**: We add the two partial products to get a final product of 6,336. Since this is a word problem, we add our label or write a sentence telling our answer.

**Part B**:

Now that we have the answer to Part A, we can complete Part B. Note that the answer to Part A must be correct in order to correctly solve Part B.

Determine what operation is needed to solve the problem. Set up your numbers according to the standard algorithm for that operation. Solve following the steps of the standard algorithm (as shown in Part A).

#### Example 4: Division

*Aligned to CCSS 6.NS.2*

A librarian received a large shipment of new books. The shipment contained 569 books. She wants to divide the books among 18 empty shelves. Any leftover books will go on display on top of the shelf. How many books will go on each shelf? How many books will go on top of the shelf?

**Step one**: Determine what operation is needed to solve the problem. Set up your numbers according to the standard algorithm for that operation.

**Step two**: Start by determining how many groups of 18 you can make out of 56. I know that 18 x 3 = 54, so we can make 3 groups of 18 out of 56. We write the 3 above the 6. Then we multiply 3 x 18 = 54. We subtract 54 from 56, which equals 2. Then, we bring down the next digit: 9.

**Step three**: We repeat the “Divide, Multiply, Subtract, Bring Down” Process. We find how many groups of 18 we can make out of 29, which is 1 group. We place the 1 above the line and above the 9. Then, we multiply 1 x 18 = 18. We subtract 18 from 29, which equals 11. There are no more digits to bring down, so 11 is our remainder.

**Step four**: We will now need to reason about our answer and be sure to answer the questions in the word problem.

Question: How many books will go on each shelf?

Answer: 31 books will go on each shelf.

Question: How many books will go on top of the shelf?

Answer: There will be 11 books on top of the shelf.

### Standard algorithm practice questions

1.

*Answer: 9,541*

2.

*Answer: 4,137*

3.

*Answer: $42.11*

4.

*Answer: 55.044*

5.

*Answer: 191,660*

6.

*Answer: 220 r37*

7.

*Answer: 63*

### Standard algorithm FAQs

### What is an example of a standard algorithm?

Example of standard algorithm multiplication:

### Why is it called standard algorithm?

It is called a “standard algorithm” because it is the general, widely-used step-by-step process used to add, subtract, multiply, or divide.

### When should standard algorithms be taught?

According to Common Core State Standards, the standard algorithm of each operation is taught in the following grade levels:

3rd grade: Addition & subtraction within 1,000

4th grade: Addition & subtraction within 1,000,000

5th grade: Multiplication

6th grade: Division

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