Whole Number Operations

Properties of Operations are special math rules that help us add and multiply numbers in easier ways.

These properties do not change the value of an expression. They just help us rearrange or break apart numbers so the math is simpler to do.

In 5th grade, the main properties of operations you will use are:

• Commutative Property
• Associative Property
• Identity Property
• Distributive Property

Learning these properties helps you solve problems faster, check your work, and understand why math works.

1. Commutative Property

The commutative property means you can switch the order of numbers when you add or multiply, and the answer stays the same.

For addition:

$$a+b=b+a$$

For multiplication:

$$a\times b=b\times a$$

Examples:

• \(7+5=5+7\)
• \(3\times 4=4\times 3\)

This works for addition and multiplication.

It does not work for subtraction or division.

• \(9-4\neq 4-9\)
• \(12\div 3\neq 3\div 12\)

2. Associative Property

The associative property means you can change the grouping of numbers when you add or multiply.

Grouping means moving the parentheses.

For addition:

$$\left(a+b\right)+c=a+\left(b+c\right)$$

For multiplication:

$$\left(a\times b\right)\times c=a\times \left(b\times c\right)$$

Examples:

• \((2+6)+4=2+(6+4)\)
• \((5\times 2)\times 3=5\times (2\times 3)\)

The numbers stay in the same order. Only the grouping changes.

This property works for addition and multiplication.

It does not work for subtraction or division.

3. Identity Property

The identity property tells us that some numbers keep a value the same.

For addition, adding 0 does not change a number.

$$a+0=a$$

Examples:

• \(15+0=15\)
• \(0+42=42\)

For multiplication, multiplying by 1 does not change a number.

$$a\times 1=a$$

Examples:

• \(9\times 1=9\)
• \(1\times 27=27\)

These are helpful because they remind us which numbers leave other numbers unchanged.

4. Distributive Property

The distributive property helps you multiply a number by a sum.

It means you can multiply the outside number by each number inside the parentheses, and then add the products.

$$a\times (b+c)=(a\times b)+(a\times c)$$

This property is very useful for mental math and for breaking apart larger numbers.

Example:

$$4\times (10+3)=(4\times 10)+(4\times 3)=40+12=52$$

You can also use it when a number is written in expanded form.

For example, since \(23=20+3\), you can write:

$$5\times 23=5\times (20+3)=(5\times 20)+(5\times 3)=100+15=115$$

Why These Properties Matter

Properties of operations help you:

• Choose an easier way to solve a problem
• Break apart numbers into friendly parts
• Check whether expressions are equal
• Understand number patterns

Instead of always solving a problem the same way, you can use properties to make the work simpler.

Worked Example 1: Using the Commutative Property

Simplify: \(18+25\)

You could switch the order:

$$18+25=25+18$$

The answer is still:

$$25+18=43$$

Why this helps: Sometimes one order is easier to think about than the other.

Worked Example 2: Using the Associative Property

Find the value of \((6+4)+9\).

Use the associative property to regroup:

$$ (6+4)+9=6+(4+9) $$

Now solve:

$$6+(4+9)=6+13=19$$

You could also solve \(6+4\) first:

$$10+9=19$$

Both ways give the same answer.

Worked Example 3: Using the Identity Property

Find the value of \(347\times 1\).

The identity property of multiplication says multiplying by 1 keeps the number the same.

$$347\times 1=347$$

Also, for addition:

$$347+0=347$$

Worked Example 4: Using the Distributive Property

Find \(7\times 16\).

Break apart 16 into \(10+6\):

$$7\times 16=7\times (10+6)$$

Distribute 7 to both parts:

$$7\times (10+6)=(7\times 10)+(7\times 6)$$

$$=70+42=112$$

So, \(7\times 16=112\).

Tips for Telling the Properties Apart

• Commutative: the order changes.
• Associative: the grouping changes.
• Identity: adding 0 or multiplying by 1 keeps the number the same.
• Distributive: multiply one number by each part inside parentheses.

Watch Out for These Mistakes

• Do not use the commutative property for subtraction or division.
• Do not confuse changing order with changing grouping.
• In the distributive property, multiply the outside number by every number inside the parentheses.

Quick Practice to Think About

1. Which property is shown by \(8+12=12+8\)?
2. Which property is shown by \((3\times 5)\times 2=3\times (5\times 2)\)?
3. What is \(64+0\)?
4. Use the distributive property to find \(3\times 14\).

Answers:

1. Commutative property
2. Associative property
3. \(64\)
4. $$3\times 14=3\times (10+4)=(3\times 10)+(3\times 4)=30+12=42$$

Summary

Properties of operations are rules that help us work with numbers in smart and efficient ways.

The commutative property changes order, the associative property changes grouping, the identity property keeps numbers the same when adding 0 or multiplying by 1, and the distributive property breaks apart numbers to make multiplication easier.

When you understand these properties, you can simplify problems and become a stronger math thinker.

Addition and Subtraction Algorithms

When we add or subtract large whole numbers, we use a step-by-step method called an algorithm. An algorithm helps us keep our work organized and accurate.

In 5th grade, you will work with multi-digit numbers. That means numbers with many digits, such as hundreds, thousands, or even more. To solve these problems, it is important to understand place value.

Each digit in a number has a value based on its place. For example, in the number \(4,582\):

• \(4\) is in the thousands place, so it means \(4,000\)
• \(5\) is in the hundreds place, so it means \(500\)
• \(8\) is in the tens place, so it means \(80\)
• \(2\) is in the ones place, so it means \(2\)

Understanding place value helps us line numbers up correctly and know when to regroup.

Main Idea for Addition

In addition, we line up digits by place value. Then we add starting from the ones place and move left to the tens, hundreds, and beyond.

If the sum in one place is \(10\) or more, we regroup. This means we write the ones digit in that place and carry the extra ten to the next place.

Steps for the Standard Addition Algorithm

1. Write the numbers in a column.
2. Line up ones under ones, tens under tens, hundreds under hundreds, and so on.
3. Add from right to left, starting with the ones place.
4. If a column totals \(10\) or more, regroup to the next place.
5. Continue until every place has been added.

Worked Example 1: Addition Without Regrouping

Find \(243 + 356\).

Add the ones: \(3 + 6 = 9\)

Add the tens: \(4 + 5 = 9\)

Add the hundreds: \(2 + 3 = 5\)

So, \(243 + 356 = 599\).

Worked Example 2: Addition With Regrouping

Find \(587 + 476\).

Start with the ones: \(7 + 6 = 13\). Write down \(3\) ones and regroup \(1\) ten.

Next, add the tens: \(8 + 7 + 1 = 16\). Write down \(6\) tens and regroup \(1\) hundred.

Now add the hundreds: \(5 + 4 + 1 = 10\).

After regrouping in all places, the answer is:

So, \(587 + 476 = 1,063\).

Main Idea for Subtraction

In subtraction, we also line up digits by place value and start from the ones place.

Sometimes the top digit is smaller than the bottom digit. When that happens, we regroup from the place to the left. Regrouping means taking \(1\) from the next place value and turning it into \(10\) of the current place.

For example:

• \(1\) ten can become \(10\) ones
• \(1\) hundred can become \(10\) tens
• \(1\) thousand can become \(10\) hundreds

Steps for the Standard Subtraction Algorithm

1. Write the larger number on top.
2. Line up digits by place value.
3. Subtract from right to left, starting with the ones place.
4. If the top digit is smaller, regroup from the place to the left.
5. Continue until every place has been subtracted.

Worked Example 3: Subtraction With One Regrouping

Find \(532 - 178\).

Start with the ones: \(2 - 8\) cannot be done, so regroup \(1\) ten from the tens place.

The \(3\) tens becomes \(2\) tens, and the \(2\) ones becomes \(12\) ones.

Now subtract the ones: \(12 - 8 = 4\)

Subtract the tens: \(2 - 7\) cannot be done, so regroup \(1\) hundred.

The \(5\) hundreds becomes \(4\) hundreds, and the \(2\) tens becomes \(12\) tens.

Now subtract the tens: \(12 - 7 = 5\)

Subtract the hundreds: \(4 - 1 = 3\)

So, \(532 - 178 = 354\).

Worked Example 4: Subtraction Across Zeros

Find \(4,002 - 587\).

This problem is tricky because there are zeros in the middle.

Start with the ones: \(2 - 7\) cannot be done. The tens digit is \(0\), so we cannot borrow from there yet.

We look to the hundreds digit. It is also \(0\), so we look to the thousands digit.

Take \(1\) thousand from \(4\) thousands. Now there are \(3\) thousands, and the hundreds place becomes \(10\) hundreds.

Then take \(1\) hundred from the \(10\) hundreds. Now there are \(9\) hundreds, and the tens place becomes \(10\) tens.

Then take \(1\) ten from the \(10\) tens. Now there are \(9\) tens, and the ones place becomes \(12\) ones.

Now subtract each place:

• Ones: \(12 - 7 = 5\)
• Tens: \(9 - 8 = 1\)
• Hundreds: \(9 - 5 = 4\)
• Thousands: \(3 - 0 = 3\)

So, \(4,002 - 587 = 3,415\).

Helpful Tips

• Always line up numbers by place value.
• Start from the right.
• In addition, regroup when a sum is \(10\) or more.
• In subtraction, regroup when the top digit is smaller than the bottom digit.
• Write neatly so your digits stay in the correct columns.

How to Check Your Work

You can check addition with subtraction. For example, if \(587 + 476 = 1,063\), then \(1,063 - 476\) should equal \(587\).

You can check subtraction with addition. For example, if \(532 - 178 = 354\), then \(354 + 178\) should equal \(532\).

Common Mistakes to Avoid

• Not lining up digits correctly
• Forgetting to regroup
• Carrying or borrowing to the wrong place
• Starting from the left instead of the right
• Skipping a digit when writing the answer

Summary

Addition and subtraction algorithms are organized ways to solve problems with large numbers. They work best when you pay close attention to place value.

In addition, add from right to left and regroup when needed. In subtraction, subtract from right to left and regroup when the top digit is too small.

With practice, these steps will become easier and faster. Neat work and careful regrouping will help you get correct answers.

Multiplication Area Models and Partial Products

When we multiply larger numbers, it helps to break the numbers apart into tens, hundreds, and ones. Then we can multiply the smaller parts and put the answers together.

Two very helpful tools for this are the area model and partial products. These tools show why multiplication works, not just how to get the answer.

In this lesson, you will learn how to use area models and partial products to multiply whole numbers with more than one digit.

1. What is an area model?

An area model is a rectangle split into smaller parts. Each side of the rectangle is broken into place-value parts, like tens and ones. Then we multiply each part to find the area of each smaller rectangle.

For example, to multiply \(23 \times 4\), we can break \(23\) into \(20 + 3\).

Now multiply each part by 4:

• \(20 \times 4 = 80\)
• \(3 \times 4 = 12\)

Then add the partial products:

So, \(23 \times 4 = 92\).

2. What are partial products?

Partial products are the smaller multiplication answers you get after breaking apart a factor. Then you add those smaller answers together to get the final product.

In the example \(23 \times 4\), the partial products are:

• \(80\)
• \(12\)

Adding them gives the total product, \(92\).

3. Why do we break numbers apart?

Breaking numbers apart makes multiplication easier because we can use place value.

For example:

• \(34\) can be broken into \(30 + 4\)
• \(126\) can be broken into \(100 + 20 + 6\)
• \(58\) can be broken into \(50 + 8\)

This helps us multiply by tens and hundreds, which is often easier to do.

4. How to use an area model

Here are the steps:

1. Break one or both factors into place-value parts.
2. Draw a rectangle.
3. Label the sides with the broken-apart numbers.
4. Multiply to find the area of each smaller rectangle.
5. Add all the smaller areas.

If both factors have more than one digit, both numbers can be broken apart.

For example, \(14 \times 12\):

This creates 4 smaller rectangles.

Worked Example 1: Multiply \(23 \times 4\)

Break apart 23:

Multiply each part by 4:

Add the partial products:

Answer: \(23 \times 4 = 92\)

Worked Example 2: Multiply \(36 \times 7\)

Break apart 36:

Now multiply each part by 7:

Add the partial products:

Answer: \(36 \times 7 = 252\)

Worked Example 3: Multiply \(14 \times 12\)

Now both factors have two digits, so break both apart:

Use an area model with 4 smaller parts:

• \(10 \times 10 = 100\)
• \(10 \times 2 = 20\)
• \(4 \times 10 = 40\)
• \(4 \times 2 = 8\)

These are the partial products. Now add them:

Answer: \(14 \times 12 = 168\)

Worked Example 4: Multiply \(23 \times 15\)

Break apart both factors:

Find each smaller product:

• \(20 \times 10 = 200\)
• \(20 \times 5 = 100\)
• \(3 \times 10 = 30\)
• \(3 \times 5 = 15\)

Add all the partial products:

Answer: \(23 \times 15 = 345\)

5. Connecting area models and partial products

An area model and partial products are really showing the same idea.

• The area model shows the multiplication with rectangles.
• The partial products show the multiplication with numbers.

For \(23 \times 15\), the area model has 4 smaller rectangles. Each rectangle matches one partial product:

• \(20 \times 10 = 200\)
• \(20 \times 5 = 100\)
• \(3 \times 10 = 30\)
• \(3 \times 5 = 15\)

Then we add them all to get \(345\).

6. Place value matters

Be careful with place value when multiplying.

For example, in \(30 \times 7\), the answer is not 21. Since 30 means 3 tens,

Also,

It helps to think about what each number really means:

• \(20\) means 2 tens
• \(10\) means 1 ten
• 2 tens times 1 ten = 2 hundreds

7. Common mistakes to avoid

• Forgetting to break numbers apart correctly. Example: \(34 = 30 + 4\), not \(3 + 4\).
• Ignoring place value. Example: \(40 \times 6 = 240\), not 24.
• Leaving out a partial product. If both factors are broken apart, make sure every part is multiplied.
• Adding incorrectly at the end. Check your addition carefully.

8. Quick practice thinking

Try thinking through these:

• \(42 \times 3 = (40 \times 3) + (2 \times 3)\)
• \(21 \times 13 = (20 \times 10) + (20 \times 3) + (1 \times 10) + (1 \times 3)\)

You can solve them by finding each partial product and adding.

Summary

Area models and partial products help you multiply by breaking numbers into tens, hundreds, and ones. This makes large multiplication problems easier to understand.

Remember:

• Break numbers apart by place value.
• Multiply each part.
• Add the partial products.

If you use place value carefully, area models and partial products can help you solve multi-digit multiplication with confidence.

Standard Multiplication Algorithm is a quick and organized way to multiply larger whole numbers.

You may already know how to multiply using partial products. The standard algorithm does the same job, but it lines up the work in a neat way so it is faster to use.

In this lesson, you will learn how to multiply multi-digit whole numbers step by step. You will also see why place value is so important.

What is the standard multiplication algorithm?

The standard multiplication algorithm is the usual written method for multiplication. You multiply one digit at a time, starting with the digit in the ones place. Then you move to the next place value.

As you multiply, you may need to regroup. Regrouping means carrying extra tens, hundreds, or thousands to the next place.

Why place value matters

Every digit has a value based on its place. In the number 34, the 3 means 3 tens, or 30, and the 4 means 4 ones.

When multiplying, the place of each digit changes the size of the product. That is why it is important to line up numbers carefully.

For example, when multiplying by a digit in the tens place, you are really multiplying by tens, not just ones.

Steps for multiplying a 2-digit number by a 1-digit number

1. Write the numbers in vertical form.
2. Start multiplying from the ones place.
3. If the product is 10 or more, write the ones digit and regroup the tens digit above the next column.
4. Multiply the next digit and add any regrouped amount.

Example 1: Multiply \(34 \times 2\)

Write the problem vertically:

$$\begin{array}{r} 34 \\ \times\ 2 \\ \hline \end{array}$$

Start with the ones place: \(2 \times 4 = 8\). Write 8 in the ones place.

Next multiply the tens place: \(2 \times 3 = 6\). Since the 3 means 3 tens, this is really 6 tens, or 60.

So the product is:

$$\begin{array}{r} 34 \\ \times\ 2 \\ \hline 68 \end{array}$$

Therefore, \(34 \times 2 = 68\).

Steps for multiplying a 2-digit number by a 2-digit number

1. Write the numbers in vertical form.
2. Multiply the top number by the ones digit of the bottom number.
3. Then multiply the top number by the tens digit of the bottom number.
4. Because the tens digit really means tens, start that second row in the tens place.
5. Add the two partial products.

Example 2: Multiply \(23 \times 14\)

Write the problem vertically:

$$\begin{array}{r} 23 \\ \times\ 14 \\ \hline \end{array}$$

First multiply by the ones digit, 4.

\(4 \times 3 = 12\). Write 2 and regroup 1 ten.

\(4 \times 2 = 8\). Add the regrouped 1 to get 9.

The first partial product is 92.

$$\begin{array}{r} 23 \\ \times\ 14 \\ \hline 92 \end{array}$$

Now multiply by the tens digit, 1. This 1 really means 10, so the new row starts in the tens place.

\(1 \times 3 = 3\), but really 3 tens, so write it in the tens place.

\(1 \times 2 = 2\), but really 2 hundreds.

The second partial product is 230.

$$\begin{array}{r} 23 \\ \times\ 14 \\ \hline 92 \\ 230 \end{array}$$

Add the partial products:

$$\begin{array}{r} 92 \\ 230 \\ \hline 322 \end{array}$$

So, \(23 \times 14 = 322\).

Understanding the zero in the second row

When you multiply by a tens digit, the product is really tens, so the second row starts one place to the left.

Some students place a 0 in the ones place as a reminder.

For example, in \(23 \times 14\), the 1 in 14 means 10. So multiplying by 1 in the tens place is really multiplying by 10.

Example 3: Multiply \(46 \times 27\)

First multiply by the ones digit, 7.

\(7 \times 6 = 42\). Write 2 and regroup 4.

\(7 \times 4 = 28\). Add 4 to get 32.

The first partial product is 322.

$$\begin{array}{r} 46 \\ \times\ 27 \\ \hline 322 \end{array}$$

Now multiply by the tens digit, 2. This really means 20.

Start in the tens place.

\(2 \times 6 = 12\). Write 2 in the tens place and regroup 1.

\(2 \times 4 = 8\). Add 1 to get 9.

The second partial product is 920.

$$\begin{array}{r} 46 \\ \times\ 27 \\ \hline 322 \\ 920 \end{array}$$

Add:

$$\begin{array}{r} 322 \\ 920 \\ \hline 1242 \end{array}$$

So, \(46 \times 27 = 1242\).

Multiplying numbers with more digits

The same steps work for larger numbers too. You still multiply one digit at a time, regroup when needed, and line up each new row by place value.

Example 4: Multiply \(324 \times 15\)

First multiply by the ones digit, 5.

\(5 \times 4 = 20\). Write 0 and regroup 2.

\(5 \times 2 = 10\). Add 2 to get 12. Write 2 and regroup 1.

\(5 \times 3 = 15\). Add 1 to get 16.

The first partial product is 1620.

$$\begin{array}{r} 324 \\ \times\ 15 \\ \hline 1620 \end{array}$$

Now multiply by the tens digit, 1. This really means 10, so begin in the tens place.

\(1 \times 4 = 4\), really 4 tens.

\(1 \times 2 = 2\), really 2 hundreds.

\(1 \times 3 = 3\), really 3 thousands.

The second partial product is 3240.

$$\begin{array}{r} 324 \\ \times\ 15 \\ \hline 1620 \\ 3240 \end{array}$$

Add the partial products:

$$\begin{array}{r} 1620 \\ 3240 \\ \hline 4860 \end{array}$$

So, \(324 \times 15 = 4860\).

Common mistakes to watch for

• Not lining up digits correctly. Make sure each row matches the correct place value.
• Forgetting to regroup. If a product is 10 or more, carry the extra value to the next column.
• Forgetting the zero or place shift in the second row. When multiplying by tens, start one place to the left.
• Adding partial products incorrectly. Check your addition carefully at the end.

Helpful tips

• Work from right to left.
• Write neatly so the place values line up.
• Say the place value to yourself: ones, tens, hundreds.
• After solving, estimate to see if your answer makes sense.

For example, to estimate \(46 \times 27\), you can think of \(50 \times 30 = 1500\). The exact answer, 1242, is close to that estimate, so it makes sense.

Brief summary

The standard multiplication algorithm is a fast way to multiply multi-digit whole numbers. You multiply one digit at a time, regroup when needed, and use place value to line up each row correctly.

Remember: multiply the ones first, then the tens, then add the partial products. If you keep your work neat and pay attention to place value, this method becomes much easier.

Division Foundations and Partial Quotients

Division helps us find out how many equal groups we can make or how many items will be in each group. In 5th grade, we divide larger numbers, and one helpful method is called partial quotients.

The partial quotients method breaks a big division problem into smaller, easier steps. Instead of trying to do everything at once, we subtract large groups of the divisor until we reach 0 or a smaller leftover amount called a remainder.

In this lesson, you will learn what division means, how partial quotients work, and how to solve multi-digit division problems step by step.

1. What division means

Division can mean:

• Sharing equally: If 24 cookies are shared among 6 children, each child gets 4 cookies.
• Making groups: If you put 24 cookies into groups of 6, you can make 4 groups.

Both ideas can be written as:

$$24 \div 6 = 4$$

In a division problem:

• The dividend is the number being divided.
• The divisor is the number you divide by.
• The quotient is the answer.
• A remainder is what is left over if the division is not exact.

For example, in \(35 \div 6 = 5\) remainder \(5\):

• 35 is the dividend
• 6 is the divisor
• 5 is the quotient
• 5 is the remainder

2. What are partial quotients?

A partial quotient is a piece of the final quotient. We find how many groups of the divisor we can subtract at one time. Then we keep going until there is not enough left to subtract another full group.

This method works well because you can use numbers that are easy for you. You might subtract 10 groups, 5 groups, 2 groups, or 1 group at a time.

At the end, you add the partial quotients together to find the total quotient.

3. The basic steps for partial quotients

1. Write the division problem.
2. Choose an easy multiple of the divisor to subtract from the dividend.
3. Record how many groups you subtracted. This is a partial quotient.
4. Subtract and find what is left.
5. Repeat until the leftover number is smaller than the divisor.
6. Add all the partial quotients.
7. If there is anything left, write it as a remainder.

4. Why this method makes sense

Suppose you are solving \(84 \div 4\). You are really asking, “How many 4s are in 84?” Instead of counting by 4s one at a time, you can remove larger chunks.

For example:

• 10 groups of 4 is 40
• Another 10 groups of 4 is 40
• 1 more group of 4 is 4

That means:

$$84 = 40 + 40 + 4$$

So the number of groups is:

$$10 + 10 + 1 = 21$$

Therefore:

$$84 \div 4 = 21$$

5. Worked Example 1: Two-digit dividend

Solve \(84 \div 4\).

We will subtract easy multiples of 4.

• Subtract \(40\), because \(10 \times 4 = 40\)
• Subtract another \(40\), because \(10 \times 4 = 40\)
• Subtract \(4\), because \(1 \times 4 = 4\)

Here is the work:

$$ 84 \div 4 $$

$$ 84 - 40 = 44 \qquad (10) $$

$$ 44 - 40 = 4 \qquad (10) $$

$$ 4 - 4 = 0 \qquad (1) $$

Now add the partial quotients:

$$10 + 10 + 1 = 21$$

So:

$$84 \div 4 = 21$$

6. Worked Example 2: Three-digit dividend

Solve \(156 \div 6\).

Think of easy multiples of 6:

• \(10 \times 6 = 60\)
• \(20 \times 6 = 120\)
• \(5 \times 6 = 30\)
• \(1 \times 6 = 6\)

We can subtract 120 first.

$$156 - 120 = 36 \qquad (20)$$

Then subtract 30.

$$36 - 30 = 6 \qquad (5)$$

Then subtract 6.

$$6 - 6 = 0 \qquad (1)$$

Add the partial quotients:

$$20 + 5 + 1 = 26$$

So:

$$156 \div 6 = 26$$

7. Worked Example 3: Division with a remainder

Solve \(178 \div 8\).

We use easy multiples of 8.

• \(20 \times 8 = 160\)
• \(2 \times 8 = 16\)

Subtract 160 first.

$$178 - 160 = 18 \qquad (20)$$

Now subtract 16.

$$18 - 16 = 2 \qquad (2)$$

The leftover 2 is smaller than 8, so we stop.

Add the partial quotients:

$$20 + 2 = 22$$

So:

$$178 \div 8 = 22 \text{ remainder } 2$$

This can also be written as:

$$178 \div 8 = 22\ R2$$

8. Worked Example 4: Using bigger chunks

Solve \(432 \div 12\).

We can use bigger chunks to make the work faster.

• \(30 \times 12 = 360\)
• \(5 \times 12 = 60\)
• \(1 \times 12 = 12\)

Now subtract step by step:

$$432 - 360 = 72 \qquad (30)$$

$$72 - 60 = 12 \qquad (5)$$

$$12 - 12 = 0 \qquad (1)$$

Add the partial quotients:

$$30 + 5 + 1 = 36$$

So:

$$432 \div 12 = 36$$

9. Important ideas to remember

• You do not have to subtract the same size chunk each time.
• You can choose any easy multiple of the divisor.
• The partial quotients must be added at the end.
• If the leftover amount is smaller than the divisor, it is the remainder.

10. How to check your answer

You can check a division answer by using multiplication.

If there is no remainder:

$$\text{quotient} \times \text{divisor} = \text{dividend}$$

Example:

To check \(156 \div 6 = 26\):

$$26 \times 6 = 156$$

If there is a remainder:

$$\text{quotient} \times \text{divisor} + \text{remainder} = \text{dividend}$$

Example:

To check \(178 \div 8 = 22\ R2\):

$$22 \times 8 + 2 = 176 + 2 = 178$$

11. Common mistakes to avoid

• Forgetting to add the partial quotients. The numbers on the side must be added together.
• Subtracting a number that is not a multiple of the divisor. If the divisor is 6, subtract numbers like 6, 12, 18, 24, 60, or 120.
• Stopping too soon. Keep going until what is left is smaller than the divisor.
• Writing a remainder that is too large. A remainder must always be less than the divisor.

12. Quick practice ideas

Try these on your own using partial quotients:

• \(96 \div 3\)
• \(245 \div 5\)
• \(203 \div 9\)

As you solve, ask yourself:

• What easy multiples of the divisor do I know?
• How much can I subtract first?
• Did I add all my partial quotients?
• Is my remainder smaller than the divisor?

Summary

Partial quotients is a division method that breaks a large problem into smaller, easier parts. You subtract easy multiples of the divisor, keep track of how many groups you removed, and then add those groups together.

This method helps you understand what division means instead of only memorizing steps. With practice, you will get faster and more confident dividing multi-digit whole numbers.

Long Division Algorithm helps us divide larger whole numbers step by step. It is a neat way to find how many equal groups we can make, and whether anything is left over.

When we use long division, we are answering questions like: How many times does one number fit into another? For example, if 84 stickers are shared equally among 4 students, long division helps us find how many stickers each student gets.

Long division may look big at first, but it becomes much easier when you follow the same steps in order every time.

The 4 main steps are:

1. Divide – Ask: how many times does the divisor fit into the number?
2. Multiply – Multiply the divisor by the number you chose.
3. Subtract – Subtract to find what is left.
4. Bring down – Bring down the next digit and repeat.

Many students remember this with the phrase Divide, Multiply, Subtract, Bring Down.

Let’s name the parts of a division problem:

• Dividend: the number being divided
• Divisor: the number you are dividing by
• Quotient: the answer
• Remainder: the amount left over, if any

In the problem \(84 \div 4 = 21\):

• 84 is the dividend
• 4 is the divisor
• 21 is the quotient

Important idea: In long division, you work from left to right, one digit at a time.

We also use multiplication facts a lot in division. Knowing facts like \(6 \times 4 = 24\) helps you decide how many times 6 goes into 24.

Example 1: One-digit divisor, no remainder

Find \(84 \div 4\).

We write it like this:

$$4\overline{)84}$$

Step 1: Divide. Ask: how many times does 4 go into 8? It goes into 8 exactly 2 times.

Write 2 above the 8.

$$\begin{array}{r}2\\4\overline{)84}\end{array}$$

Step 2: Multiply. Multiply \(2 \times 4 = 8\).

Write 8 under the 8.

$$\begin{array}{r}2\\4\overline{)84}\\\underline{8}\end{array}$$

Step 3: Subtract. \(8 - 8 = 0\).

Step 4: Bring down. Bring down the 4.

Now ask: how many times does 4 go into 4? It goes 1 time.

Write 1 above the 4.

$$\begin{array}{r}21\\4\overline{)84}\\\underline{8}\\04\end{array}$$

Multiply: \(1 \times 4 = 4\).

Subtract: \(4 - 4 = 0\).

There is nothing left, so the answer is:

$$84 \div 4 = 21$$

Example 2: One-digit divisor with a remainder

Find \(95 \div 4\).

Set up the problem:

$$4\overline{)95}$$

Step 1: Divide. How many times does 4 go into 9? It goes 2 times, because \(2 \times 4 = 8\), and \(3 \times 4 = 12\) is too big.

Write 2 above the 9.

Step 2: Multiply. \(2 \times 4 = 8\).

Step 3: Subtract. \(9 - 8 = 1\).

Step 4: Bring down. Bring down the 5, making 15.

Now ask: how many times does 4 go into 15? It goes 3 times, because \(3 \times 4 = 12\), and \(4 \times 4 = 16\) is too big.

Write 3 above the 5.

Multiply: \(3 \times 4 = 12\).

Subtract: \(15 - 12 = 3\).

There are no more digits to bring down, so 3 is the remainder.

$$95 \div 4 = 23\text{ R }3$$

This means 4 fits into 95 a total of 23 times, with 3 left over.

Example 3: When the first digit is too small

Find \(156 \div 12\).

Set it up:

$$12\overline{)156}$$

Look at the first digit, 1. Can 12 go into 1? No. So we look at the first two digits, 15.

Step 1: Divide. How many times does 12 go into 15? It goes 1 time.

Write 1 above the 5 in 156.

Step 2: Multiply. \(1 \times 12 = 12\).

Step 3: Subtract. \(15 - 12 = 3\).

Step 4: Bring down. Bring down the 6, making 36.

Now ask: how many times does 12 go into 36? It goes 3 times because \(3 \times 12 = 36\).

Write 3 above the 6.

Multiply: \(3 \times 12 = 36\).

Subtract: \(36 - 36 = 0\).

So the answer is:

$$156 \div 12 = 13$$

Example 4: Two-digit divisor with a remainder

Find \(389 \div 15\).

Set it up:

$$15\overline{)389}$$

First, check 3. Since 15 cannot go into 3, look at 38.

Step 1: Divide. How many times does 15 go into 38? It goes 2 times because \(2 \times 15 = 30\), and \(3 \times 15 = 45\) is too big.

Write 2 above the 8.

Step 2: Multiply. \(2 \times 15 = 30\).

Step 3: Subtract. \(38 - 30 = 8\).

Step 4: Bring down. Bring down the 9, making 89.

Now ask: how many times does 15 go into 89? It goes 5 times because \(5 \times 15 = 75\), and \(6 \times 15 = 90\) is too big.

Write 5 above the 9.

Multiply: \(5 \times 15 = 75\).

Subtract: \(89 - 75 = 14\).

There are no more digits to bring down, so the remainder is 14.

$$389 \div 15 = 25\text{ R }14$$

How to check your answer

You can check division with multiplication.

If there is no remainder, multiply the quotient by the divisor.

For Example 3:

$$13 \times 12 = 156$$

So \(156 \div 12 = 13\) is correct.

If there is a remainder, use this rule:

$$\text{divisor} \times \text{quotient} + \text{remainder} = \text{dividend}$$

For Example 4:

$$15 \times 25 + 14 = 375 + 14 = 389$$

So the answer is correct.

Tips for success

• Go one step at a time. Do not skip steps.
• Use multiplication facts. They help you decide how many times the divisor fits.
• Choose the greatest number that is not too big. For example, when dividing by 15, if 6 times is too much, try 5 times.
• Keep digits lined up carefully. Good spacing helps prevent mistakes.
• Remember that the remainder must be smaller than the divisor. For example, when dividing by 15, the remainder must be less than 15.

Common mistakes to avoid

• Forgetting to bring down the next digit
• Using a multiplication number that is too large
• Placing the quotient digit above the wrong digit
• Forgetting to subtract after multiplying
• Writing a remainder that is larger than the divisor

What if the divisor does not go into the first digit?

This happens often with two-digit divisors. Just look at more digits from the dividend until the divisor can fit.

For example, in \(156 \div 12\), 12 cannot go into 1, so we look at 15 instead.

Let’s review the pattern:

1. Look at the first digit or digits.
2. Divide.
3. Multiply.
4. Subtract.
5. Bring down the next digit.
6. Repeat until there are no more digits left.

Brief Summary

Long division is a step-by-step way to divide large whole numbers. The steps are Divide, Multiply, Subtract, Bring Down. You can use long division with one-digit and two-digit divisors, and your answer may have a remainder. Always check your work by multiplying the quotient by the divisor and adding the remainder if needed.

Interpreting Remainders means deciding what a remainder means in a real-world division problem.

Sometimes the remainder is just extra that is left over. Sometimes it means you need one more group. Other times, the remainder should be written as part of an answer, like a fraction or decimal.

In this lesson, you will learn how to divide, find the remainder, and then decide what to do with it based on the situation.

First, what is a remainder?

When one whole number does not divide evenly by another whole number, there is an amount left over. That amount is called the remainder.

For example, in $$17 \div 5 = 3 \text{ remainder } 2$$ the number 5 fits into 17 three times, and 2 are left over.

Why does interpreting the remainder matter?

In math class, you might stop after finding the quotient and remainder. But in word problems, you must think about what the answer means.

Ask yourself:

• Does the leftover amount get ignored?
• Do I need to round up because one more group is needed?
• Should I write the remainder as a fraction or decimal because the leftover part matters?

There are 3 common ways to interpret a remainder:

1. Ignore the remainder
2. Round up
3. Use the remainder as part of the answer

1. Ignore the remainder

Ignore the remainder when only the number of full groups matters.

For example, if 26 students are making teams of 4, you can make:

$$26 \div 4 = 6 \text{ remainder } 2$$

This means there are 6 full teams, with 2 students left over. If the question asks, “How many full teams can be made?” the answer is 6 teams. The remainder is ignored because it does not make another full team.

2. Round up

Round up when the remainder means you need one more group, item, or trip.

For example, if 26 students need to ride in vans that hold 4 students each:

$$26 \div 4 = 6 \text{ remainder } 2$$

Six vans hold only 24 students. There are still 2 students left. So you need 7 vans.

Even though the quotient is 6 remainder 2, the real answer is 7 because everyone must fit.

3. Use the remainder as part of the answer

Sometimes the leftover part is important, so you do not ignore it or round it away.

You can write the remainder as a fraction:

$$17 \div 5 = 3 \frac{2}{5}$$

because the remainder 2 is part of another group of 5.

You can also write it as a decimal:

$$17 \div 5 = 3.4$$

This is useful when the answer can be shared exactly, such as money, distance, or length.

How do I know which one to use?

Use the story in the problem to decide.

• Ignore the remainder when only complete groups count.
• Round up when you need enough groups to cover everything.
• Use a fraction or decimal when the leftover part is still part of the answer.

A helpful thinking plan:

1. Divide.
2. Find the remainder.
3. Ask, “What does the remainder mean in this problem?”
4. Choose whether to ignore it, round up, or keep it as part of the answer.

Worked Example 1: Ignore the remainder

Problem: A teacher has 23 markers. She puts them into boxes that hold 5 markers each. How many full boxes can she make?

Step 1: Divide.

$$23 \div 5 = 4 \text{ remainder } 3$$

Step 2: Interpret the remainder.

The question asks for full boxes. The 3 leftover markers do not fill another box.

Answer: She can make 4 full boxes.

Worked Example 2: Round up

Problem: There are 23 markers. Boxes hold 5 markers each. How many boxes are needed to hold all the markers?

Step 1: Divide.

$$23 \div 5 = 4 \text{ remainder } 3$$

Step 2: Interpret the remainder.

Four boxes can hold only 20 markers. There are still 3 markers left, so one more box is needed.

Answer: 5 boxes are needed.

Worked Example 3: Use the remainder as a fraction

Problem: Three granola bars are shared equally among 4 children. How much does each child get?

Step 1: Write the division.

$$3 \div 4$$

Step 2: Think about the situation.

The bars can be split into equal parts, so the leftover amount should be part of the answer.

$$3 \div 4 = \frac{3}{4}$$

Answer: Each child gets \(\frac{3}{4}\) of a granola bar.

Worked Example 4: Use the remainder as a decimal

Problem: A rope that is 14 meters long is cut into 4 equal pieces. How long is each piece?

Step 1: Divide.

$$14 \div 4 = 3 \text{ remainder } 2$$

Step 2: Interpret the remainder.

The leftover part is still part of each piece, so we should not ignore it or round up.

The remainder 2 out of 4 is:

$$\frac{2}{4} = \frac{1}{2} = 0.5$$

So:

$$14 \div 4 = 3.5$$

Answer: Each piece is 3.5 meters long.

Look at the question carefully

Sometimes two problems use the same division, but the answers are different because the questions are different.

For example:

$$31 \div 6 = 5 \text{ remainder } 1$$

• If 31 cookies are packed into bags of 6, how many full bags? Answer: 5
• If 31 cookies are packed into bags of 6, how many bags are needed? Answer: 6
• If 31 feet of ribbon are shared equally into 6 parts, how much in each part? Answer: \(5 \frac{1}{6}\) feet

Clue words can help

• Full groups, complete rows, or whole teams often mean ignore the remainder.
• Needed, enough, all, or everyone fits often mean round up.
• Shared equally, each piece, or exact amount often mean use a fraction or decimal.

Be careful of these common mistakes:

• Do not always round up just because there is a remainder.
• Do not always ignore the remainder just because the quotient is a whole number.
• Do not forget to read what the question is really asking.

Try these quick thinking checks:

• 27 chairs are put into rows of 5. How many full rows? 5 rows because $$27 \div 5 = 5 \text{ remainder } 2$$ and only full rows count.
• 27 chairs are put into rows of 5. How many rows are needed for all the chairs? 6 rows because the 2 extra chairs still need a row.
• 27 liters of juice are poured equally into 5 containers. How much in each container? \(5 \frac{2}{5}\) liters or 5.4 liters.

Summary

A remainder is the amount left over after division.

To interpret a remainder, think about the real-life meaning of the problem. You may need to ignore the remainder, round up, or keep the remainder as a fraction or decimal.

The most important step is to ask, “What does the leftover amount mean here?”

Order of Operations tells us the correct order to solve a math expression when it has more than one operation.

This is important because if people solve the same expression in different orders, they can get different answers. Order of Operations helps everyone get the same correct answer.

An expression is a math sentence with numbers and operation signs like addition, subtraction, multiplication, and division. For example, \(8 + 3 \times 2\) is an expression.

When solving expressions, follow these steps in order:

1. Grouping symbols first: solve anything inside parentheses \(( )\) or brackets \([ ]\).
2. Multiplication and division: work from left to right.
3. Addition and subtraction: work from left to right.

You do not always go left to right for everything. First, you must check for grouping symbols. Then do multiplication and division before addition and subtraction.

Here is the order in a short form:

$$\text{Grouping Symbols} \rightarrow \text{Multiplication/Division} \rightarrow \text{Addition/Subtraction}$$

Let’s look closely at each part.

1. Grouping symbols

Grouping symbols tell you what to solve first. In 5th grade, the most common grouping symbols are:

• Parentheses: \(( )\)
• Brackets: \([ ]\)

If there are brackets and parentheses together, solve the innermost part first.

2. Multiplication and division

After solving inside grouping symbols, do all multiplication and division from left to right.

This means if division comes before multiplication, do the division first because it is farther left.

3. Addition and subtraction

Last, do addition and subtraction from left to right.

Just like multiplication and division, you do whichever one comes first as you move from left to right.

Important reminders

• Addition does not always come before subtraction.
• Multiplication does not always come before division.
• For those pairs, work left to right.
• Always solve inside grouping symbols first.

Worked Example 1

Solve:

$$8 + 3 \times 2$$

First, look for grouping symbols. There are none.

Next, do multiplication before addition.

$$3 \times 2 = 6$$

Now the expression becomes:

$$8 + 6$$

Then add:

$$8 + 6 = 14$$

Answer: \(14\)

If someone added first, they would get \(8 + 3 = 11\), then \(11 \times 2 = 22\), which is incorrect. That is why order matters.

Worked Example 2

Solve:

$$\left(12 - 4\right) \div 2$$

First, solve inside the parentheses:

$$12 - 4 = 8$$

Now the expression becomes:

$$8 \div 2$$

Now divide:

$$8 \div 2 = 4$$

Answer: \(4\)

Worked Example 3

Solve:

$$18 \div 3 \times 2$$

There are no grouping symbols.

Now do division and multiplication from left to right.

First:

$$18 \div 3 = 6$$

Now the expression becomes:

$$6 \times 2$$

Then multiply:

$$6 \times 2 = 12$$

Answer: \(12\)

Notice that we did not multiply \(3 \times 2\) first. We followed left to right for division and multiplication.

Worked Example 4

Solve:

$$[6 + (10 - 2)] \times 3$$

First, solve inside the parentheses:

$$10 - 2 = 8$$

Now the expression becomes:

$$[6 + 8] \times 3$$

Next, solve inside the brackets:

$$6 + 8 = 14$$

Now the expression becomes:

$$14 \times 3$$

Then multiply:

$$14 \times 3 = 42$$

Answer: \(42\)

How to check your work

• Did you solve inside parentheses or brackets first?
• Did you do multiplication and division before addition and subtraction?
• Did you move left to right when operations were in the same group?
• Did you rewrite the expression carefully after each step?

Common mistakes to avoid

• Mistake: Always working left to right, no matter what.
  Fix: Check for grouping symbols first, then multiplication/division, then addition/subtraction.
• Mistake: Doing addition before multiplication.
  Fix: Multiplication and division come before addition and subtraction.
• Mistake: Forgetting to solve everything inside brackets or parentheses.
  Fix: Finish the grouped part before moving on.
• Mistake: Not following left to right for multiplication and division or for addition and subtraction.
  Fix: In the same group, go from left to right.

Try these on your own

1. \(7 + 4 \times 5\)
2. \((15 - 9) + 6\)
3. \(20 \div 5 \times 3\)
4. \([9 + (8 \div 2)] - 5\)

Answers

1. \(7 + 4 \times 5 = 7 + 20 = 27\)
2. \((15 - 9) + 6 = 6 + 6 = 12\)
3. \(20 \div 5 \times 3 = 4 \times 3 = 12\)
4. \([9 + (8 \div 2)] - 5 = [9 + 4] - 5 = 13 - 5 = 8\)

Summary

Order of Operations is the rule for solving expressions in the correct order. First solve inside grouping symbols, then do multiplication and division from left to right, and last do addition and subtraction from left to right.

When you follow these steps carefully, you can solve expressions correctly and confidently.

Mental Math and Compensation is a smart way to do math in your head.

Instead of solving a problem the long way, you can change a number to a friendly number first. Friendly numbers are numbers that are easy to work with, like 10, 20, 50, 100, or 1,000.

When you change a number to make the math easier, you must also compensate. That means you fix the answer so it stays correct.

Compensation is like keeping a balance. If you add a little extra at the beginning, you take that extra away at the end. If you take away too much at first, you put it back later.

Why use compensation?

• It helps you solve problems faster.
• It makes mental math easier.
• It helps you notice number patterns.
• It builds confidence with larger numbers.

1. Compensation with Addition

In addition, you can change one addend to a nearby friendly number. Then adjust the total to make up for the change.

For example, if you change 39 to 40, you added 1 extra. So after adding, you must subtract 1.

Here is the idea:

$$39 + 26 = 40 + 26 - 1$$

This works because 40 is 1 more than 39.

Worked Example 1

Find \(39 + 26\) mentally.

1. Change 39 to the friendly number 40.
2. Add: \(40 + 26 = 66\)
3. Since 39 was really 1 less than 40, subtract 1.

$$40 + 26 = 66$$ $$66 - 1 = 65$$

So, \(39 + 26 = 65\).

2. Compensation with Subtraction

In subtraction, you can also change a number to a friendly number. But you must be careful to adjust correctly.

Suppose you solve \(52 - 19\). The number 19 is close to 20, which is easier to subtract.

If you subtract 20 instead of 19, you subtract 1 too much. So you need to add 1 back.

Here is the idea:

$$52 - 19 = 52 - 20 + 1$$

Worked Example 2

Find \(52 - 19\) mentally.

1. Change 19 to 20.
2. Subtract: \(52 - 20 = 32\)
3. You subtracted 1 too much, so add 1 back.

$$52 - 20 = 32$$ $$32 + 1 = 33$$

So, \(52 - 19 = 33\).

3. Compensation with Larger Addition

Compensation is very helpful with bigger numbers too.

If a number is close to 100 or 1,000, you can use that friendly number to make the work quick and neat.

Worked Example 3

Find \(298 + 47\) mentally.

1. Change 298 to 300.
2. Add: \(300 + 47 = 347\)
3. But 298 is 2 less than 300, so subtract 2.

$$300 + 47 = 347$$ $$347 - 2 = 345$$

So, \(298 + 47 = 345\).

4. Compensation with Larger Subtraction

You can use the same idea with larger subtraction problems.

Worked Example 4

Find \(604 - 297\) mentally.

1. Change 297 to 300.
2. Subtract: \(604 - 300 = 304\)
3. Since 300 is 3 more than 297, you subtracted 3 too much.
4. Add 3 back.

$$604 - 300 = 304$$ $$304 + 3 = 307$$

So, \(604 - 297 = 307\).

How to know whether to add back or subtract back

• Addition: If you rounded a number up, subtract the extra at the end. If you rounded a number down, add the missing amount at the end.
• Subtraction: If you subtracted a bigger number than the original, add back the extra. If you subtracted a smaller number than the original, subtract the missing amount.

Quick examples

• \(48 + 27 = 50 + 27 - 2 = 75\)
• \(73 + 19 = 73 + 20 - 1 = 92\)
• \(81 - 29 = 81 - 30 + 1 = 52\)
• \(400 - 198 = 400 - 200 + 2 = 202\)

Helpful steps for using compensation

1. Look for a number close to a friendly number.
2. Change it to the friendly number.
3. Do the easier math problem in your head.
4. Adjust the answer to make up for the change.

When is compensation useful?

• When a number ends in 9, 8, 19, 29, 99, or 98
• When a number is close to 10, 20, 50, 100, or 1,000
• When you want to solve a problem quickly without writing it down

Be careful!

The most common mistake is forgetting to adjust the answer after changing the number.

Another common mistake is adjusting in the wrong direction. Always ask yourself:

• Did I make the number bigger or smaller?
• Did I add too much or subtract too much?
• What do I need to do to fix it?

Summary

Mental math and compensation help you solve problems quickly by changing numbers into friendly numbers.

After using the friendly number, you must adjust the answer so it matches the original problem.

With practice, compensation makes addition and subtraction with whole numbers faster and easier to do in your head.