5.2.1: Written Retrieval and Problem Solving Practice
- Page ID
- 185695
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)Student Learning Objectives
SLO 5.3 Students will compute the greatest common factor (GCF) and least common multiple (LCM) of pairs of numbers using various methods, including prime factorization, set intersection, and algorithms.
SLO 5.4 Students will evaluate the relationships between the GCF, LCM, and the product of two numbers, demonstrating a deeper understanding of number relationships and their practical applications.
The mild assignment ๐ถ️ is a shorter and less challenging assignment. The medium assignment๐ถ️๐ถ️ is right in the middle in terms of length and level of difficulty. Finally, the spicy assignment ๐ถ️๐ถ️๐ถ️ is a longer assignment that is the most challenging.
(Spicy ๐ถ️๐ถ️๐ถ️ Problems are OPTIONAL, but go for it when you have the time to challenge yourself)
๐ถ️๐ถ
Use the set intersection method to find the GCFs.
a) GCF(42, 28) b) GCF(60, 84)
- Answer
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a)
Factors of 42:
1,2,3,6,7,14,21,42Factors of 28:
1,2,4,7,14,28Common factors:
1,2,7,14✅ GCF = 14
b)
Factors of 60:
1,2,3,4,5,6,10,12,15,20,30,60Factors of 84:
1,2,3,4,6,7,12,14,21,28,42,84Common factors:
1,2,3,4,6,121, 2, 3, 4, 6, 121,2,3,4,6,12✅ GCF = 12
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Use the prime factorization method to find the GCFs.
a) GCF(36, 42) b) GCF(24, 66)
- Answer
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a)
Prime Factorization:
36=22 x 32
42=2 x 3 x 7
✅ GCF = 2 x 3 = 6
b)
Prime Factorization:
24=23 x 3
66=2 x 3 x 11
✅ GCF = 2 x 3 = 6
๐ถ️๐ถ️
Using the Euclidean algorithm, find the following GCFs.
a) GCF(39, 91) b) GCF(72, 160)
- Answer
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a) We’ll start with the larger number first.
91÷39=2 remainder 13
→ 91=2⋅39+1339÷13=3 remainder 0
→ 39=3⋅13+0✅ GCF = 13
b)
160÷72=2 remainder 16
→ 160=2⋅72+1672÷16=4 remainder 8
→ 72=4⋅16+816÷8=2 remainder 0
→ 16=2⋅8+0
๐ถ️๐ถ️
Use the set intersection method to find the following LCMs.
a) LCM(24, 30) b) LCM(42, 28)
- Answer
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a)
Multiples of 24:
24,48,72,96,120,144,...24, 48, 72, 96, 120, 144, ...24,48,72,96,120,144,...Multiples of 30:
30,60,90,120,150,...30, 60, 90, 120, 150, ...30,60,90,120,150,...Common multiples:
→ 120,240,…120, 240✅ LCM = 120
b)
Multiples of 42:
42,84,126,168,210,252,294,...Multiples of 28:
28,56,84,112,140,168,...Common multiples:
→ 84,168,…84, 168,✅ LCM = 84
๐ถ️๐ถ️
Find the following LCMs using the prime factorization method.
a) LCM(60, 72) b) LCM(35, 110)
- Answer
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a)
Prime Factorizations:
60=22⋅3⋅5
72=23⋅32
LCM:
LCM=23⋅32⋅5=8⋅9⋅5=72⋅5=360b)
Prime Factorizations:
35=5⋅7
110=2⋅5⋅11
LCM:
LCM=2⋅5⋅7⋅11=770
๐ถ️๐ถ️
Try some alternative method or one you want to practice to find the LCM and GCF.
GCF(125, 210) and LCM(125, 210)
- Answer
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GCF(125, 210) = 5
LCM(125, 210) = 5250
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Two counting numbers are relatively prime if the greatest common factor of the two numbers is 1. Which of the following pairs of numbers are relatively prime?
a) 4 and 9 b) 24 and 123
- Answer
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a) 4 and 9 → Relatively prime
b) 24 and 123 → Not relatively prime
๐ถ️๐ถ️๐ถ️
For each of the pairs of numbers in parts a − c below
i. sketch a Venn diagram with the prime factors of a and b in the appropriate locations.
ii. find GCF(a, b) and LCM(a, b).
a) a = 63, b = 90
b) a = 16, b = 49
- Answer
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a) GCF(63, 90) = 9, LCM(63, 90) = 630
b) GCF(16, 49) = 1, LCM(16, 49) = 784
๐ถ️๐ถ️
How many factors do each of the following numbers have?
a) 360
b) 102
c) 33 x 52 x 7
- Answer
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a) 24 Factors b) 8 Factors c) 24 factors
๐ถ️๐ถ️
Find the LCM(33, 110, 484).
- Answer
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- 33 = 3 · 11
110 = 2 · 55 = 2 · 5 · 11
484 = 2 · 242 = 2 · 2 · 121 = 2 · 2 · 11 · 11 = \(2^2\) · \(11^2\) - The bases that appear in the prime factorizations are 2, 3, 5, and 11.
- The largest exponents that appear on 2, 3, 5, and 11 are, respectively, 2, 1, 1, and 2.
\(2^2\) from 484
\(3^1\) from 33
\(5^1\) from 110
\(11^2\) from 484 - The LCM is the product of these numbers.
LCM = \(2^2\) · 3 · 5 · \(11^2\)
= 4 · 3 · 5 · 121
= 7260 ; Thus, 7260 is the smallest number that 33, 110, and 484 divide into without remainders.
- 33 = 3 · 11
Problem Solving Problems
1) A recipe for peanut butter cookies will make 15 cookies. A recipe for chocolate cookies will make two dozen cookies. If you want to have the same number of each type of cookie, what is the least number of each that you will need to make using complete recipes?
2) What is the least number of cards that could satisfy the following three conditions?
- If all the coins are put in two equal piles, there is one coin left over.
- If all the coins are put in three equal piles, there is one coin left over.
- If all the coins are put in five equal piles, there is one coin left over.
3) Two numbers are said to be betrothed if the sum of all proper factors greater than 1 of one number equals the other, and vice versa. For example, 48 and 75 are betrothed, since 48 = 3 + 5 + 15 + 25 proper factors of 75 except for 1, and 75 = 2 + 3 + 4 + 6 + 8 + 12 + 16 + 24, proper factors of 48 except for 1.
Determine which of the following pairs are betrothed.
a) (140, 195) b) (1575, 1648)
- Proper Factor (or proper divisor)
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A positive factor of the number that is less than the number itself. Proper factors of 12: 1,2,3,4,6 (We do not include 12 itself.)
4) Hilda’s car gets 34 miles per gallon and Naomi’s gets 8 miles per gallon. When traveling from Washington, D.C., to Philadelphia, they both used a whole number of gallons of gasoline. How far is it from Philadelphia to Washington, D.C.?
5) Problem:
Ms. Rivera is organizing school supply kits for students.
- She has 84 pencils and 60 erasers.
- She wants to divide them into kits so that each kit has the same number of pencils and the same number of erasers, with no supplies left over.
What is the greatest number of kits she can make? How many pencils and how many erasers will each kit contain?
6) Problem:
Two friends, Mia and Jaden, go jogging around a park that forms a loop.
- Mia finishes one loop every 12 minutes.
- Jaden finishes one loop every 18 minutes.
If they both start jogging at the same time and keep running at their own paces, after how many minutes will they be at the starting point together again? If the park opens at 6:00 a.m., what is the next time they will meet at the starting point?