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| Student Learning Objectives: |
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| | 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.
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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?
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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 volume V of a rectangular box is given by the product of its length l, width w, and height h. V=l* w* h A box has an volume of 1025 cubic centimeters.
If the dimensions of the box are whole numbers greater than 1, what are their possible dimensions?
The volume of a rectangular box with length l, width w, and height h is given by the following formula. V=l* w* h We are told that a box has a volume of 1025 cubic centimeters. Let's substitute this information into the formula for the volume of a rectangular box. V&=l* w* h &⇓ 1025&=l* w* h We need to find 3 factors whose product is 1025, so let's find the prime factorization of 1025. This information can help us to know what the factors might be.
We have 5 factors to choose from: 1, 5, 5, 41, and 1025. We need to see what combinations of 3 of these factors can make 1025.
| Combinations | Simplification |
|---|---|
| 1*1*1025= 1025 | 1*1*1025= 1025 |
| 5*5*41= 1025 | 5*5*41= 1025 |
| 1*(5*5)*41= 1025 | 1*25*41= 1025 |
| 1*5*(5*41)= 1025 | 1*5*205= 1025 |
We found all the combinations of these factors that make 1025. However, recall that all the dimensions of the box must all be greater than 1. Let's reconsider our table with this in mind.
| Combinations | Valid Dimensions? |
|---|---|
| 1* 1*1025=1025 | * |
| 5*5*41=1025 | ✓ |
| 1*25*41=1025 | * |
| 1*5*205=1025 | * |
We can then say that the possible dimensions of the box are 41, 5, and 5 centimeters.
Notice that it does not matter which dimension any of the values is assigned to — the box can be 41 centimeters in height, length, or width, so long as the other two dimensions are 5 centimeters!
Dominika is creating identical fruit baskets using 56 bananas, 40 apples, and 65 oranges.
What is the greatest number of fruit arrangements she can make with the smallest amount of leftover fruit?
We are asked to find the greatest number of identical fruit baskets that Dominika can make with the least amount of fruit left over. Consider the amount of fruit she has. &Bananas: &56 &Apples: &40 &Oranges:& 65 The greatest common factor of these numbers is the number of identical baskets that Dominika can make without having any leftover fruit. Let's list the factors of each of these numbers to find the common factors and choose the greatest one.
| Factors | |
|---|---|
| 56 | 1, 2, 4, 7, 8, 14, 28, 56 |
| 40 | 1, 2, 4, 5, 8, 10, 20, 40 |
| 65 | 1, 5, 13, 65 |
The only common factor between these numbers is 1, but it does not really make sense for Dominika to make only one arrangement with all the fruit in it. Since there are no other factors common to all three three numbers, any other number of arrangements will end up with some fruit left over. Let's identify any factors common to at least two of the numbers to minimize the amount of fruit leftover.
| Factors | |
|---|---|
| 56 | 1, 2, 4, 7, 8, 14, 28, 56 |
| 40 | 1, 2, 4, 5, 8, 10, 20, 40 |
| 65 | 1, 5, 13, 65 |
Dominika could make 5 and 8 arrangements. If she makes five arrangements, there will be some bananas left over. If she makes eight baskets, there will be some oranges left over. Let's find out how much fruit Dominika has on hand at the start of the basket production process. Total Fruit: 56 + 40 + 65 = 161 Now let's see which option produces the least amount of leftover fruit.
| Fruit Per Basket | Total Fruit Used | Fruit Left Over | |
|---|---|---|---|
| 5 Baskets | 11 bananas, 8 apples, 13 oranges | 11* 5+8* 5+13* 5= 160 | 161- 160=1 |
| 8 Baskets | 7 bananas, 5 apples, 8 oranges | 7* 8+5* 8+11* 8= 160 | 161- 160=1 |
Notice that both options have 1 piece of fruit left over. However, since 8 is greater than 5, the most amount of baskets that Dominika can make with the least amount of fruit leftover is 8 baskets.
We can use the applet to determine the amount of fruit for each basket option.
Consider the following statement.
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The product of two different prime numbers is their least common multiple. |
Is this statement always, sometimes, or never true?
We are asked if the product of two different prime numbers gives the least common multiple (LCM) of these numbers. Let's begin by finding the LCM of some pairs of prime numbers. This will give us an idea about the truthfulness of the given statement. Let's consider the factor trees of 13 and 17.
Notice that the prime factors of each of these numbers are the numbers themselves and that these factors appear only once. This means that the product of these numbers is their least common multiple. LCM(13,17)&= 13* 17 &⇕ LCM(13,17)&=221 Now let's look at the prime factors of 47 and 97.
The situation for this pair of numbers is the same — the only prime factors are the numbers themselves. Let's multiply them to find their LCM. LCM(47,97)&= 47* 97 &⇕ LCM(47,97)&=4559 Since the only factors of a prime number are 1 and the number itself, the only prime factor of a prime number is that number itself. This means that the product of two different prime numbers gives us their LCM. This means that the statement is always true.
The product of two different prime numbers is their least common multiple. ✓