Divisibility Rules, Factor Pairs, and Prime Numbers | Grade 5 Math | LevelKite

You already know that a factor of a number divides it with nothing left over. For example, $4$ is a factor of $20$ because $20 = 4 \times 5$ .

There is a second way to say this: $20$ is divisible by $4$ . A number is divisible by $4$ when you can divide it by $4$ and get a whole number, with no remainder. The number you divide by is called the divisor.

So these are the same question, asked two ways:

Is $4$ a factor of $20$ ?
Is $20$ divisible by $4$ ?

In this lesson you will learn quick checks that answer this kind of question. Then you will use them to find every pair of whole numbers that multiply to make a number, and to sort numbers by how many factors they have.

A divisibility rule is a quick check on the digits of a whole number. It tells you whether a certain divisor is a factor, and you do not have to divide. Three ideas make the rules work.

The last digit. $74$ is $7$ tens and $4$ ones. Every ten is divisible by $2$ , by $5$ and by $10$ , so the tens never leave a remainder. Only the ones digit matters for these three divisors.

The sum of the digits. Each ten is $9 + 1$ , and $9$ is divisible by $3$ and by $9$ . So each ten leaves $1$ over. In $74$ , the seven tens leave $7$ over and the ones add $4$ more: $74 = 7 \times 9 + 7 + 4$ . All that is left to check is $7 + 4 = 11$ , the sum of the digits. $11$ is not divisible by $3$ or by $9$ , so $74$ is not either. Hundreds work the same way, because $100 = 99 + 1$ and $99 = 9 \times 11$ .

Two rules together. $6 = 2 \times 3$ , so every number divisible by $6$ is divisible by $2$ and by $3$ . It works the other way too. Count by threes: $3, 6, 9, 12, 15, 18$ . The numbers in that list that are also divisible by $2$ are $6$ , $12$ and $18$ . Those are exactly the numbers you say when you count by sixes, and the pattern keeps going.

Divisibility rules for whole numbers. Pass the check and the divisor is a factor. Fail and it is not.

Divisor	It is a factor of a whole number exactly when	A number that passes	A number that fails
$2$	the last digit is $0$ , $2$ , $4$ , $6$ or $8$ . These numbers are even. All other whole numbers are odd.	$58$ : last digit $8$ , and $58 = 2 \times 29$	$57$ : last digit $7$
$5$	the last digit is $0$ or $5$	$85$ : last digit $5$ , and $85 = 5 \times 17$	$58$ : last digit $8$
$10$	the last digit is $0$	$90$ : last digit $0$ , and $90 = 10 \times 9$	$85$ : last digit $5$
$3$	the sum of the digits is divisible by $3$	$57$ : $5 + 7 = 12$ , and $57 = 3 \times 19$	$58$ : $5 + 8 = 13$
$9$	the sum of the digits is divisible by $9$	$72$ : $7 + 2 = 9$ , and $72 = 9 \times 8$	$57$ : $5 + 7 = 12$
$6$	the number passes both the rule for $2$ and the rule for $3$	$78$ : even, $7 + 8 = 15$ , and $78 = 6 \times 13$	$57$ : passes for $3$ , but it is odd

Worked example

Quick checks on 135

Which of the divisors $2$ , $3$ , $5$ , $6$ , $9$ and $10$ are factors of $135$ ?

Start with the last digit, because it decides 2, 5 and 10. The last digit of 135 is 5. It is not 0, 2, 4, 6 or 8, so 135 is odd and 2 is not a factor. The rule for 5 needs a last digit of 0 or 5, so 5 is a factor. The rule for 10 needs a last digit of 0, so 10 is not a factor.
Add the digits, because their sum decides 3 and 9. The sum is 9. Since 9 is divisible by 3 and by 9, both 3 and 9 are factors of 135.
$1 + 3 + 5 = 9$
Check 6 last. Its rule needs the rule for 2 and the rule for 3 to both pass. The rule for 3 passed but the rule for 2 failed, so 6 is not a factor.
Match each yes to a multiplication sentence. When a divisor is a factor, you can multiply it by a whole number to make 135.
$135 = 3 \times 45 = 5 \times 27 = 9 \times 15$

Answer

Of the six divisors, $3$ , $5$ and $9$ are factors of $135$ . The divisors $2$ , $6$ and $10$ are not.

Now put the quick checks to work. Take a whole number that is $1$ or more, such as $12$ . Each of its factors has a partner: the number you multiply it by to make $12$ . A factor pair is two whole numbers that multiply to make the number. $3$ and $4$ are a factor pair of $12$ because $3 \times 4 = 12$ .

To find every factor pair, test the divisors in order: $1$ , then $2$ , then $3$ , and so on. Skip none. Each time a divisor is a factor, divide to find its partner and write the pair down.

Here is what you get for $12$ if you test every divisor from $1$ all the way to $12$ :

\underbrace{1 \times 12 \quad 2 \times 6 \quad 3 \times 4}_{\text{new pairs}} \qquad \underbrace{4 \times 3 \quad 6 \times 2 \quad 12 \times 1}_{\text{repeats}}

The first number climbs while its partner drops. After $3 \times 4$ they cross over, and the pairs begin to repeat. Every pair from then on is one you already have, turned around: $4 \times 3$ is the same pair as $3 \times 4$ .

So you can stop early. Here is the stopping rule for a whole number that is $1$ or more: when you test in order from $1$ and skip nothing, stop as soon as the divisor times itself is more than the number. That divisor is your stopping point. For $12$ the stopping point is $4$ , because $4 \times 4 = 16$ and $16 > 12$ .

Why is that safe? At the stopping point, and at every divisor after it, any partner must be smaller than the stopping point. You have already tested all of those smaller numbers, so that pair is already on your list.

One special case: a divisor can be its own partner, as in $6 \times 6 = 36$ . That still counts as a factor pair, but it gives just one factor, $6$ .

Worked example

All the factor pairs of 48

Find all the factor pairs of $48$ .

Test 1, 2 and 3. Every whole number has 1 as a factor. The last digit of 48 is 8, so 48 is even and 2 is a factor. The digits add to 12, which is divisible by 3, so 3 is a factor. Divide to find each partner: 48 divided by 2 is 24, and 48 divided by 3 is 16.
$1 \times 48 \qquad 2 \times 24 \qquad 3 \times 16$
Test 4, 5 and 6. Dividing 48 by 4 gives 12 with no remainder, so 4 is a factor. The last digit is not 0 or 5, so 5 is not a factor. The rules for 2 and 3 both passed, so 6 is a factor. Its partner is 8, because 48 divided by 6 is 8.
$4 \times 12 \qquad 6 \times 8$
The next divisor is 7. Check the stopping rule before you test it: 7 times itself is 49, which is more than 48. So 7 is the stopping point. If you kept going, the next factor you would find is 8 with partner 6, and that is the pair 6 and 8 again.
$7 \times 7 = 49 > 48$
Write the pairs you found as one line of number sentences. The five pairs hold ten different factors of 48: 1, 2, 3, 4, 6, 8, 12, 16, 24 and 48.
$48 = 1 \times 48 = 2 \times 24 = 3 \times 16 = 4 \times 12 = 6 \times 8$

Answer

The factor pairs of $48$ are $1$ and $48$ , $2$ and $24$ , $3$ and $16$ , $4$ and $12$ , and $6$ and $8$ .

Factor pairs also tell you what kind of number you have.

Every whole number greater than $1$ has a factor pair made of $1$ and the number itself. That already gives it two different factors.

Some numbers stop there. Test the divisors for $7$ : it is odd, so $2$ is not a factor, and $3 \times 3 = 9$ is more than $7$ , so $3$ is the stopping point. The only factor pair of $7$ is $1 \times 7$ .

Other numbers have more. $6$ has the pair $1 \times 6$ and also the pair $2 \times 3$ . That makes four factors: $1$ , $2$ , $3$ and $6$ .

These two kinds of numbers have names.

Key concept

The factors decide: prime, composite or neither

Find the factor pairs of a whole number, then count its factors.

A prime number is a whole number greater than $1$ that has exactly two factors: $1$ and itself.
A composite number is a whole number greater than $1$ that has more than two factors.
The number $1$ is neither prime nor composite. Its only factor pair is $1 \times 1$ , so it has just one factor. That is too few to be prime, and too few to be composite.

Sorting numbers by counting their factors

Number	Factor pairs	Factors	How many factors	Kind of number
$1$	$1 \times 1$	$1$	$1$	Neither
$2$	$1 \times 2$	$1$ , $2$	$2$	Prime
$9$	$1 \times 9$ , $3 \times 3$	$1$ , $3$ , $9$	$3$	Composite
$12$	$1 \times 12$ , $2 \times 6$ , $3 \times 4$	$1$ , $2$ , $3$ , $4$ , $6$ , $12$	$6$	Composite
$13$	$1 \times 13$	$1$ , $13$	$2$	Prime
$15$	$1 \times 15$ , $3 \times 5$	$1$ , $3$ , $5$ , $15$	$4$	Composite

Worked example

Is 97 prime or composite?

Is $97$ prime or composite?

The number 97 is greater than 1, so it has the factor pair 1 and 97. Look for a second pair. Find the stopping point first: 10 times itself is 100, which is more than 97. So the divisors to test are 2 through 9.
$10 \times 10 = 100 > 97$
Test 2, 3, 4 and 5. The last digit is 7, so 97 is odd and 2 is not a factor. The digits add to 16, which is not divisible by 3, so 3 is not a factor. Any whole number times 4 is even and 97 is odd, so 4 is not a factor. The last digit is not 0 or 5, so 5 is not a factor.
$9 + 7 = 16$
Test 6, 7, 8 and 9. The rule for 6 needs 2 and 3 to be factors, and neither is. For 7, use two facts: 7 times 13 is 91 and 7 times 14 is 98. The answers jump from 91 straight to 98 and skip 97, so 7 is not a factor. Any whole number times 8 is even, so 8 is not a factor. The digits add to 16, which is not divisible by 9, so 9 is not a factor.
$7 \times 13 = 91 \qquad 7 \times 14 = 98$
None of the divisors from 2 through 9 is a factor, so 97 has no second factor pair. It has exactly two factors, 1 and 97.
$97 = 1 \times 97$

Answer

$97$ is prime, because its only factors are $1$ and $97$ .

The big idea: the factors of a whole number tell you what kind of number it is.

Divisibility rules test $2$ , $3$ , $5$ , $6$ , $9$ and $10$ by looking at digits.
To find every factor pair of a whole number that is $1$ or more, test divisors in order from $1$ . Stop when the divisor times itself is more than the number.
For a whole number greater than $1$ , exactly two factors means prime and more than two means composite. The number $1$ has one factor, so it is neither.

Next, you will split composite numbers into prime numbers multiplied together, using a diagram called a factor tree. You will also find the least common multiple of two numbers: the first number that shows up in both lists when you count by each of them.