# 4.2 Factors and Divisibility

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4.2 Factors and Divisibility
Numbers that are multiplied together are called factors. Factors of a number, a, are numbers that when multiplied together produce a product of a. The number 12 has 6 possible factors: 1 x 12 = 12 2 x 6 = 12 3 x 4 = 12 So the factors are 1, 2, 3, 4, 6, and 12. A factor pair table can be written for 12 to find all the factors. Just start from 1 and keep incrementing and checking if 12 is divisible by it. When a factor you choose has already appeared in the table. You know you are finished. Note that a number is always a factor of itself because a x 1 = a When two whole numbers, such as 3 and 2, multiply to get a product, 6 (3 x 2 = 6), then we can say these facts: 6 is a multiple of 3 and 2 6 is divisble by 3 and 2 3 and 2 divide evenly into 6. 3 and 2 are factors of 6. Example: 12 is a multiple of 4 and a multiple of 3 12 is divisible by 3 and 4 3 and 4 divide evenly into 12 3 and 4 are factors of 12. 12

600÷10=60 8,000÷200 = 80÷2 = 40 Division Review Division with 0
Division Properties: Division is the “inverse” of multiplication. That is, does the opposite of what multiplying does. Quotient Dividend Divisor Quotient Divisor Dividend If 4 times 5 is 20, then 20÷5 answers the question, “5 times what equals 20?” and 20÷4 answers the question, “4 times what equals 20?” 4(5) = 20 4 = 20÷5 5 = 20÷4 Division with 0 0 divided by any nonzero number is 0. 0÷5 = 0 Any nonzero number divided by 0 is undefined. 5÷0 is undefined. 0/0 is undetermined. More Division Properties Any number divided by 1 is itself. 5÷1 = 5 Any number divided by itself is 1 (provided that number ≠0) 5÷5 = 1 When dividing by multiples of 10, you may cancel an equal number of “right-hand” zeroes. 600÷10=60 8,000÷200 = 80÷2 = 40

-9 -12 -14 6 27 40 -6 -24 -35 3 59 -56 3 There are no more digits left to carry down, so 3 is the remainder and 147 is not exactly divisible by 4. 147÷4 = 36 R3 “36 with 3 left over” No remainder. 96 is exactly divisible by 3. 96÷3 = 32 There are no more digits left to carry down, so 3 is the remainder and 1809 is not exactly divisible by 7. 147÷7 = 36 R3 “36 with 3 left over” Check: 32 x 3 = 96 Check: 36 x = 147 Don’t forget to add the remainder if there is one. Check: 258 x = 1809 Don’t forget to add the remainder if there is one. Checking Division: Quotient X Divisor + Remainder = Dividend Translation: The answer to the division problem times what you divided by, plus the remainder, should equal the number you divided into.

Div. By Divisibility Tests Example
2 A number is divisible by 2  if the last digit is 0, 2, 4, 6 or 8. 168 is divisible by 2 since the last digit is 8. 3 A number is divisible by 3  if the sum of the digits is divisible by 3. 168 is divisible by 3 since the sum of the digits is 15 (1+6+8=15), and 15 is divisible by 3. 4 A number is divisible by 4  if the number formed by the last two digits is divisible by 4. (also, a number if divisible by 4 if after being halved, it is still even). 316 is divisible by 4 since 16 is divisible by 4. 72 is divisible by 4 because half of 72 is 36 and 36 is even. 5 A number is divisible by 5  if the last digit is either 0 or 5. 195 is divisible by 5 since the last digit is 5. 6 A number is divisible by 6  if it is divisible by 2 AND it is divisible by 3. 168 is divisible by 6 since it is divisible by 2 AND it is divisible by 3. 7 Take the last digit, double it, and subtract it from the rest of the number; if the answer is divisible by 7 (including 0), then the number is also. If you had 203, you would double the last digit,3, to get 6, and subtract that from 20 to get 14, which is divisible by 7. 8 A number is divisible by 8  if the number formed by the last three digits is divisible by 8. 7,120 is divisible by 8 since 120 is divisible by 8. 9 A number is divisible by 9  if the sum of the digits is divisible by 9. 549 is divisible by 9 since the sum of the digits is 18 (5+4+9=18), and 18 is divisible by 9. 10 A number is divisible by 10  if the last digit is 0. 1,470 is divisible by 10 since the last digit is 0.

1)All the teams must have the same number of players.
Example A total of 216 girls tried out for a city volleyball program. How many girls should be put on the team roster if the following requirements must be met? 1)All the teams must have the same number of players. (find a number that goes exactly into 216, so there is no remainder) 2) A reasonable number of players on a team is 7 to 10 (divide 216 players by 7 players per team, then 8, then 9, then 10). But don’t bother with 10 because we know 10 doesn’t go exactly into 216. 3) There must be an even number of teams. (The quotient must be EVEN). 3 27 24 8 goes exactly into 216, but the quotient is 27, which is ODD. (Does not meet 3rd req.) This meets all 3 requirements: 9 players per team leaves no remainder, 9 is an acceptable number for a team (which is a number between 7 and 10), and the number of teams is 24, which is an even number. 6<7, and there are no more digits left to carry down, so 6 is the remainder and 216 is not exactly divisible by 7. 21 16 56 18 36 6

Prime Numbers A prime number is a whole number, greater than 1, that has only 1 an itself as factors. Composite Numbers A composite number is a whole number, greater than 1, that are not prime.

Any composite number has exactly one set of prime factors.
Prime Factorization To find the prime factorization of a whole number means to write it as the product of only prime numbers. This is can be useful when finding things like the Greatest Common Factor or Least Common Multiple between two numbers (we’ll get into that later). Example: Factor 90 into its prime factors. 90 Choose any two factors of 90 (besides 1 and 90) Then do the same with each of those factors. Keep going until you have only prime factors as the bottom “roots” of the “factor tree.” 9 10 3 3 2 5 90 = 3●3●2●5 Putting these factors in numerical order and then re-writing repeated factors as powers gives: 90 = 2●32●5 Theorem: Any composite number has exactly one set of prime factors. Example 5 Find the prime factorization of 210 First, pick any two factors of 210. For instance 21 and 10. We could have also picked 7 and 30 as the factors. 210 210 21 10 7 d 30 d d 3 7 2 5 6 5 d Notice that either method gives us 210 = 2●3●5●7 3 2 d

Alternate Factoring Method – I call it the “Pyramid Division Method”
For more info: With this method, you start and the bottom of the pyramid and move up, so you have to leave lots of room at the top of your problem. Example: Find the prime factorization of 90. Start with the first prime factor you can think of that goes into 90. We can’t choose 9 or 10 because they aren’t prime. Since 90 is even, start with the factor 2. is 3 prime? Yes! You are done. is 9 prime? No. Keep dividing. is 45 prime? No. Keep dividing. Prime Factorization of 90 is 2 x 3 x 3 x 5 Write the factors in numerical order and use exponents where factors are repeated =

Upside-Down Division Method
Example: Find the prime factorization of 60 Is 30 prime? No. Is 10 prime? No. 5 Is 5 prime? Yes. Stop Prime Factorization of 60 is

The Greatest Common Factor is the largest number that goes into two numbers.
What is the GCF of 20 and 24? The largest number that goes into both 20 and 24 (the largest divisor of both 20 and 24) Both numbers are divisible by 2. Is 2 the GCF? 20÷2 = 10, and 24÷2 = 12. But 10 and 12 have common factors, so 2 is not the GCF of 20 and 24. A bigger number can go into both 20 and 24. 4 is the GCF of 20 and ÷4 = 5, and 24÷4 = 6. 5 and 6 have no common factors, so 4 is the GCF.

Finding the GCF with Prime Factorization
Find the GCF of 24 and 32 24 32 4 6 4 8 2 2 2 3 2 2 2 4 2 2 24= 2● 2 ● 2 ● 3 32 = 2● 2 ● 2● 2 ● 2 Cancel out factors that are in COMMON in both factorizations. The factors that you crossed out make up the GCF. GCF =