Sign In
| Student Learning Objectives: |
|---|
|
| | 16 Theory slides |
| | 16 Exercises - Grade E - A |
| | Each lesson is meant to take 1-2 classroom sessions |
The following applet creates arrangements using the factors of 12 and 18 as the width of an arrangement of blocks. It considers vertical and horizontal arrangements as different arrangements.
An essential property of any whole number greater than 1 is that it can be expressed as a product of prime numbers. This process is called prime factorization.
Prime factorization, also called complete factorization, is the decomposition of a whole number into a product of its prime factors. 12 &= 2*2*3 &⇕ 12 &= 2^2* 3 The prime factors are found by dividing the number by the smallest prime number that is a factor of that number. This process is repeated with the quotient until the resulting quotient is a prime number. The following table shows the prime factorization of 80.
| Number | Smallest Prime Factor | Quotient |
|---|---|---|
| 80 | 2 | 80/2 = 40 |
| 40 | 2 | 40/2 = 20 |
| 20 | 2 | 20/2 = 10 |
| 10 | 2 | 10/2 = 5 |
| Prime Factorization | ||
| 80 = 2* 2* 2* 2* 5 | ||
A factor tree is a diagram that shows the prime factors of a number. The tree begins with a root node that contains the number whose prime factorization is needed. Two branches extend from the root node and connect to a factor pair of the number. This process continues by breaking each factor into its factors until only prime factors appear at the end of the branches.
Magdalena loves puzzles. She is currently solving a special edition of a Sudoku, a game that consists of a 9×9 grid with some cells containing numbers and others blank. The purpose of the game is to find the missing values by using the numbers 1-9 only once in each row and column and in each of the nine 3×3 boxes.
In this special version of the game, each cell contains a particular math challenge whose solution helps find the corresponding missing number. Magdalena is now focusing on filling in the red and yellow cells.
With this information in mind, the factor tree of 54 can now be created.
Now, consider that the missing value on the red cell is the most repeated prime factor of the prime factorization of 54. Prime Factorization of54 54=2* 3* 3* 3 Here, the number 3 is repeated three times, while the number 2 appears only once. This means that the missing value on the red cell is 3.
In this case, the prime factorization of 64 contains only the number 2 six times. Prime Factorization of64 64=2*2*2*2*2*2 Therefore, the missing value on the yellow cell of the Sudoku game is 2. Magdalena can now fill the yellow and red cells in the Sudoku board!
Factors are used to divide a set of items into equal amounts. However, it may be difficult when two or more different sets are to be divided into a certain number of groups with equal amounts of each item. The common factors of the sets, and the greatest common factor in particular, can help find a solution to this type of problem.
Factors that are shared by two or more numbers are called common factors. The greatest of these common factors is called the greatest common factor (GCF). Consider, for example, the factors of 4 and 8. Factors of4: & 1, 2, 4 Factors of8: & 1, 2, 4,8 The common factors of 4 and 8 are 1, 2, and 4. The GCF(4,8)= 4. The GCF can also be determined by multiplying the prime factors shared in the prime factorization of the numbers in question.
greatest common divisorbecause a factor of a number divides that number evenly.
The greatest common factor (GCF) of two numbers can be determined by finding the prime factorization of the numbers. Next, the common prime factors of the prime factorizations are identified. The GCF is then given by the product of the common prime factors. Consider the following pair of numbers. 42,24 The GCF(42,24) will be found by following these four steps.
Magdalena continues with the process of solving this special version of the Sudoku game. This time, she wants to know which number goes in the purple cells.
The clue is that this purple cells must be filled with the greatest common factor (GCF) of 35 and 84. What number goes into the purple cells?
Each of these steps will now be applied.
The prime factorization of 35 can be found by using a factor tree. The root node of the tree is 35 and then factor pairs of 35 will extend from the root node.
The prime factorization of 35 is 5* 7.
A similar process can be followed to find the prime factorization of 84.
The prime factorization of 84 is 2* 2* 3* 7.
Now that the two prime factorizations are stated, find the common prime factor shared by these factorizations. &35=5* 7 &84=2*2*3* 7 The only common prime factors shared by these factorization is the number 7.
In this case, only one factor is shared by the prime factorizations of 35 and 84, so the greatest common factor of 35 and 84 is 7. This number goes into the purple cells.
What a great achievement! Magdalena has made great progress on her puzzle.
Consider the following applet that creates different arrangements of blocks using multiples of 12 and 18.
Similar to factors, it may be of interest to find the smallest multiple of two or more different numbers. This number is called the least common multiple.
The least common multiple (LCM) of two whole numbers a and b is the smallest whole number that is a multiple of both a and b. It is denoted as LCM(a,b). The least common multiple of a and b is the smallest whole number that is divisible by both a and b. Some examples can be seen in the table below.
| Numbers | Multiples of Numbers | Common Multiples | Least Common Multiple |
|---|---|---|---|
| 2 and 3 | Multiples of2:& 2, 4, 6, 8, 10, 12, ... Multiples of3:& 3, 6, 9, 12, 15, ... | 6, 12, 18, 24, ... | LCM(2,3)= 6 |
| 8 and 12 | Multiples of 8:& 8, 16, 24, 32, 40, 48, ... Multiples of12:& 12, 24, 36, 48, ... | 24, 48, 72, 96, ... | LCM(8,12)= 24 |
To determine the least common multiple (LCM) of two or more numbers, begin by finding the prime factorization of each number. Then, highlight all the instances of each prime factor in the prime factorization that is repeated most. Finally, the LCM is given by the product of the highlighted factors. This process will be illustrated with this pair of numbers. 54 and 60 The LCM(54,60) will be found by following these four steps.
The product of these prime factors is the least common multiple of the numbers.
| Prime Factors | Multiplication | LCM(54,60) |
|---|---|---|
| 2, 2, 3, 3, 3 and 5 | 2*2*3*3*3*5 | 540 |
The least common multiple of 54 and 60, which can also be written as LCM(54,60), is 540.
Magdalena is having a great time solving her special Sudoku puzzle but the next clue looks scary.
The clue claims that to find the numbers to fill in the orange and pink cells, first find the least common multiple (LCM) of 10 and 16. Next, these conditions must be met.
What are the numbers that Magdalena will write in the orange and pink cells? Write the number for the orange cells first.
Next, each of these steps will be performed.
Both prime factorizations will be found using factor trees. One tree will have 10 as its root, while the root of the second tree will be 16. Two branches extend from each root node to connect a factor pair. The process is repeated with the factors until only prime factors are on each branch.
Now that the prime factorizations of 10 and 16 are found write them in a table. In the first row, place the prime factorization of 10 and the factorization of 16 in the second row. The table's columns will contain each factor while matching them vertically when possible.
Lastly, bring down the factors of 10 and 16 in each column of the table created previously.
The product of these prime factors is the least common multiple of 10 and 16.
| Factors | Multiplication | LCM(10,16) |
|---|---|---|
| 2, 2, 2, 2, and 5 | 2*2*2*2*5 | 80 |
The least common multiple of 10 and 16 is 80.
Magdalena found that the LCM(10,16) is 80 but she is not ready to fill in the cells in the puzzle yet. For the orange cells, she needs to find the number that gives 80 when multiplied by 10. orange* 10=80 ⇒ orange= 8 The number that goes in the pink cells can be found in a similar way. pink*16=80 ⇒ pink= 5 This means that the number that goes in the orange cells is 8 and the number that goes in the pink cells is 5. What a big step Magdalena has made now in her solution process!
Find the greatest common factor (GCF) or the least common multiple (LCM) of the given numbers, as requested.
An important use of the greatest common factor is that it can help simplify numeric expressions. Consider, for example, the following sum. 48+30 The greatest common factor of these numbers is 6. This means that each number can be rewritten using this common factor. 48+30 ⇔ 6* 8+ 6* 5 The greatest common factor can be pulled out of each addend. This factor will be multiplied by the sum of the numbers left after pulling it out. 6* 8+ 6* 5 ⇔ 6( 8+ 5) The property applied to pull the greatest common factor out a sum is called the Distributive Property. This property states that multiplying a number by the sum of two or more addends produces the same result as multiplying the number by each addend individually and then adding all the products together.
Exhausted from solving math problems to solve the puzzle, Magdalena goes to the kitchen to have something to eat. She chooses a delicious slice of Italian pizza.
She is just about to return to her puzzle when her mother asks her for some help. Magdalena's mother works as a volunteer in a retirement home. She plans to gift some bouquets to the ladies in the retirement home next weekend.
She has 49 roses and 84 tulips. Each bouquet will have the same number of flowers and contain only roses or only tulips.
GCF(49,89)=? First, find the prime factorizations of these numbers using a factor tree.
Now that the prime factorizations have been found, look for the common prime factors between them. &49=7* 7 &84 = 2*2*3* 7 The only common factor between the factorizations is 7, so the GCF(49,84) is 7. This means that each bouquet will contain exactly 7 flowers!
49 roses + 84 tulips= 133 flowers Let c be the greatest possible number of flowers per bouquet. Next, let a be the number of rose bouquets and b the number of tulip bouquets. Note that the sum of a and b represents the total number of bouquets. Total Number of Bouquets: a+ b Additionally, multiplying the total number of bouquets by the number of flowers per bouquet c will equal the number of flowers in total. It was previously determined that there are 7 flowers per bouquet. Total Number of Flowers: c( a+ b) ⇓ 7( a+ b) The Distributive Property can be applied to this expression. 7( a+ b) ⇕ 7 a+ 7 b Now, consider that the number of rose bouquets a times seven will equal 49, the total number of roses. The missing value of a can be found using this information. Number of Rose Bouquets: 7* a=49 ⇓ a= 7 The number of tulip bouquets b can be determined by following a similar procedure. Number of Tulip Bouquets: 7* b=84 ⇓ b= 12 This means that Magdalena and her mother can create 7 rose bouquets and 12 tulip bouquets. Total Number of Bouquets: 7+ 12=19 bouquets
Notice that this is equivalent to multiplying the total number of bouquets, 7+12=19, by the total number of flowers per bouquet, 7. 19* 7= 133
After helping her mom and having a snack, Magdalena returns to her Sudoku puzzle. The board is almost done, but she thinks the next riddles look extremely difficult.
Solve the following situations to help Magdalena complete that challenging puzzle.
|
A citywide high school soccer club has 100 juniors, 88 sophomores, and 76 seniors. The head coach wants to divide the students into groups of the same size. Each team must have the same numbers of juniors, sophomores, and seniors. What is the greatest possible number of groups the coach can make? |
What is the number to fill in the blue cells?
|
Davontay, Vincenzo, and Tadeo saw each other at the cinema today. If Davontay goes to the cinema every 6^(th) day, Vincenzo every 10^(th) day, and Tadeo every 5^(th) day, how many times must Tadeo go before the three friends will meet at the cinema again? |
What is the number to fill in the green cells?
There are 100 juniors, 88 sophomores, and 76 seniors in the club. The coach wants to divide each grade level evenly to create the greatest number of groups possible. In other words, he wants to find the greatest common factor of these numbers. First, list the factors of each number.
| Number | Factors |
|---|---|
| 100 | 1, 2, 4, 5, 10, 20, 25, 50, 100 |
| 88 | 1, 2, 4, 8, 11, 22, 44, 88 |
| 76 | 1, 2, 4, 19, 38, 76 |
Notice that these numbers have the common factors 1, 2, and 4. Their greatest common factor is then 4. This means that the greatest number of groups that the coach can make is 4. This is also the number for the blue cells of Magdalena's puzzle — 4.
Davontay's Visits Multiples of6: 6, 12, 18, 24, 30, ... The same thought can be applied to Vincenzo's and Tadeo's situations. This means that the least common multiple of 6, 10, and 5 will give the number of days until the boys visit the cinema on the same day again. List the multiples of each to find the LCM(6,10,5).
| Number | Multiples |
|---|---|
| 6 | 6, 12, 18, 24, 30, 36, 42, ... |
| 10 | 10, 20, 30, 40, 50, 60, 70, ... |
| 5 | 5, 10, 15, 20, 25, 30, 35, ... |
The least common multiple of 6, 10, and 5 is 30. This means it will take 30 days for the boys to visit the cinema on the same day. Now, recall that Tadeo goes to the cinema every five days. This means that 5 times a certain number must equal 30. 5* ?= 30 ? = 6 This number is 6. This means that Tadeo will go to the cinema 6 times before meeting his friends. This is the solution to the riddle, so Magdalena can fill in the green cells with the number 6.
Great news — the remaining numbers are easy to decipher and Magdalena has now filled in the rest of the board. What an outstanding achievement!
The choir consists of 80 sixth-grade students and 96 seventh-grade students. 80 students in sixth grade 96 students in seventh grade We want to form groups with the same number of students in each grade. This means the number of students in each grade must be divided evenly by the total number of groups. This means that we can find the possible numbers of groups by finding the factors of 80 and 96. Let's list these factors! Factors of80: & 1, 2, 4, 5, 8, 10, 16, &20, 40, 80 Factors of96: & 1, 2, 3, 4, 6, 8, 12, & 16, 24, 32, 48, 96 The common factors of these numbers are the numbers of possible groups the music teacher could form. Number of Possible Groups: 1, 2, 4, 8, 16 The greatest of these numbers, 16, is the greatest common factor of 80 and 96. This also represents the greater number of groups the teacher could create.
We found previously that the greatest number possible is 16. Let's a be the number of sixth-grade students and b the number of students of seventh-grade that form each group. Note that the sum of a and b represents the total number of students each group will have.
Total Students per Group:
a+ b
Multiplying this sum by the number of groups will result in the total number of students in the choir.
Total Number of Students in the Choir:
16( a+ b)
We can now apply the Distributive Property to this expression.
16( a+ b)
⇕
16* a+ 16* b
The product of 16 and a will equal 80 because there are 80 sixth-grade students.
Number of Sixth-Grade Students Per Group:
16* a=80
⇓
a= 5
We can follow a similar procedure to calculate the number of seventh-grade students per group.
Number of Seventh-Grade Students Per Group:
16* b=96
⇓
b= 6
This means that there will be six seventh-grade students in each group.
We are told that Tiffaniqua has 26 blue marbles and 39 red marbles. Blue Marbles: 26 Red Marbles: 39 Tiffaniqua wants to split these marbles into equal groups. Each group will contain the same number of red marbles and the same number of blue marbles without any marbles left over. The possible groups are given by the common factors of 26 and 39. Let's list the factors of each! Factors of26: & 1, 2, 13, 26 Factors of39: & 1, 3, 13, 39 The common factors are 1 and 13. The greater of these common factors is 13. This means that the greatest common factor of 26 and 39 is 13. The highest number of friends Tiffaniqua could give a bag to without having any leftover marbles is 13.
We found that the highest number of friends Tiffaniqua could give bags to is 13. She has 26 blue marbles. We want to find the number of blue marbles that she should put in each bag. We can do this by finding what number times 13 is 26. Let's let b be this number. 13* b &= 26 &⇓ b&= 2 This means that each of the 13 bags Tiffaniqua will give her friends should contain two blue marbles.
Davontay is helping his sister make ribbon ornaments for their Christmas tree. They have three ribbons to make these ornaments. The ribbons are green, blue, and orange.
The green ribbon is 96 inches long, the blue one 72 inches, and the orange 60 inches. Davontay's sister asks him to cut the ribbons into strips of equal length and to make sure they are as long as possible. How long should Davontay cut the ribbons to be?
We are told that the siblings have three ribbons of different lengths. The green ribbon is 96 inches long, 72 inches for the blue one, and 60 inches for the orange ribbon. These ribbons must be divided into equal lengths while keeping them as long as possible.
This means we should find the greatest number that divides these three measures evenly, without any remainder. We can do this by finding the greatest common factor of 96, 72, and 60. Let's list the factors of these three numbers.
| Factors | |
|---|---|
| 96 | 1, 2, 3, 4, 6, 8, 12, 16, 24, 32, 48, 96 |
| 72 | 1, 2, 3, 4, 6, 8, 9, 12, 18, 24, 36, 72 |
| 60 | 1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30, 60 |
We can see from the table that the greatest common factors is 12. This means that Davontay will cut each ribbon into 12-inch pieces. We can also find how many pieces he can cut from each ribbon if we divide their lengths by 12. Let's do it!
| Size (Inches) | Number of Strips Produced | |
|---|---|---|
| Green | 96 | 96/12=8 |
| Blue | 72 | 72/12=6 |
| Orange | 60 | 60/12=5 |
This was a big step in the decoration process. Davontay and her sister created the most beautiful Christmas tree they had ever made.
At the supermarket, they notice that hot dogs come in packs of 10, while buns come in packs of eight. Zain's mother says that she has to make 80 hot dogs to have no hot dogs or buns left over.
We want find the smallest number divisible by the numbers of hot dogs and buns in each pack. This means we should find the least common multiple (LCM) of 10 and 8. This will help us determine whether Zain's mother is right. Let's list the first multiples of 10 and 8. Multiples of10:& 10, 20, 30, 40, 50, 60, & 70, 80, 90, 100, ... Multiples of8:& 8, 16, 24, 32, 40, 48, & 56, 64, 72, 80, ... We can see from these lists that two common multiples of 10 and 8 are 40 and 80. The LCM( 10, 8) is 40. This means that Zain's mother is not right. The least amount of hot dogs she can make and not have any hot dogs or buns left over is 40.
Let's now find the smallest number of packs of buns and hot dogs that Zain's mother must buy and not have any leftovers. We know that hot dogs come in packs of 10 and buns in packs of 8. We also know that the LCM( 10, 8)= 40. Let's divide 40 by 10 and 8 to find how many packs of each are needed. Packs of Hot Dogs: 40/10=4 packs [0.8em] Packs of Buns: 40/8=5 packs This means that Zain's mother should buy 4 packs of hot dogs and five packs of buns in order to make the smallest number of hot dogs without any leftovers.
Ali is helping his grandma plant sunflowers in her garden.
They have enough sunflowers to plant rows of either 5 or 6 flowers if every row has the same number of flowers. What is the least number of sunflowers they could have?
We are told that Ali and his grandma have enough sunflowers to plant 5 or 6 in each row of a garden.
This means that the total number of sunflowers they have is a multiple of both 5 and 6. Let's list the first multiples of 5 and 6 to give us an idea of the possible number of sunflowers they could have. Multiples of5:& 5, 10, 15, 20, 25, 30, 35, 40, & 45, 50, 55, 60, ... Multiples of6:& 6, 12, 18, 24, 30, 36, 42, 48, & 54, 60, 66, 72, ... Two possible options are 30 and 60. The least common multiple LCM of 5 and 6 is 30, so the least number of sunflowers Ali and his grandma could have is 30.
We can also use the prime factorization to find this number. Let's look at the prime factorizations of 5 and 6.
| Number | Prime Factorization |
|---|---|
| 5 | 5 |
| 6 | 2*3 |
We can see that we have three different factors in the prime factorizations. Since these prime factors only appear once, their product will give us the LCM( 5, 6). LCM( 5, 6)=2*3*5 ⇕ LCM( 5, 6)= 30 This matches our previous result. Ali and his grandma decide to plant six sunflowers per row.
Diego and his mother are waiting for his father to arrive at a subway station. Diego notices that three subway lines arrived at the same time. The following table shows the arrival schedule of these three lines.
| Subway Line | Arrival Time |
|---|---|
| A | Every 12 minutes |
| B | Every 15 minutes |
| C | Every 10 minutes |
How long will Diego have to wait until all three lines arrive at the station together again?
We are told that the three subways line arrived at the station at the same time. We want to know how long Diego will have to wait until the three lines meet at the station again. Let's look at the arrival times.
| Subway Line | Arrival Time |
|---|---|
| A | Every 12 minutes |
| B | Every 15 minutes |
| C | Every 10 minutes |
The next arrival of line A will be in 12 minutes. The train after that will arrive 12 minutes after that, and so on. The arrivals of line A are given by multiples of 12. Subway A: 12, 24, 36, 48, 60, 72, 84, ... We can apply this thought process to the other lines. Let's list the first multiples of each arrival time.
| Arrival Time | Multiples | |
|---|---|---|
| Line A | Every 12 minutes | 12, 24, 36, 48, 60, 72, 84, ... |
| Line B | Every 15 minutes | 15, 30, 45, 60, 75, 90, 105, ... |
| Line C | Every 10 minutes | 10, 20, 30, 40, 50, 60, 70, ... |
We can see from the table that the least common multiple of the three arrival times is 60. This means that Diego would have to wait for 60 minutes for the three lines arrive at the station together again. Wow! But before that, his father arrives and they can go home to have a great time together!