 Factors and Primes
 Learning Objective(s)
 · Recognize (by using the divisibility rule) if a number is divisible by 2, 3, 4, 5, 6, 9, or 10.
 · Find the factors of a number.
 · Determine whether a number is prime, composite, or neither.
 · Find the prime factorization of a number.
Natural numbers, also called counting numbers(1, 2, 3, and so on), can be expressed as a product of their factors. When working with a fraction, you often need to make the fraction as simple as possible.
This means that the numerator and the denominator have no common factors other than 1. It will help to find factors, so that later you can simplify and compare fractions.
When a natural number is expressed as a product of two other natural numbers, those other numbers are factors of the original number. For example, two factors of 12 are 3 and 4, because 3 • 4 = 12.
When one number can be divided by another number with no remainder, we say the first number is divisible by the other number. For example, 20 is divisible by 4 (). If a number is divisible by another number, it is also a multiple of that number. For example, 20 is divisible by 4, so 20 is a multiple of 4.
Divisibility tests are rules that let you quickly tell if one number is divisible by another. There are many divisibility tests. Here are some of the most useful and easy to remember:
 A number is divisible by 2 if the last (ones) digit is divisible by 2. That is, the last digit is 0, 2, 4, 6, or 8. (We then say the number is an even number.) For example, in the number 236, the last digit is 6. Since 6 is divisible by 2 (6 ÷ 2 = 3), 236 is divisible by 2.
 A number is divisible by 3 if the sum of all the digits is divisible by 3. For example, the sum of the digits of 411 is 4 + 1 + 1 = 6. Since 6 is divisible by 3 (), 411 is divisible by 3.
 A number is divisible by 5 if the last digit is 0 or 5. For example, 275 and 1,340 are divisible by 5 because the last digits are 5 and 0.
 A number is divisible by 10 if the last digit is 0. For example, 520 is divisible by 10 (last digit is 0).

Here is a summary of the most commonly used divisibility rules.
A number is divisible by  Condition  Example 
2  The last digit is even (0, 2, 4, 6, 8).  426 yes 273 no 
3  The sum of the digits is divisible by 3.  642 yes (6 + 4 + 2 = 12, 12 is divisible by 3) 721 no (7 + 2 + 1 = 10, 10 is not divisible by 3) 
4  The last two digits form a number that is divisible by 4.  164 yes (64 is divisible by 4) 135 no (35 is not divisible by 4) 
5  The last digit is 0 or 5.  685 yes 432 no 
6  The number is divisible by 2 and 3.  324 yes (it is even and 3 + 2 + 4 = 9) 411 no (although divisible by 3, it is not even) 
9  The sum of the digits is divisible by 9.  279 yes (2 + 7 + 9 = 18) 512 no (5 + 1 + 2 =8) 
10  The last digit is a 0.  62 yes 238 no 
If you need to check for divisibility of a number without a rule, divide (either using a calculator or by hand). If the result is a number without any fractional part or remainder, then the number is divisible by the divisor. If you forget a rule, you can also use this strategy.
Show/Hide Answer A) 2 and 3 only Incorrect. Although 522 is divisible by 2 (the last digit is even) and 3 (5 + 2 + 2 = 9, which is a multiple of 3), it is also divisible by 6 and 9. The correct answer is 2, 3, 6, and 9 only. B) 4 only Incorrect. The last two digits (22) are not divisible by 4, so 522 is not divisible by 4. 522 has a last digit divisible by 2, so 522 is divisible by 2. The sum of the digits is divisible by 3 (5 + 2 + 2 = 9) and by 9, so 522 is divisible by 3 and 9. Since 522 is divisible by 2 and by 3, it is divisible by 6. Since the last digit is not 0 or 5, 522 is not divisible by 5 or 10. The correct answer is 2, 3, 6, and 9 only. C) 2, 3, 6, and 9 only Correct. 522 is divisible by 2 (the last digit is even) and 3 (5 + 2 + 2 = 9, which is a multiple of 3). Since it is divisible by 2 and 3, it is also divisible by 6. Also, the sum of the digits is divisible by 9, so 522 is divisible by 9. Since the last digit is not 0 or 5, 522 is not divisible by 5 or 10. The number formed by the last two digits, 22, is not divisible by 4, so 522 is not divisibly by 4. D) 4, 5, and 10 only Incorrect. The last two digits are not divisible by 4, so 522 is not divisible by 4. The last digit is not 0 or 5 so 522 is not divisible by 5. The last digit is not 0, so 522 is not divisible by 10. However, 522 is divisible by 2 (the last digit is even) and 3 (5 + 2 + 2 = 9, which is a multiple of 3). Since it is divisible by 2 and 3, it is also divisible by 6. Also, the sum of the digits is divisible by 9, so 522 is divisible by 9. The correct answer is 2, 3, 6, and 9 only. 
To find all the factors of a number, you need to find all numbers that can divide into the original number without a remainder. The divisibility rules from above will be extremely useful!
Suppose you need to find the factors of 30. Since 30 is a number you are familiar with, and small enough, you should know many of the factors without applying any rules. You can start by listing the factors as they come to mind:
 2 • 15
 3 • 10
 5 • 6
Is that it? Not quite. All natural numbers except 1 also have 1 and the number itself as factors:
1 • 30
The factors of 30 are 1, 2, 3, 5, 6, 10, 15, and 30.
When you find one factor of a number, you can easily find another factor—it is the quotient using that first factor as the divisor. For example, once you know 2 is a factor of 30, then 30 ¸ 2 is another factor. A pair of factors whose product is a given number is a factor pair of the original number. So, 2 and 15 are a factor pair for 30.
What do you do if you need to factor a greater number and you can’t easily see its factors? That’s where the divisibility rules will come in quite handy. Here is a general set of steps that you may follow:
 Begin with 1 and check the numbers sequentially, using divisibility rules or division.
 When you find a factor, find the other number in the factor pair.
 Keep checking sequentially, until you reach the second number in the last factor pair you found, or until the result of dividing gives a number less than the divisor.
Note that you can stop checking when the result of dividing is less than the number you’re checking. This means that you have already found all factor pairs, and continuing the process would find pairs that have been previously found.
Example  
Problem  Find factors of 165.  Factors  Explanation  Divisible? 
divisible by 1?  1 • 165 = 165  All numbers are divisible by 1.  Yes  
divisible by 2?  The last digit, 5, is not even, so 165 is not divisible by 2.  No  
divisible by 3?  1 + 6 + 5 = 12, which is divisible by 3, so 165 is divisible by 3.  Yes  
3 • 55 = 165  Use division to find the other factor.  
divisible by 4?  Since 165 is not an even number, it will not be divisible by any even number. The divisibility test for 4 also applies: 65 is not divisible by 4, so 165 is not divisible by 4.  No  
divisible by 5?  Since the last digit is 5, 165 is divisible by 5.  Yes  
5 • 33 = 165  Use division to find the other factor.  Yes  
divisible by 6?  Since 165 is not divisible by 2, it is not divisible by 6.  No  
divisible by 7?  There is no divisibility test for 7, so you have to divide. is not a whole number, so it is not divisible by 7.  No  
divisible by 8?  Since 165 is not divisible by 2, we know that it cannot be divisible by any other even number. Note also that dividing 165 by 8 would not result in a whole number.  No  
divisible by 9?  1 + 6 + 5 = 12, which is not a multiple of 9. 165 is not divisible by 9.  No  
divisible by 10?  The ending digit is a 5 not a 0. 165 is not divisible by 10.  No  
divisible by 11?  with no remainder, so 165 is divisible by 11.  Yes  
11 • 15 = 165  We already performed the division to find the other factor that pairs with 11.  Yes  
divisible by 12?  165 cannot be divisible by an even number and 12 is even. Also, dividing 165 by 12 would not result in a whole number.  No  
divisible by 13?  is not a whole number, so 165 is not divisible by 13.  No  
Done checking numbers  Since the result of 165 ÷ 13 is less than 13, you can stop. Any factor greater than 13 would already have been found as the pair of a factor less than 13.  
Answer  The factors of 165 are 1, 3, 5, 11, 15, 33, 55, 165. 
If a number has exactly two factors, 1 and itself, the number is a prime number. A number that has more factors than itself and 1 is called a composite number. The number 1 is considered neither prime nor composite, as its only factor is 1. To determine whether a number is prime, composite, or neither, check factors. Here are some examples.
Number  Composite, Prime, or Neither?  Explanation 
1  Neither  1 does not have two different factors, so it is not prime. 
2  Prime  2 has only the factors 2 and 1. 
3  Prime  3 has only the factors 3 and 1. 
4  Composite  4 has more than two factors: 1, 2, and 4, so it is composite. 
5, 7,11,13  Prime  Each number has only two factors: 1 and itself. 
6, 8, 9,10, 50, 63  Composite  Each number has more than two factors. 
Find all the factors of 48. Show/Hide Answer1, 2, 3, 4, 6, 8, 12, 16, 24, 48 
A composite number written as a product of only prime numbers is called the prime factorization of the number. One way to find the prime factorization of a number is to begin with the prime numbers 2, 3, 5, 7, 11 and so on, and determine whether the number is divisible by the primes.
For example, if you want to find the prime factorization of 20, start by checking if 20 is divisible by 2. Yes, 2 • 10 = 20.
Then factor 10, which is also divisible by 2 (2 • 5 = 10).
Both of those factors are prime, so you can stop. The prime factorization of 20 is 2 • 2 • 5, which you can write using exponential notation as 22 • 5.
One way to find the prime factorization of a number is to use successive divisions.
Divide 20 by 2 to get 10. 2 is being used because it is a prime number and a factor of 20. You could also have started with 5. 
Then divide 10 by 2 to get 5. 
Multiplying these divisors forms the prime factorization of 20. 
To help you organize the factoring process, you can create a factor tree. This is a diagram that shows a factor pair for a composite number.
Then, each factor that isn’t prime is also shown as a factor pair. You can continue showing factor pairs for composite factors, until you have only prime factors.
When a prime number is found as a factor, circle it so you can find it more easily later.
Written using exponential notation, the prime factorization of 20 is again 22 • 5.
Notice that you don’t have to start checking the number using divisibility of prime numbers. You can factor 20 to 4 • 5, and then factor 4 to 2 • 2, giving the same prime factorization: 2 • 2 • 5.
Now look at a more complicated factorization.
Notice that there are two different trees, but they both produce the same result: five 2s and one 3. Every number will only have one, unique prime factorization. You can use any sets of factor pairs you wish, as long as you keep factoring composite numbers.
When you rewrite the prime factorization of 96 (2 • 2 • 2 • 2 • 2 • 3) in exponential notation, the five 2s can be written as 25. So, 96 = 25 • 3.
When finding the prime factorization of 72, Marie began a tree diagram using the two factors 9 and 8. Which of the following statements are true? 1. Marie started the diagram incorrectly and should have started the tree diagram using the factors 2 and 36. 2. Marie’s next set of factor pairs could be 3, 3 and 2, 4. 3. Marie’s next set of factor pairs could be 3, 3 and 9, 8. 4. Marie didn’t have to use a tree diagram.
Show/Hide Answer A) 1 only Incorrect. Marie could have started her tree diagram with the factors 2 and 36, but she does not have to start with those factors. Starting with 9 and 8 is fine. The correct answer is D. B) 2 only Incorrect. Yes, Marie’s next set of factor pairs could contain 3, 3, and 2, 4, but statement 4 is also correct. The correct answer is D. C) 3 and 4 only Incorrect. Statement 4 is correct: Marie does not have to use a tree diagram, but 3 is not true. When creating a prime factorization, the factors are not combined with the previous composite. The correct answer is D. D) 2 and 4 only Correct. Marie’s next set of factor pairs could read 3, 3, and 2, 4, as 3 • 3 is a factorization of 9 and 2 • 4 is a factorization of 8. Marie could also find the prime factorization by using successive divisions. 
Finding the factors of a natural number means that you find all the possible numbers that will divide into the given number without a remainder.
There are many rules of divisibility to help you to find factors more quickly. A prime number is a number that has exactly two factors. A composite number is a number that has more than two factors.
The prime factorization of a number is the product of the number’s prime factors.
Tests for Even Divisibility
Tests for Even Divisibility
Division by 2
Any number that ends in 0,2,4,6 or 8 is evenly divisible by 2.
Division by 3
Add the number's digits. If the sum is evenly divisible by 3, then so is the number. Example: does 3 divide evenly into 2,169,252? Yes it does, because the sum of the digits is 27, and 27 is divisble by 3. If you prefer, you can keep adding numbers until one digit remains. For example, continuing with 27, (2 + 7 = 9), which is also evenly divisible by 3.
Division by 4
If the number's last 2 digits are 00 or if they form a 2digit number evenly divisible by 4, then number itself is divisible by 4. How about 56,789,000,000? Last 2 digits are 00, so it's divisible by 4. Try 786,565,544. Last 2 digits, 44, are divisible by 4 so, yes, the whole number is divisible by 4.
Division by 5
Any number that ends in a 0 or 5 is evenly divisible by 5.
Division by 6
The number has to be even. If it's not, forget it. Otherwise, add up the digits and see if the sum is evenly divisible by 3. It it is, the number is evenly divisible by 6. Try 108,273,288.
The digits sum to 39 which divides evenly into 13 by 3, so the number is evenly divisible by 6. If you want, you can keep adding numbers until only one digit remains and do the same thing.
So in this case, 3 + 9 = 12 and 1 + 2 = 3, and 3 is evenly divisible by 3.
Division by 7
Multiply the last digit by 2. Subtract this answer from the remaining digits. Is this number evenly divisible by 7? If it is, then your original number is evenly divisible by 7. Try 364. 4, the last digit, multiplied by 2 = 8. 36, the remaining digits, minus 8 = 28. 28 is evenly divisble by 7, and thus, so is 364
Division by 8
If the number's last 3 digits are 000 or if they form a 3digit number evenly divisible by 8, then the number itself is divisible by 8. How about 56,789,000,000? Last 3 digits are 000, so it's divisible by 8. Try 786,565,120. The last 3 digits, 120, divide by 8 into 15, so yes, the whole number is divisible by 8.
Division by 9
Sum the number's digits. If it divides by 9, it is evenly divisble by 9. As with the tests for 3 and 6, you can keep adding numbers until you're left with only one digit.
Division by 10
Any number that ends in 0 is evenly divisible by 10.
Division by 11
Here are four methods:
 If the sum of every other digit, starting with the first, is equal to the sum of every other digit starting with the second, then the number is evenly divisible by 11. Try 13057. (1 + 0 + 7 = 3 + 5), therefore it should divide evenly by 11. And indeed it does: ( 13057 div 11 = 1187).
 If all the digits are the same and there's an even number of digits, then the number is evenly divisible by 11. 333,333 – Yes. 3,333,333 – No.
 If the number is a 3 digit number with different digits, add the two outside digits. If the difference between the sum and the middle digit is 11 then 11 divides evenly into the 3 digit number. If the sum is the same as the middle digit, then 11 will also divide evenly into the number. Try 484. (4 + 4 = 8) which equals the middle digit so 11 divides into 484 evenly. How about 913? (9 + 3 = 12) and (12 – 1 = 11) so 913 is evenly divisible by 11.
 If the digits are different, count from the right and then add the numbers in the odd positions and the even positions. Subtract the smaller number from the larger. If the difference is evenly divisible by 11, so is your original number. Take the number 181,907. The numbers 8, 9, and 7 are in the odd positions. They sum to 24. The numbers 1, 1, and 0 are in the even positions. They sum to 2. Subtract 2 from 24 to get 22. 22 divides by 11 into 2, so 181,907 is evenly divisible by 11.
Division by 12
If the number can be evenly divided by 3 and 4, the same can also be said for 12. Use the methods for Division by 3 and Division by 4 above. If they both work, your number is also evenly divisible by 12.
Division by 15
If the number can be evenly divided by 3 and 5, the same can also be said for 15. Use the methods for Division by 3 and Division by 5 above. If they both work, your number is also evenly divisible by 15.
Division by 24
If the number can be evenly divided by 3 and 8, the same can also be said for 24. Use the methods for Division by 3 and Division by 8 above. If they both work, your number is also evenly divisible by 24.
Division by 33
If the number can be evenly divided by 3 and 11, the same can also be said for 33. Use the methods for Division by 3 and Division by 11 above. If they both work, your number is also evenly divisible by 33.
Division by 36
If the number can be evenly divided by 4 and 9, the same can also be said for 36. Use the methods for Division by 4 and Division by 9 above. If they both work, your number is also evenly divisible by 36.
How to Calculate If a Number Is Evenly Divisible by Another Single Digit Number

1
Divide any number by 1. Every number has 1 as a factor.[1] This is because any number (x{displaystyle x}), is equal to 1×x{displaystyle 1 imes x}.
 For example, 168,293 is divisible by 1, since 1×168,293=168,293{displaystyle 1 imes 168,293=168,293}.

2
Divide even numbers by 2. By definition, an even number is one that is divisible by 2.[2]. So to check if any number, no matter how long, is divisible by 2, look at the last digit. If the last digit is even, the entire number is divisible by 2.[3]
 Remember that 0 is an even number.[4]

3
Check for divisibility by 3. To do this, add up all the digits in the number. If the sum of all digits is divisible by 3, the number is divisible by 3.[5]
 You can repeat the addition of digits if the original sum is too long for you to gauge divisibility by 3.[6] For example, the digits in 3,989,978,579,968,769,877 add up to 141. You can then add again: 1+4+1=6{displaystyle 1+4+1=6}. Since 6 is divisible by 3, you know the entire number is divisible by 3.

4
Check for divisibility by 4. Look at the last two digits in the number. Is the number made by the last two digits divisible by 4? If so, the entire number is divisible by 4.[7] Note that only even numbers are divisible by 4. Multiples of 100 are always divisible by 4.[8]
 Another way to check for divisibility by 4 is to divide the number by 2 twice. If the quotient is still a whole number, the original number is divisible by 4.[9]
 For example, 8762=438{displaystyle {frac {876}{2}}=438}, and then 4382=219{displaystyle {frac {438}{2}}=219}. Since 219 is a whole number, you know that 876 is divisible by 4.
 Another way to check for divisibility by 4 is to divide the number by 2 twice. If the quotient is still a whole number, the original number is divisible by 4.[9]

5
Check for divisibility of numbers for 5. Since any number ending in 0 or 5 is a multiple of 5, any number whose last digit is 0 or 5 is divisible by 5.[10]

6
Check for divisibility by 6. If a number is even, and the sum of its digits are divisible by 3, then the number is divisible by 6. In other words, if a number is divisible by 2 and 3, it is divisible by 6.[11]

7
Check for divisibility by 7. Separate the last digit from the rest of the number. Double the last digit. Then, subtract that product from the number made by the remaining digits. If the difference is divisible by 7, then the whole number is divisible by 7.[12]
 For example, to find out if 567 is divisible by 7, first separate the last digit from the number. This gives you 56 and 7. Double the last digit, 7: 7×2=14{displaystyle 7 imes 2=14}. Then, subtract 14 from 56: 56−14=42{displaystyle 5614=42}. Since 42 is divisible by 7, you know that 567 is divisible by 7.

8
Check for divisibility by 8. Look at the last three digits in the number. If the number they make is divisible by 8, then the entire number is divisible by 8.[13]
 Another way to do this is to halve the last three digits 3 times. If the final quotient is a whole number, then the entire number is divisible by 8.[14]
 For example, 1282=64{displaystyle {frac {128}{2}}=64}, then 642=32{displaystyle {frac {64}{2}}=32}, then 322=16{displaystyle {frac {32}{2}}=16}. Since 16 is a whole number, you know that the number 11,128 is divisible by 8.
 Another way to do this is to halve the last three digits 3 times. If the final quotient is a whole number, then the entire number is divisible by 8.[14]

9
Check for divisibility by 9. A number is divisible by 9 if the sum of its digits is divisible by 9.[15].
 If, after adding up the sum of all the component parts of a number which comes out to another two digit or bigger number the sum is exposed, take that number and add it's component parts. (Take for example 189: 1+8+9=27…if you then take 2+7 you will get 9. Therefore, 189 is evenly divisible by 9.)
 You can repeat the addition of digits if the original sum is too long for you to gauge divisibility by 9.[16] For example, the digits in 3,989,978,579,968,769,877 add up to 141. You can then add again: 1+4+1=6{displaystyle 1+4+1=6}. Since 6 is not divisible by 9, you know the entire number is not divisible by 9.

10
Check for divisibility by 10. Divisibility by ten can happen when the last digit of that number ends in 0 – that's the only way.
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