Unlocking the Secrets of Greatest Common Factor and Least Common Multiple with Factor Trees

Number	Smallest Prime Factor	Quotient
$80$	$2$	$\frac{8 0}{2} = 40$
$40$	$2$	$\frac{4 0}{2} = 20$
$20$	$2$	$\frac{2 0}{2} = 10$
$10$	$2$	$\frac{1 0}{2} = 5$
Prime Factorization
$80 = 2 \cdot 2 \cdot 2 \cdot 2 \cdot 5$

Number

Smallest Prime Factor

Quotient

80

2

\frac{8 0}{2} = 40

40

2

\frac{4 0}{2} = 20

20

2

\frac{2 0}{2} = 10

10

2

\frac{1 0}{2} = 5

Prime Factorization

80 = 2 \cdot 2 \cdot 2 \cdot 2 \cdot 5

Numbers	Multiples of Numbers	Common Multiples	Least Common Multiple
$2$ and $3$	$Multiples of 2 : Multiples of 3 : 2, 4, 6, 8, 10, 12, \dots 3, 6, 9, 12, 15, \dots$	$\underline{6}, 12, 18, 24, \dots$	$LCM (2, 3) = 6$
$8$ and $12$	$Multiples of 8 : Multiples of 12 : 8, 16, 24, 32, 40, 48, \dots 12, 24, 36, 48, \dots$	$\underline{24}, 48, 72, 96, \dots$	$LCM (8, 12) = 24$

Numbers

Multiples of Numbers

Common Multiples

Least Common Multiple

2

and

3

Multiples of 2 : Multiples of 3 : 2, 4, 6, 8, 10, 12, \dots 3, 6, 9, 12, 15, \dots

\underline{6}, 12, 18, 24, \dots

LCM (2, 3) = 6

8

and

12

Multiples of 8 : Multiples of 12 : 8, 16, 24, 32, 40, 48, \dots 12, 24, 36, 48, \dots

\underline{24}, 48, 72, 96, \dots

LCM (8, 12) = 24

Prime Factors	Multiplication	$LCM (54, 60)$
$2,$ $2,$ $3,$ $3,$ $3$ and $5$	$2 \cdot 2 \cdot 3 \cdot 3 \cdot 3 \cdot 5$	$540$

Prime Factors

Multiplication

LCM (54, 60)

2,

2,

3,

3,

3

and

5

2 \cdot 2 \cdot 3 \cdot 3 \cdot 3 \cdot 5

540

Factors	Multiplication	$LCM (10, 16)$
$2,$ $2,$ $2,$ $2,$ and $5$	$2 \cdot 2 \cdot 2 \cdot 2 \cdot 5$	$80$

Factors

Multiplication

LCM (10, 16)

2,

2,

2,

2,

and

5

2 \cdot 2 \cdot 2 \cdot 2 \cdot 5

80

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?

Davontay, Vincenzo, and Tadeo saw each other at the cinema today. If Davontay goes to the cinema every $6^{th}$ day, Vincenzo every $1 0^{th}$ day, and Tadeo every $5^{th}$ day, how many times must Tadeo go before the three friends will meet at the cinema again?

Number	Factors
$100$	$1,$ $2,$ $4,$ $5,$ $10,$ $20,$ $25,$ $50,$ $100$
$88$	$1,$ $2,$ $4,$ $8,$ $11,$ $22,$ $44,$ $88$
$76$	$1,$ $2,$ $4,$ $19,$ $38,$ $76$

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

Number	Multiples
$6$	$6,$ $12,$ $18,$ $24,$ $30,$ $36,$ $42,$ $\dots$
$10$	$10,$ $20,$ $30,$ $40,$ $50,$ $60,$ $70,$ $\dots$
$5$	$5,$ $10,$ $15,$ $20,$ $25,$ $30,$ $35,$ $\dots$

Number

Multiples

6

6,

12,

18,

24,

30,

36,

42,

\dots

10

10,

20,

30,

40,

50,

60,

70,

\dots

5

5,

10,

15,

20,

25,

30,

35,

\dots

The Washington Junior High School choir has $80$ students in sixth grade and $96$ in seventh grade. The music teacher wants to create groups with the same combination of sixth- and seventh-grade students in each group to perform in the upcoming school festival.

He wants to create as many groups as possible.

What is the greatest number of groups the teacher can create?

The teacher creates as many groups as possible. How many seventh-grade students are in each group?

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.

Tiffaniqua plans to give gift bags with blue and red marbles to her friends. She has $26$ blue marbles and $39$ red marbles. Each bag will have the same number of each marbles with none left over.

What is the highest number of friends Tiffaniqua could give bags to?

If Tiffaniqua gave a bag to the highest number of friends possible, how many blue marbles should each bag contain?

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.

Zain's mother is going to make hot dogs for a picnic.

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.

Is Zain's mother right?

Select the option that represents the smallest number of packs of hot dog packs and buns Zain's mother would have to buy in order to not have any leftovers.

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

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!

	16 Theory slides
	16 Exercises - Grade E - A
	Each lesson is meant to take 1-2 classroom sessions

Greatest Common Factor and Least Common Multiple

Catch-Up and Review

Hint

Solution

Hint

Solution

Prime Factorization of 35

Prime Factorization of 84

Identify Common Prime Factors

Multiply

Hint

Solution

Prime Factorization of 10 and 16

Math Prime Factors Vertically

Bring Down the Primes in Each Column and Multiply

Finding the Numbers to Fill in the Orange and Pink Cells

Hint

Solution

Checking Our Answer

Hint

Solution

Greatest Common Factor and Least Common Multiple

Recommended exercises

Prime Factorization of $35$

Prime Factorization of $84$

Prime Factorization of $10$ and $16$