#set document(title: "2.5 Prime Factorization and the Least Common Multiple", author: "OpenStax / XYZ Homework") #set page(width: 8.5in, height: auto, margin: 1in) #import "@preview/cetz:0.5.2" #set text(font: ("STIX Two Text", "Libertinus Serif", "New Computer Modern"), size: 10.5pt, lang: "en") #show math.equation: set text(font: ("STIX Two Math", "New Computer Modern Math")) #set par(justify: true, leading: 0.62em, spacing: 0.9em) #set enum(spacing: 1.1em) // room between list items so tall inline fractions don't collide #set list(spacing: 1.1em) #set table(stroke: 0.5pt + rgb("#c7ccd3")) #let BLUE = rgb("#183B6F") // brand navy — section bars + example/solution labels (white on navy 11.09:1) #let ORANGE = rgb("#A94509") // brand primary-700 — AA-safe deep orange for TEXT (5.93:1 on white; raw brand #F37021 is 2.94:1 and must never carry text) #let RED = rgb("#DC2626") // brand error-600 #let GREEN = rgb("#059669") // brand success-600 (decoration only; small green text uses green-text #007942) #show heading.where(level: 1): it => block(width: 100%, above: 0pt, below: 16pt, fill: gradient.linear(BLUE, rgb("#2C5AA0")), inset: (x: 14pt, y: 12pt), radius: 3pt, text(fill: white, weight: "bold", size: 19pt, it.body)) #show heading.where(level: 2): it => block(width: 100%, above: 18pt, below: 10pt, fill: BLUE, inset: (x: 10pt, y: 6pt), radius: 2pt, text(fill: white, weight: "bold", size: 12pt, it.body)) #show heading.where(level: 3): it => text(fill: ORANGE, weight: "bold", size: 12.5pt, it.body) #show heading.where(level: 4): it => text(fill: BLUE, weight: "bold", size: 10.5pt, it.body) #let examplebox(label, title, body) = block(width: 100%, breakable: true, fill: rgb("#EFF1F5"), stroke: 0.5pt + rgb("#CFDDF0"), radius: 4pt, inset: 10pt, above: 12pt, below: 12pt)[ #block(below: 6pt)[#box(fill: BLUE, inset: (x: 6pt, y: 2pt), radius: 2pt, text(fill: white, weight: "bold", size: 8.5pt, label)) #h(0.4em) #strong[#title]] #body] // rail = decorative left rule (raw brand token); labelcolor = AA-safe label text shade #let notebox(label, rail, labelcolor, tint, body) = block(width: 100%, breakable: true, fill: tint, stroke: (left: 3pt + rail), inset: (left: 10pt, rest: 8pt), radius: (right: 4pt), above: 11pt, below: 11pt)[ #text(fill: labelcolor, weight: "bold", size: 7.5pt, tracking: 0.5pt)[#upper(label)] #linebreak() #body] #let solutionbox(body) = block(above: 4pt, below: 8pt)[ #text(fill: BLUE, weight: "bold", size: 8.5pt)[Solution] #linebreak() #body] #let figph(msg) = block(width: 100%, height: 60pt, fill: rgb("#f6f7f9"), stroke: (paint: rgb("#c7ccd3"), dash: "dashed"), radius: 4pt, inset: 10pt)[ #align(center + horizon, text(fill: rgb("#889"), style: "italic", size: 9pt, msg))] // Standardize inlined figure sizes: measure the natural CeTZ canvas, then scale to a // consistent envelope (aspect-aware; see build_typst.py FIG_* constants). Unlike the // print preamble, dimensions are FLOORED: in an editor a user can trim a figure to a // degenerate 1-D shape (a bare line), and w/h or tw/w would then divide by zero. #let _STD_W = 3.5 #let _WIDE_W = 5.6 #let _MAX_H = 3.4 #let _ASPECT_WIDE = 2.2 #let _UPSCALE_MAX = 1.15 #let stdfig(body) = context { let m = measure(body) let w = calc.max(m.width / 1in, 0.01) let h = calc.max(m.height / 1in, 0.01) let tw = if w / h > _ASPECT_WIDE { _WIDE_W } else { _STD_W } let s = calc.min(tw / w, _MAX_H / h, _UPSCALE_MAX) align(center, box(scale(x: s * 100%, y: s * 100%, reflow: true, body))) } #show figure: set block(breakable: false) #set figure(gap: 8pt) #show figure.caption: set text(size: 8.5pt, fill: rgb("#555")) == 2.5#h(0.6em)Prime Factorization and the Least Common Multiple #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Before you get started, take this readiness quiz. Is #math.equation(block: false, alt: "810")[$810$] divisible by #math.equation(block: false, alt: "2 , 3 , 5 , 6 , or 10 ?")[$2 , 3 , 5 , 6 , "or" #h(0.2em) 10 ?$] #linebreak() If you missed this problem, review . #solutionbox[ #math.equation(block: true, alt: "2 , 3 , 5 , 6 , 10")[$2 , 3 , 5 , 6 , 10$] ] ] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Is #math.equation(block: false, alt: "127")[$127$] prime or composite? #linebreak() If you missed this problem, review . #solutionbox[ prime ] Write #math.equation(block: false, alt: "2 times 2 times 2 times 2")[$2 ⋅ 2 ⋅ 2 ⋅ 2$] in exponential notation. #linebreak() If you missed this problem, review . #solutionbox[ #math.equation(block: true, alt: "2 to the power 4")[$2^(4)$] ] ] === Find the Prime Factorization of a Composite Number In the previous section, we found the factors of a number. Prime numbers have only two factors, the number #math.equation(block: false, alt: "1")[$1$] and the prime number itself. Composite numbers have more than two factors, and every composite number can be written as a unique product of primes. This is called the #strong[prime factorization] of a number. When we write the prime factorization of a number, we are rewriting the number as a product of primes. Finding the prime factorization of a composite number will help you later in this course. #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #emph[Prime Factorization] The prime factorization of a number is the product of prime numbers that equals the number. ] You may want to refer to the following list of prime numbers less than #math.equation(block: false, alt: "50")[$50$] as you work through this section. #math.equation(block: true, alt: "2 , 3 , 5 , 7 , 11 , 13 , 17 , 19 , 23 , 29 , 31 , 37 , 41 , 43 , 47")[$2 , 3 , 5 , 7 , 11 , 13 , 17 , 19 , 23 , 29 , 31 , 37 , 41 , 43 , 47$] ==== Prime Factorization Using the Factor Tree Method One way to find the prime factorization of a number is to make a #strong[factor tree]. We start by writing the number, and then writing it as the product of two factors. We write the factors below the number and connect them to the number with a small line segment—a “branch” of the factor tree. If a factor is prime, we circle it (like a bud on a tree), and do not factor that “branch” any further. If a factor is not prime, we repeat this process, writing it as the product of two factors and adding new branches to the tree. We continue until all the branches end with a prime. When the factor tree is complete, the circled primes give us the prime factorization. For example, let’s find the prime factorization of #math.equation(block: false, alt: "36 .")[$36 .$] We can start with any factor pair such as #math.equation(block: false, alt: "3")[$3$] and #math.equation(block: false, alt: "12 .")[$12 .$] We write #math.equation(block: false, alt: "3")[$3$] and #math.equation(block: false, alt: "12")[$12$] below #math.equation(block: false, alt: "36")[$36$] with branches connecting them. #figure(figph[The figure shows a factor tree with the number 36 at the top. Two branches are splitting out from under 36. The right branch has a number 3 at the end with a circle around it. The left branch has the number 12 at the end.], alt: "The figure shows a factor tree with the number 36 at the top. Two branches are splitting out from under 36. The right branch has a number 3 at the end with a circle around it. The left branch has the number 12 at the end.", caption: none) The factor #math.equation(block: false, alt: "3")[$3$] is prime, so we circle it. The factor #math.equation(block: false, alt: "12")[$12$] is composite, so we need to find its factors. Let’s use #math.equation(block: false, alt: "3")[$3$] and #math.equation(block: false, alt: "4 .")[$4 .$] We write these factors on the tree under the #math.equation(block: false, alt: "12 .")[$12 .$] #figure(figph[The figure shows a factor tree with the number 36 at the top. Two branches are splitting out from under 36. The right branch has a number 3 at the end with a circle around it. The left branch has the number 12 at the end. Two more branches are splitting out from under 12. The right branch has the number 4 at the end and the left branch has the number 3 at the end.], alt: "The figure shows a factor tree with the number 36 at the top. Two branches are splitting out from under 36. The right branch has a number 3 at the end with a circle around it. The left branch has the number 12 at the end. Two more branches are splitting out from under 12. The right branch has the number 4 at the end and the left branch has the number 3 at the end.", caption: none) The factor #math.equation(block: false, alt: "3")[$3$] is prime, so we circle it. The factor #math.equation(block: false, alt: "4")[$4$] is composite, and it factors into #math.equation(block: false, alt: "2 times 2 .")[$2 · 2 .$] We write these factors under the #math.equation(block: false, alt: "4 .")[$4 .$] Since #math.equation(block: false, alt: "2")[$2$] is prime, we circle both #math.equation(block: false, alt: "2 s .")[$2 "s" .$] #figure(figph[The figure shows a factor tree with the number 36 at the top. Two branches are splitting out from under 36. The right branch has a number 3 at the end with a circle around it. The left branch has the number 12 at the end. Two more branches are splitting out from under 12. The right branch has the number 4 at the end and the left branch has the number 3 at the end with a circle around it. Two more branches are splitting out from under 4. Both the left and right branch have the number 2 at the end with a circle around it.], alt: "The figure shows a factor tree with the number 36 at the top. Two branches are splitting out from under 36. The right branch has a number 3 at the end with a circle around it. The left branch has the number 12 at the end. Two more branches are splitting out from under 12. The right branch has the number 4 at the end and the left branch has the number 3 at the end with a circle around it. Two more branches are splitting out from under 4. Both the left and right branch have the number 2 at the end with a circle around it.", caption: none) The prime factorization is the product of the circled primes. We generally write the prime factorization in order from least to greatest. #math.equation(block: true, alt: "2 times 2 times 3 times 3")[$2 ⋅ 2 ⋅ 3 ⋅ 3$]In cases like this, where some of the prime factors are repeated, we can write prime factorization in exponential form. #math.equation(block: true, alt: "2 times 2 times 3 times 3; 2 squared times 3 squared")[$2 ⋅ 2 ⋅ 3 ⋅ 3 \ \ 2^(2) ⋅ 3^(2)$]Note that we could have started our factor tree with any factor pair of #math.equation(block: false, alt: "36 .")[$36 .$] We chose #math.equation(block: false, alt: "12")[$12$] and #math.equation(block: false, alt: "3 ,")[$3 ,$] but the same result would have been the same if we had started with #math.equation(block: false, alt: "2")[$2$] and #math.equation(block: false, alt: "18 , 4")[$18 , 4$] and #math.equation(block: false, alt: "9 , or 6 and 6 .")[$9 , "or" #h(0.2em) 6 #h(0.2em) "and" #h(0.2em) 6 .$] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #emph[Find the prime factorization of a composite number using the tree method.] + Find any factor pair of the given number, and use these numbers to create two branches. + If a factor is prime, that branch is complete. Circle the prime. + If a factor is not prime, write it as the product of a factor pair and continue the process. + Write the composite number as the product of all the circled primes. ] #examplebox("Example 1")[][ Find the prime factorization of #math.equation(block: false, alt: "48")[$48$] using the factor tree method. #solutionbox[ #figure(table( columns: 2, align: left, inset: 6pt, table.header([We can start our tree using any factor pair of 48. Let's use 2 and 24.We circle the 2 because it is prime and so that branch is complete.], [#figure(figph[A factor tree showing the number 48 branching into its factors, with 2 and 24 as its immediate children, and the number 2 circled.], alt: "A factor tree showing the number 48 branching into its factors, with 2 and 24 as its immediate children, and the number 2 circled.", caption: none)]), [Now we will factor 24. Let's use 4 and 6.], [#figure(figph[A partially completed factor tree illustrating the prime factorization of the number 48, with the prime factor 2 circled, showing branches for 24, 4, and 6.], alt: "A partially completed factor tree illustrating the prime factorization of the number 48, with the prime factor 2 circled, showing branches for 24, 4, and 6.", caption: none)], [Neither factor is prime, so we do not circle either. #linebreak() We factor the 4, using 2 and 2. #linebreak() We factor 6, using 2 and 3.We circle the 2s and the 3 since they are prime. Now all of the branches end in a prime.], [#figure(figph[A factor tree demonstrates the prime factorization of 48, revealing its prime factors as 2, 2, 2, 2, and 3, which multiply to 48. This method breaks a number into its fundamental prime components.], alt: "A factor tree demonstrates the prime factorization of 48, revealing its prime factors as 2, 2, 2, 2, and 3, which multiply to 48. This method breaks a number into its fundamental prime components.", caption: none)], [Write the product of the circled numbers.], [#math.equation(block: false, alt: "2 times 2 times 2 times 2 times 3")[$2 ⋅ 2 ⋅ 2 ⋅ 2 ⋅ 3$]], [Write in exponential form.], [#math.equation(block: false, alt: "2 to the power 4 times 3")[$2^(4) ⋅ 3$]], )) Check this on your own by multiplying all the factors together. The result should be #math.equation(block: false, alt: "48 .")[$48 .$] ] ] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Find the prime factorization using the factor tree method: #math.equation(block: false, alt: "80")[$80$] #solutionbox[ 2 ⋅ 2 ⋅ 2 ⋅ 2 ⋅ 5, or 2#super[4] ⋅ 5 ] ] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Find the prime factorization using the factor tree method: #math.equation(block: false, alt: "60")[$60$] #solutionbox[ 2 ⋅ 2 ⋅ 3 ⋅ 5, or 2#super[2] ⋅ 3 ⋅ 5 ] ] #examplebox("Example 2")[][ Find the prime factorization of 84 using the factor tree method. #solutionbox[ #figure(table( columns: 2, align: left, inset: 6pt, table.header([We start with the factor pair 4 and 21.Neither factor is prime so we factor them further.], [#figure(figph[A factor tree shows the number 84 branching into its factors 4 and 21.], alt: "A factor tree shows the number 84 branching into its factors 4 and 21.", caption: none)]), [Now the factors are all prime, so we circle them.], [#figure(figph[A prime factorization tree illustrating how the number 84 breaks down into its prime factors: 2, 2, 3, and 7.], alt: "A prime factorization tree illustrating how the number 84 breaks down into its prime factors: 2, 2, 3, and 7.", caption: none)], [Then we write 84 as the product of all circled primes.], [#math.equation(block: false, alt: "2 times 2 times 3 times 7")[$2 ⋅ 2 ⋅ 3 ⋅ 7$] #linebreak() #math.equation(block: false, alt: "2 squared times 3 times 7")[$2^(2) ⋅ 3 ⋅ 7$]], )) Draw a factor tree of #math.equation(block: false, alt: "84 .")[$84 .$] ] ] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Find the prime factorization using the factor tree method: #math.equation(block: false, alt: "126")[$126$] #solutionbox[ 2 ⋅ 3 ⋅ 3 ⋅ 7, or 2 ⋅ 3#super[2] ⋅ 7 ] ] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Find the prime factorization using the factor tree method: #math.equation(block: false, alt: "294")[$294$] #solutionbox[ 2 ⋅ 3 ⋅ 7 ⋅ 7, or 2 ⋅ 3 ⋅ 7#super[2] ] ] ==== Prime Factorization Using the Ladder Method The #strong[ladder method] is another way to find the prime factors of a composite number. It leads to the same result as the factor tree method. Some people prefer the ladder method to the factor tree method, and vice versa. To begin building the “ladder,” divide the given number by its smallest prime factor. For example, to start the ladder for #math.equation(block: false, alt: "36 ,")[$36 ,$] we divide #math.equation(block: false, alt: "36")[$36$] by #math.equation(block: false, alt: "2 ,")[$2 ,$] the smallest prime factor of #math.equation(block: false, alt: "36 .")[$36 .$] #figure(figph[The image shows the division of 2 into 36 to get the quotient 18. This division is represented using a division bracket with 2 on the outside left of the bracket, 36 inside the bracket and 18 above the 36, outside the bracket.], alt: "The image shows the division of 2 into 36 to get the quotient 18. This division is represented using a division bracket with 2 on the outside left of the bracket, 36 inside the bracket and 18 above the 36, outside the bracket.", caption: none) To add a “step” to the ladder, we continue dividing by the same prime until it no longer divides evenly. #figure(figph[The image shows the division of 2 into 36 to get the quotient 18. This division is represented using a division bracket with 2 on the outside left of the bracket, 36 inside the bracket and 18 above the 36, outside the bracket. Another division bracket is written around the 18 with a 2 on the outside left of the bracket and a 9 above the 18, outside of the bracket.], alt: "The image shows the division of 2 into 36 to get the quotient 18. This division is represented using a division bracket with 2 on the outside left of the bracket, 36 inside the bracket and 18 above the 36, outside the bracket. Another division bracket is written around the 18 with a 2 on the outside left of the bracket and a 9 above the 18, outside of the bracket.", caption: none) Then we divide by the next prime; so we divide #math.equation(block: false, alt: "9")[$9$] by #math.equation(block: false, alt: "3 .")[$3 .$] #figure(figph[The image shows the division of 2 into 36 to get the quotient 18. This division is represented using a division bracket with 2 on the outside left of the bracket, 36 inside the bracket and 18 above the 36, outside the bracket. Another division bracket is written around the 18 with a 2 on the outside left of the bracket and a 9 above the 18, outside of the bracket. Another division bracket is written around the 9 with a 3 on the outside left of the bracket and a 3 above the 9, outside of the bracket.], alt: "The image shows the division of 2 into 36 to get the quotient 18. This division is represented using a division bracket with 2 on the outside left of the bracket, 36 inside the bracket and 18 above the 36, outside the bracket. Another division bracket is written around the 18 with a 2 on the outside left of the bracket and a 9 above the 18, outside of the bracket. Another division bracket is written around the 9 with a 3 on the outside left of the bracket and a 3 above the 9, outside of the bracket.", caption: none) We continue dividing up the ladder in this way until the quotient is prime. Since the quotient, #math.equation(block: false, alt: "3 ,")[$3 ,$] is prime, we stop here. Do you see why the ladder method is sometimes called stacked division? The prime factorization is the product of all the primes on the sides and top of the ladder. #math.equation(block: true, alt: "2 times 2 times 3 times 3; 2 squared times 3 squared")[$2 ⋅ 2 ⋅ 3 ⋅ 3 \ \ 2^(2) ⋅ 3^(2)$]Notice that the result is the same as we obtained with the factor tree method. #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #emph[Find the prime factorization of a composite number using the ladder method.] + Divide the number by the smallest prime. + Continue dividing by that prime until it no longer divides evenly. + Divide by the next prime until it no longer divides evenly. + Continue until the quotient is a prime. + Write the composite number as the product of all the primes on the sides and top of the ladder. ] #examplebox("Example 3")[][ Find the prime factorization of #math.equation(block: false, alt: "120")[$120$] using the ladder method. #solutionbox[ #figure(table( columns: 2, align: left, inset: 6pt, table.header([Divide the number by the smallest prime, which is 2.], [#figure(figph[A long division problem showing 120 divided by 2, with the quotient 60 written above the division bar. The divisor is 2, the dividend is 120, and the result of the division is 60.], alt: "A long division problem showing 120 divided by 2, with the quotient 60 written above the division bar. The divisor is 2, the dividend is 120, and the result of the division is 60.", caption: none)]), [Continue dividing by 2 until it no longer divides evenly.], [#figure(figph[A series of long division calculations demonstrating repeated division by 2: 120 divided by 2 equals 60, then 60 divided by 2 equals 30, and 30 divided by 2 equals 15.], alt: "A series of long division calculations demonstrating repeated division by 2: 120 divided by 2 equals 60, then 60 divided by 2 equals 30, and 30 divided by 2 equals 15.", caption: none)], [Divide by the next prime, 3.], [#figure(figph[Decomposing 120 into its prime factors using repeated division, resulting in 2, 2, 2, 3, and 5.], alt: "Decomposing 120 into its prime factors using repeated division, resulting in 2, 2, 2, 3, and 5.", caption: none)], [The quotient, 5, is prime, so the ladder is complete. Write the prime factorization of 120.], [#math.equation(block: false, alt: "2 times 2 times 2 times 3 times 5")[$2 ⋅ 2 ⋅ 2 ⋅ 3 ⋅ 5$] #linebreak() #math.equation(block: false, alt: "2 cubed times 3 times 5")[$2^(3) ⋅ 3 ⋅ 5$]], )) Check this yourself by multiplying the factors. The result should be #math.equation(block: false, alt: "120 .")[$120 .$] ] ] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Find the prime factorization using the ladder method: #math.equation(block: false, alt: "80")[$80$] #solutionbox[ 2 ⋅ 2 ⋅ 2 ⋅ 2 ⋅ 5, or 2#super[4] ⋅ 5 ] ] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Find the prime factorization using the ladder method: #math.equation(block: false, alt: "60")[$60$] #solutionbox[ 2 ⋅ 2 ⋅ 3 ⋅ 5, or 2#super[2] ⋅ 3 ⋅ 5 ] ] #examplebox("Example 4")[][ Find the prime factorization of #math.equation(block: false, alt: "48")[$48$] using the ladder method. #solutionbox[ #figure(table( columns: 2, align: left, inset: 6pt, table.header([Divide the number by the smallest prime, 2.], [#figure(figph[A long division problem showing 48 divided by 2, with the quotient 24 written above the divisor, indicating that 24 is the result of 48 divided by 2.], alt: "A long division problem showing 48 divided by 2, with the quotient 24 written above the divisor, indicating that 24 is the result of 48 divided by 2.", caption: none)]), [Continue dividing by 2 until it no longer divides evenly.], [#figure(figph[A step-by-step illustration of repeated division by 2, showing 48 divided by 2 repeatedly until it reaches 3. The divisions shown are 48/2=24, 24/2=12, 12/2=6, and 6/2=3.], alt: "A step-by-step illustration of repeated division by 2, showing 48 divided by 2 repeatedly until it reaches 3. The divisions shown are 48/2=24, 24/2=12, 12/2=6, and 6/2=3.", caption: none)], [The quotient, 3, is prime, so the ladder is complete. Write the prime factorization of 48.], [#math.equation(block: false, alt: "2 times 2 times 2 times 2 times 3")[$2 ⋅ 2 ⋅ 2 ⋅ 2 ⋅ 3$] #linebreak() #math.equation(block: false, alt: "2 to the power 4 times 3")[$2^(4) ⋅ 3$]], )) ] ] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Find the prime factorization using the ladder method. #math.equation(block: false, alt: "126")[$126$] #solutionbox[ 2 ⋅ 3 ⋅ 3 ⋅ 7, or 2 ⋅ 3#super[2] ⋅ 7 ] ] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Find the prime factorization using the ladder method. #math.equation(block: false, alt: "294")[$294$] #solutionbox[ 2 ⋅ 3 ⋅ 7 ⋅ 7, or 2 ⋅ 3 ⋅ 7#super[2] ] ] === Find the Least Common Multiple (LCM) of Two Numbers One of the reasons we look at multiples and primes is to use these techniques to find the least common multiple of two numbers. This will be useful when we add and subtract fractions with different denominators. ==== Listing Multiples Method A common multiple of two numbers is a number that is a multiple of both numbers. Suppose we want to find common multiples of #math.equation(block: false, alt: "10")[$10$] and #math.equation(block: false, alt: "25 .")[$25 .$] We can list the first several multiples of each number. Then we look for multiples that are common to both lists—these are the common multiples. #math.equation(block: true, alt: "10 : 10 , 20 , 30 , 40 , 50 , 60 , 70 , 80 , 90 , 100 , 110 , …; 25 : 25 , 50 , 75 , 100 , 125 , …")[$10 ":" 10 , 20 , 30 , 40 , #h(0.2em) 50 , 60 , 70 , 80 , 90 , 100 , 110 , "…" \ 25 ":" 25 , #h(0.2em) 50 , 75 , #h(0.2em) 100 , 125 , "…"$]We see that #math.equation(block: false, alt: "50")[$50$] and #math.equation(block: false, alt: "100")[$100$] appear in both lists. They are common multiples of #math.equation(block: false, alt: "10")[$10$] and #math.equation(block: false, alt: "25 .")[$25 .$] We would find more common multiples if we continued the list of multiples for each. The smallest number that is a multiple of two numbers is called the #strong[least common multiple] (LCM). So the least LCM of #math.equation(block: false, alt: "10")[$10$] and #math.equation(block: false, alt: "25")[$25$] is #math.equation(block: false, alt: "50 .")[$50 .$] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #emph[Find the least common multiple (LCM) of two numbers by listing multiples.] + List the first several multiples of each number. + Look for multiples common to both lists. If there are no common multiples in the lists, write out additional multiples for each number. + Look for the smallest number that is common to both lists. + This number is the LCM. ] #examplebox("Example 5")[][ Find the LCM of #math.equation(block: false, alt: "15")[$15$] and #math.equation(block: false, alt: "20")[$20$] by listing multiples. #solutionbox[ List the first several multiples of #math.equation(block: false, alt: "15")[$15$] and of #math.equation(block: false, alt: "20 .")[$20 .$] Identify the first common multiple. #math.equation(block: true, alt: "15: 15 , 30 , 45 , 60 , 75 , 90 , 105 , 120; 20: 20 , 40 , 60 , 80 , 100 , 120 , 140 , 160")[$"15:" #h(0.2em) 15 , 30 , 45 , #h(0.2em) 60 , 75 , 90 , 105 , 120 \ "20:" #h(0.2em) 20 , 40 , #h(0.2em) 60 , 80 , 100 , 120 , 140 , 160$] The smallest number to appear on both lists is #math.equation(block: false, alt: "60 ,")[$60 ,$] so #math.equation(block: false, alt: "60")[$60$] is the least common multiple of #math.equation(block: false, alt: "15")[$15$] and #math.equation(block: false, alt: "20 .")[$20 .$] Notice that #math.equation(block: false, alt: "120")[$120$] is on both lists, too. It is a common multiple, but it is not the least common multiple. ] ] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Find the least common multiple (LCM) of the given numbers: #math.equation(block: false, alt: "9 and 12")[$9 #h(0.2em) "and" #h(0.2em) 12$] #solutionbox[ 36 ] ] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Find the least common multiple (LCM) of the given numbers: #math.equation(block: false, alt: "18 and 24")[$18 #h(0.2em) "and" #h(0.2em) 24$] #solutionbox[ 72 ] ] ==== Prime Factors Method Another way to find the least common multiple of two numbers is to use their prime factors. We’ll use this method to find the LCM of #math.equation(block: false, alt: "12")[$12$] and #math.equation(block: false, alt: "18 .")[$18 .$] We start by finding the prime factorization of each number. #math.equation(block: true, alt: "12 equals 2 times 2 times 3 18 equals 2 times 3 times 3")[$12 = 2 ⋅ 2 ⋅ 3 #h(3em) 18 = 2 ⋅ 3 ⋅ 3$]Then we write each number as a product of primes, matching primes vertically when possible. #math.equation(block: true, alt: "12 equals 2 times 2 times 3; 18 equals 2 times 3 times 3")[$12 = 2 ⋅ 2 ⋅ 3 \ 18 = 2 ⋅ #h(1.1em) 3 ⋅ 3$]Now we bring down the primes in each column. The LCM is the product of these factors. #figure(figph[The image shows the prime factorization of 12 written as the equation 12 equals 2 times 2 times 3. Below this equation is another showing the prime factorization of 18 written as the equation 18 equals 2 times 3 times 3. The two equations line up vertically at the equal symbol. The first 2 in the prime factorization of 12 aligns with the 2 in the prime factorization of 18. Under the second 2 in the prime factorization of 12 is a gap in the prime factorization of 18. Under the 3 in the prime factorization of 12 is the first 3 in the prime factorization of 18. The second 3 in the prime factorization has no factors above it from the prime factorization of 12. A horizontal line is drawn under the prime factorization of 18. Below this line is the equation LCM equal to 2 times 2 times 3 times 3. Arrows are drawn down vertically from the prime factorization of 12 through the prime factorization of 18 ending at the LCM equation. The first arrow starts at the first 2 in the prime factorization of 12 and continues down through the 2 in the prime factorization of 18. Ending with the first 2 in the LCM. The second arrow starts at the next 2 in the prime factorization of 12 and continues down through the gap in the prime factorization of 18. Ending with the second 2 in the LCM. The third arrow starts at the 3 in the prime factorization of 12 and continues down through the first 3 in the prime factorization of 18. Ending with the first 3 in the LCM. The last arrow starts at the second 3 in the prime factorization of 18 and points down to the second 3 in the LCM.], alt: "The image shows the prime factorization of 12 written as the equation 12 equals 2 times 2 times 3. Below this equation is another showing the prime factorization of 18 written as the equation 18 equals 2 times 3 times 3. The two equations line up vertically at the equal symbol. The first 2 in the prime factorization of 12 aligns with the 2 in the prime factorization of 18. Under the second 2 in the prime factorization of 12 is a gap in the prime factorization of 18. Under the 3 in the prime factorization of 12 is the first 3 in the prime factorization of 18. The second 3 in the prime factorization has no factors above it from the prime factorization of 12. A horizontal line is drawn under the prime factorization of 18. Below this line is the equation LCM equal to 2 times 2 times 3 times 3. Arrows are drawn down vertically from the prime factorization of 12 through the prime factorization of 18 ending at the LCM equation. The first arrow starts at the first 2 in the prime factorization of 12 and continues down through the 2 in the prime factorization of 18. Ending with the first 2 in the LCM. The second arrow starts at the next 2 in the prime factorization of 12 and continues down through the gap in the prime factorization of 18. Ending with the second 2 in the LCM. The third arrow starts at the 3 in the prime factorization of 12 and continues down through the first 3 in the prime factorization of 18. Ending with the first 3 in the LCM. The last arrow starts at the second 3 in the prime factorization of 18 and points down to the second 3 in the LCM.", caption: none) Notice that the prime factors of #math.equation(block: false, alt: "12")[$12$] and the prime factors of #math.equation(block: false, alt: "18")[$18$] are included in the LCM. By matching up the common primes, each common prime factor is used only once. This ensures that #math.equation(block: false, alt: "36")[$36$] is the least common multiple. #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #emph[Find the LCM using the prime factors method.] + Find the prime factorization of each number. + Write each number as a product of primes, matching primes vertically when possible. + Bring down the primes in each column. + Multiply the factors to get the LCM. ] #examplebox("Example 6")[][ Find the LCM of #math.equation(block: false, alt: "15")[$15$] and #math.equation(block: false, alt: "18")[$18$] using the prime factors method. #solutionbox[ #figure(table( columns: 2, align: left, inset: 6pt, table.header([Write each number as a product of primes.], [#figure(figph[The image displays the prime factorization of two numbers: 15 is shown as 3 times 5, and 18 is shown as 2 times 3 times 3.], alt: "The image displays the prime factorization of two numbers: 15 is shown as 3 times 5, and 18 is shown as 2 times 3 times 3.", caption: none)]), [Write each number as a product of primes, matching primes vertically when possible.], [#figure(figph[The prime factorization of 15 (3 \* 5) and 18 (2 \* 3 \* 3) is shown in a clear, concise format.], alt: "The prime factorization of 15 (3 * 5) and 18 (2 * 3 * 3) is shown in a clear, concise format.", caption: none)], [Bring down the primes in each column.], [#figure(figph[Finding the Least Common Multiple (LCM) of 15 and 18 is shown using prime factorization. 15 = 3 \* 5 and 18 = 2 \* 3 \* 3. The LCM combines these factors to 2 \* 3 \* 3 \* 5.], alt: "Finding the Least Common Multiple (LCM) of 15 and 18 is shown using prime factorization. 15 = 3 * 5 and 18 = 2 * 3 * 3. The LCM combines these factors to 2 * 3 * 3 * 5.", caption: none)], [Multiply the factors to get the LCM.], [#math.equation(block: false, alt: "LCM equals 2 times 3 times 3 times 5")[$"LCM" = 2 ⋅ 3 ⋅ 3 ⋅ 5$] #linebreak() The LCM of 15 and 18 is 90.], )) ] ] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Find the LCM using the prime factors method. #math.equation(block: false, alt: "15 and 20")[$15 #h(0.2em) "and" #h(0.2em) 20$] #solutionbox[ 60 ] ] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Find the LCM using the prime factors method. #math.equation(block: false, alt: "15 and 35")[$15 #h(0.2em) "and" #h(0.2em) 35$] #solutionbox[ 105 ] ] #examplebox("Example 7")[][ Find the LCM of #math.equation(block: false, alt: "50")[$50$] and #math.equation(block: false, alt: "100")[$100$] using the prime factors method. #solutionbox[ #figure(table( columns: 2, align: left, inset: 6pt, table.header([Write the prime factorization of each number.], [#figure(figph[The image shows the prime factorization of 50 as 2 x 5 x 5, and the prime factorization of 100 as 2 x 2 x 5 x 5.], alt: "The image shows the prime factorization of 50 as 2 x 5 x 5, and the prime factorization of 100 as 2 x 2 x 5 x 5.", caption: none)]), [Write each number as a product of primes, matching primes vertically when possible.], [#figure(figph[The prime factorization of 50 (2\*5\*5) and 100 (2\*2\*5\*5) is displayed on a white background.], alt: "The prime factorization of 50 (2*5*5) and 100 (2*2*5*5) is displayed on a white background.", caption: none)], [Bring down the primes in each column.], [#figure(figph[The image demonstrates the calculation of the Least Common Multiple (LCM) for 50 and 100 using prime factorization, showing 50 as 2 x 5 x 5 and 100 as 2 x 2 x 5 x 5, resulting in an LCM of 2 x 2 x 5 x 5.], alt: "The image demonstrates the calculation of the Least Common Multiple (LCM) for 50 and 100 using prime factorization, showing 50 as 2 x 5 x 5 and 100 as 2 x 2 x 5 x 5, resulting in an LCM of 2 x 2 x 5 x 5.", caption: none)], [Multiply the factors to get the LCM.], [#math.equation(block: false, alt: "LCM equals 2 times 2 times 5 times 5")[$"LCM" = 2 ⋅ 2 ⋅ 5 ⋅ 5$] #linebreak() The LCM of 50 and 100 is 100.], )) ] ] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Find the LCM using the prime factors method: #math.equation(block: false, alt: "55 , 88")[$55 , 88$] #solutionbox[ 440 ] ] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Find the LCM using the prime factors method: #math.equation(block: false, alt: "60 , 72")[$60 , 72$] #solutionbox[ 360 ] ] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #emph[ACCESS ADDITIONAL ONLINE RESOURCES] - #link("https://openstaxcollege.org/l/24PrimeFactor1")[Ex 1: Prime Factorization] - #link("https://openstaxcollege.org/l/24PrimeFactor2")[Ex 2: Prime Factorization] - #link("https://openstaxcollege.org/l/24PrimeFactor3")[Ex 3: Prime Factorization] - #link("https://openstaxcollege.org/l/24stackeddivis")[Ex 1: Prime Factorization Using Stacked Division] - #link("https://openstaxcollege.org/l/24stackeddivis2")[Ex 2: Prime Factorization Using Stacked Division] - #link("https://openstaxcollege.org/l/24LCM")[The Least Common Multiple] - #link("https://openstaxcollege.org/l/24LCM2")[Example: Determining the Least Common Multiple Using a List of Multiples] - #link("https://openstaxcollege.org/l/24LCMFactor")[Example: Determining the Least Common Multiple Using Prime Factorization] ] === Key Concepts - #emph[Find the prime factorization of a composite number using the tree method.] + Find any factor pair of the given number, and use these numbers to create two branches. + If a factor is prime, that branch is complete. Circle the prime. + If a factor is not prime, write it as the product of a factor pair and continue the process. + Write the composite number as the product of all the circled primes. - #emph[Find the prime factorization of a composite number using the ladder method.] + Divide the number by the smallest prime. + Continue dividing by that prime until it no longer divides evenly. + Divide by the next prime until it no longer divides evenly. + Continue until the quotient is a prime. + Write the composite number as the product of all the primes on the sides and top of the ladder. - #emph[Find the LCM by listing multiples.] + List the first several multiples of each number. + Look for multiples common to both lists. If there are no common multiples in the lists, write out additional multiples for each number. + Look for the smallest number that is common to both lists. + This number is the LCM. - #emph[Find the LCM using the prime factors method.] + Find the prime factorization of each number. + Write each number as a product of primes, matching primes vertically when possible. + Bring down the primes in each column. + Multiply the factors to get the LCM. === Section Exercises ==== Practice Makes Perfect #strong[Find the Prime Factorization of a Composite Number] In the following exercises, find the prime factorization of each number using the factor tree method. #math.equation(block: true, alt: "86")[$86$] #solutionbox[ 2 ⋅ 43 ] #math.equation(block: true, alt: "78")[$78$] #math.equation(block: true, alt: "132")[$132$] #solutionbox[ 2 ⋅ 2 ⋅ 3 ⋅ 11 ] #math.equation(block: true, alt: "455")[$455$] #math.equation(block: true, alt: "693")[$693$] #solutionbox[ 3 ⋅ 3 ⋅ 7 ⋅ 11 ] #math.equation(block: true, alt: "420")[$420$] #math.equation(block: true, alt: "115")[$115$] #solutionbox[ 5 ⋅ 23 ] #math.equation(block: true, alt: "225")[$225$] #math.equation(block: true, alt: "2475")[$2475$] #solutionbox[ 3 ⋅ 3 ⋅ 5 ⋅ 5 ⋅ 11 ] 1560 In the following exercises, find the prime factorization of each number using the ladder method. #math.equation(block: true, alt: "56")[$56$] #solutionbox[ 2 ⋅ 2 ⋅ 2 ⋅ 7 ] #math.equation(block: true, alt: "72")[$72$] #math.equation(block: true, alt: "168")[$168$] #solutionbox[ 2 ⋅ 2 ⋅ 2 ⋅ 3 ⋅ 7 ] #math.equation(block: true, alt: "252")[$252$] #math.equation(block: true, alt: "391")[$391$] #solutionbox[ 17 ⋅ 23 ] #math.equation(block: true, alt: "400")[$400$] #math.equation(block: true, alt: "432")[$432$] #solutionbox[ 2 ⋅ 2 ⋅ 2 ⋅ 2 ⋅ 3 ⋅ 3 ⋅ 3 ] #math.equation(block: true, alt: "627")[$627$] #math.equation(block: true, alt: "2160")[$2160$] #solutionbox[ 2 ⋅ 2 ⋅ 2 ⋅ 2 ⋅ 3 ⋅ 3 ⋅ 3 ⋅ 5 ] #math.equation(block: true, alt: "2520")[$2520$] In the following exercises, find the prime factorization of each number using any method. #math.equation(block: true, alt: "150")[$150$] #solutionbox[ 2 ⋅ 3 ⋅ 5 ⋅ 5 ] #math.equation(block: true, alt: "180")[$180$] #math.equation(block: true, alt: "525")[$525$] #solutionbox[ 3 ⋅ 5 ⋅ 5 ⋅ 7 ] #math.equation(block: true, alt: "444")[$444$] #math.equation(block: true, alt: "36")[$36$] #solutionbox[ 2 ⋅ 2 ⋅ 3 ⋅ 3 ] #math.equation(block: true, alt: "50")[$50$] #math.equation(block: true, alt: "350")[$350$] #solutionbox[ 2 ⋅ 5 ⋅ 5 ⋅ 7 ] #math.equation(block: true, alt: "144")[$144$] #strong[Find the Least Common Multiple (LCM) of Two Numbers] In the following exercises, find the least common multiple (LCM) by listing multiples. #math.equation(block: true, alt: "8 , 12")[$8 , 12$] #solutionbox[ 24 ] #math.equation(block: true, alt: "4 , 3")[$4 , 3$] #math.equation(block: true, alt: "6 , 15")[$6 , 15$] #solutionbox[ 30 ] #math.equation(block: true, alt: "12 , 16")[$12 , 16$] #math.equation(block: true, alt: "30 , 40")[$30 , 40$] #solutionbox[ 120 ] #math.equation(block: true, alt: "20 , 30")[$20 , 30$] #math.equation(block: true, alt: "60 , 75")[$60 , 75$] #solutionbox[ 300 ] #math.equation(block: true, alt: "44 , 55")[$44 , 55$] In the following exercises, find the least common multiple (LCM) by using the prime factors method. #math.equation(block: true, alt: "8 , 12")[$8 , 12$] #solutionbox[ 24 ] #math.equation(block: true, alt: "12 , 16")[$12 , 16$] #math.equation(block: true, alt: "24 , 30")[$24 , 30$] #solutionbox[ 120 ] #math.equation(block: true, alt: "28 , 40")[$28 , 40$] #math.equation(block: true, alt: "70 , 84")[$70 , 84$] #solutionbox[ 420 ] #math.equation(block: true, alt: "84 , 90")[$84 , 90$] In the following exercises, find the least common multiple (LCM) using any method. #math.equation(block: true, alt: "6 , 21")[$6 , 21$] #solutionbox[ 42 ] #math.equation(block: true, alt: "9 , 15")[$9 , 15$] #math.equation(block: true, alt: "24 , 30")[$24 , 30$] #solutionbox[ 120 ] #math.equation(block: true, alt: "32 , 40")[$32 , 40$] ==== Everyday Math #strong[Grocery shopping] Hot dogs are sold in packages of ten, but hot dog buns come in packs of eight. What is the smallest number of hot dogs and buns that can be purchased if you want to have the same number of hot dogs and buns? (Hint: it is the LCM!) #solutionbox[ 40 ] #strong[Grocery shopping] Paper plates are sold in packages of #math.equation(block: false, alt: "12")[$12$] and party cups come in packs of #math.equation(block: false, alt: "8 .")[$8 .$] What is the smallest number of plates and cups you can purchase if you want to have the same number of each? (Hint: it is the LCM!) ==== Writing Exercises Do you prefer to find the prime factorization of a composite number by using the factor tree method or the ladder method? Why? Do you prefer to find the LCM by listing multiples or by using the prime factors method? Why? ==== Self Check ⓐ After completing the exercises, use this checklist to evaluate your mastery of the objectives of this section. #figure(figph[A self-assessment table for math skills: prime factorization and least common multiple (LCM). Users can rate their understanding as Confidently, With some help, or No-I don't get it!], alt: "A self-assessment table for math skills: prime factorization and least common multiple (LCM). Users can rate their understanding as Confidently, With some help, or No-I don't get it!", caption: none) ⓑ Overall, after looking at the checklist, do you think you are well-prepared for the next Chapter? Why or why not? === Chapter Review Exercises ==== Use the Language of Algebra #strong[Use Variables and Algebraic Symbols] In the following exercises, translate from algebra to English. #math.equation(block: true, alt: "3 times 8")[$3 ⋅ 8$] #solutionbox[ 3 times 8, the product of three and eight. ] #math.equation(block: true, alt: "12 minus x")[$12 − x$] #math.equation(block: true, alt: "24 divided by 6")[$24 ÷ 6$] #solutionbox[ 24 divided by 6, the quotient of twenty-four and six. ] #math.equation(block: true, alt: "9 plus 2 a")[$9 + 2 a$] #math.equation(block: true, alt: "50 greater than or equal to 47")[$50 ≥ 47$] #solutionbox[ 50 is greater than or equal to 47 ] #math.equation(block: true, alt: "3 y less than 15")[$3 y < 15$] #math.equation(block: true, alt: "n plus 4 equals 13")[$n + 4 = 13$] #solutionbox[ The sum of #emph[n] and 4 is equal to 13 ] #math.equation(block: true, alt: "32 minus k equals 7")[$32 − k = 7$] #strong[Identify Expressions and Equations] In the following exercises, determine if each is an expression or equation. #math.equation(block: true, alt: "5 plus u equals 84")[$5 + u = 84$] #solutionbox[ equation ] #math.equation(block: true, alt: "36 minus 6 s")[$36 − 6 s$] #math.equation(block: true, alt: "4 y minus 11")[$4 y − 11$] #solutionbox[ expression ] #math.equation(block: true, alt: "10 x equals 120")[$10 x = 120$] #strong[Simplify Expressions with Exponents] In the following exercises, write in exponential form. #math.equation(block: true, alt: "2 times 2 times 2")[$2 ⋅ 2 ⋅ 2$] #solutionbox[ 2#super[3] ] #math.equation(block: true, alt: "a times a times a times a times a")[$a ⋅ a ⋅ a ⋅ a ⋅ a$] #math.equation(block: true, alt: "x times x times x times x times x times x")[$x ⋅ x ⋅ x ⋅ x ⋅ x ⋅ x$] #solutionbox[ #emph[x]#super[6] ] #math.equation(block: true, alt: "10 times 10 times 10")[$10 ⋅ 10 ⋅ 10$] In the following exercises, write in expanded form. #math.equation(block: true, alt: "8 to the power 4")[$8^(4)$] #solutionbox[ 8 ⋅ 8 ⋅ 8 ⋅ 8 ] #math.equation(block: true, alt: "3 to the power 6")[$3^(6)$] #math.equation(block: true, alt: "y to the power 5")[$y^(5)$] #solutionbox[ #emph[y] ⋅ #emph[y] ⋅ #emph[y] ⋅ #emph[y] ⋅ #emph[y] ] #math.equation(block: true, alt: "n to the power 4")[$n^(4)$] In the following exercises, simplify each expression. #math.equation(block: true, alt: "3 to the power 4")[$3^(4)$] #solutionbox[ 81 ] #math.equation(block: true, alt: "10 to the power 6")[$10^(6)$] #math.equation(block: true, alt: "2 to the power 7")[$2^(7)$] #solutionbox[ 128 ] #math.equation(block: true, alt: "4 cubed")[$4^(3)$] #strong[Simplify Expressions Using the Order of Operations] In the following exercises, simplify. #math.equation(block: true, alt: "10 plus 2 times 5")[$10 + 2 ⋅ 5$] #solutionbox[ 20 ] #math.equation(block: true, alt: "open parenthesis 10 plus 2 close parenthesis times 5")[$( 10 + 2 ) ⋅ 5$] #math.equation(block: true, alt: "open parenthesis 30 plus 6 close parenthesis divided by 2")[$( 30 + 6 ) ÷ 2$] #solutionbox[ 18 ] #math.equation(block: true, alt: "30 plus 6 divided by 2")[$30 + 6 ÷ 2$] #math.equation(block: true, alt: "7 squared plus 5 squared")[$7^(2) + 5^(2)$] #solutionbox[ 74 ] #math.equation(block: true, alt: "open parenthesis 7 plus 5 close parenthesis squared")[$attach(( 7 + 5 ), t: 2)$] #math.equation(block: true, alt: "4 plus 3 open parenthesis 10 minus 1 close parenthesis")[$4 + 3 ( 10 − 1 )$] #solutionbox[ 31 ] #math.equation(block: true, alt: "open parenthesis 4 plus 3 close parenthesis open parenthesis 10 minus 1 close parenthesis")[$( 4 + 3 ) ( 10 − 1 )$] ==== Evaluate, Simplify, and Translate Expressions #strong[Evaluate an Expression] In the following exercises, evaluate the following expressions. #math.equation(block: true, alt: "9 x minus 5 when x equals 7")[$9 x − 5 #h(0.2em) "when" #h(0.2em) x = 7$] #solutionbox[ 58 ] #math.equation(block: true, alt: "y cubed when y equals 5")[$y^(3) #h(0.2em) "when" #h(0.2em) y = 5$] #math.equation(block: true, alt: "3 a minus 4 b when a equals 10 , b equals 1")[$3 a − 4 b #h(0.2em) "when" #h(0.2em) a = 10 , b = 1$] #solutionbox[ 26 ] #math.equation(block: true, alt: "b h when b equals 7 , h equals 8")[$b h #h(0.2em) "when" #h(0.2em) b = 7 , h = 8$] #strong[Identify Terms, Coefficients and Like Terms] In the following exercises, identify the terms in each expression. #math.equation(block: true, alt: "12 n squared plus 3 n plus 1")[$12 n^(2) + 3 n + 1$] #solutionbox[ 12#emph[n]#super[2],3#emph[n], 1 ] #math.equation(block: true, alt: "4 x cubed plus 11 x plus 3")[$4 x^(3) + 11 x + 3$] In the following exercises, identify the coefficient of each term. #math.equation(block: true, alt: "6 y")[$6 y$] #solutionbox[ 6 ] #math.equation(block: true, alt: "13 x squared")[$13 x^(2)$] In the following exercises, identify the like terms. #math.equation(block: true, alt: "5 x squared , 3 , 5 y squared , 3 x , x , 4")[$5 x^(2) , 3 , 5 y^(2) , 3 x , x , 4$] #solutionbox[ 3 and 4; 3#emph[x] and #emph[x] ] #math.equation(block: true, alt: "8 , 8 r squared , 8 r , 3 r , r squared , 3 s")[$8 , 8 r^(2) , 8 r , 3 r , r^(2) , 3 s$] #strong[Simplify Expressions by Combining Like Terms] In the following exercises, simplify the following expressions by combining like terms. #math.equation(block: true, alt: "15 a plus 9 a")[$15 a + 9 a$] #solutionbox[ 24#emph[a] ] #math.equation(block: true, alt: "12 y plus 3 y plus y")[$12 y + 3 y + y$] #math.equation(block: true, alt: "4 x plus 7 x plus 3 x")[$4 x + 7 x + 3 x$] #solutionbox[ 14#emph[x] ] #math.equation(block: true, alt: "6 plus 5 c plus 3")[$6 + 5 c + 3$] #math.equation(block: true, alt: "8 n plus 2 plus 4 n plus 9")[$8 n + 2 + 4 n + 9$] #solutionbox[ 12#emph[n] + 11 ] #math.equation(block: true, alt: "19 p plus 5 plus 4 p minus 1 plus 3 p")[$19 p + 5 + 4 p − 1 + 3 p$] #math.equation(block: true, alt: "7 y squared plus 2 y plus 11 plus 3 y squared minus 8")[$7 y^(2) + 2 y + 11 + 3 y^(2) − 8$] #solutionbox[ 10#emph[y]#super[2] + 2#emph[y] + 3 ] #math.equation(block: true, alt: "13 x squared minus x plus 6 plus 5 x squared plus 9 x")[$13 x^(2) − x + 6 + 5 x^(2) + 9 x$] #strong[Translate English Phrases to Algebraic Expressions] In the following exercises, translate the following phrases into algebraic expressions. the difference of #math.equation(block: false, alt: "x")[$x$] and #math.equation(block: false, alt: "6")[$6$] #solutionbox[ #emph[x] − 6 ] the sum of #math.equation(block: false, alt: "10")[$10$] and twice #math.equation(block: false, alt: "a")[$a$] the product of #math.equation(block: false, alt: "3 n")[$3 n$] and #math.equation(block: false, alt: "9")[$9$] #solutionbox[ 3#emph[n] ⋅ 9 ] the quotient of #math.equation(block: false, alt: "s")[$s$] and #math.equation(block: false, alt: "4")[$4$] #math.equation(block: false, alt: "5")[$5$] times the sum of #math.equation(block: false, alt: "y")[$y$] and #math.equation(block: false, alt: "1")[$1$] #solutionbox[ 5(#emph[y] + 1) ] #math.equation(block: false, alt: "10")[$10$] less than the product of #math.equation(block: false, alt: "5")[$5$] and #math.equation(block: false, alt: "z")[$z$] Jack bought a sandwich and a coffee. The cost of the sandwich was #math.equation(block: false, alt: "$3")[$"$3"$] more than the cost of the coffee. Call the cost of the coffee #math.equation(block: false, alt: "c .")[$c .$] Write an expression for the cost of the sandwich. #solutionbox[ #emph[c] + 3 ] The number of poetry books on Brianna’s bookshelf is #math.equation(block: false, alt: "5")[$5$] less than twice the number of novels. Call the number of novels #math.equation(block: false, alt: "n .")[$n .$] Write an expression for the number of poetry books. ==== Solve Equations Using the Subtraction and Addition Properties of Equality #strong[Determine Whether a Number is a Solution of an Equation] In the following exercises, determine whether each number is a solution to the equation. #math.equation(block: true, alt: "y plus 16 equals 40")[$y + 16 = 40$] + ⓐ #math.equation(block: false, alt: "24")[$24$] + ⓑ #math.equation(block: false, alt: "56")[$56$] #solutionbox[ + ⓐ yes + ⓑ no ] #math.equation(block: true, alt: "d minus 6 equals 21")[$d − 6 = 21$] + ⓐ #math.equation(block: false, alt: "15")[$15$] + ⓑ #math.equation(block: false, alt: "27")[$27$] #math.equation(block: true, alt: "4 n plus 12 equals 36")[$4 n + 12 = 36$] + ⓐ #math.equation(block: false, alt: "6")[$6$] + ⓑ #math.equation(block: false, alt: "12")[$12$] #solutionbox[ + ⓐ yes + ⓑ no ] #math.equation(block: true, alt: "20 q minus 10 equals 70")[$20 q − 10 = 70$] + ⓐ #math.equation(block: false, alt: "3")[$3$] + ⓑ #math.equation(block: false, alt: "4")[$4$] #math.equation(block: true, alt: "15 x minus 5 equals 10 x plus 45")[$15 x − 5 = 10 x + 45$] + ⓐ #math.equation(block: false, alt: "2")[$2$] + ⓑ #math.equation(block: false, alt: "10")[$10$] #solutionbox[ + ⓐ no + ⓑ yes ] #math.equation(block: true, alt: "22 p minus 6 equals 18 p plus 86")[$22 p − 6 = 18 p + 86$] + ⓐ #math.equation(block: false, alt: "4")[$4$] + ⓑ #math.equation(block: false, alt: "23")[$23$] #strong[Model the Subtraction Property of Equality] In the following exercises, write the equation modeled by the envelopes and counters and then solve the equation using the subtraction property of equality. #figure(figph[This image is divided into two parts: the first part shows an envelope and 3 blue counters and the next to it, the second part shows five counters.], alt: "This image is divided into two parts: the first part shows an envelope and 3 blue counters and the next to it, the second part shows five counters.", caption: none) #solutionbox[ #emph[x] + 3 = 5; #emph[x] = 2 ] #figure(figph[This image is divided into two parts: the first part shows an envelope and 4 blue counters and next to it, the second part shows 9 counters.], alt: "This image is divided into two parts: the first part shows an envelope and 4 blue counters and next to it, the second part shows 9 counters.", caption: none) #strong[Solve Equations using the Subtraction Property of Equality] In the following exercises, solve each equation using the subtraction property of equality. #math.equation(block: true, alt: "c plus 8 equals 14")[$c + 8 = 14$] #solutionbox[ #emph[c] = 6 ] #math.equation(block: true, alt: "v plus 8 equals 150")[$v + 8 = 150$] #math.equation(block: true, alt: "23 equals x plus 12")[$23 = x + 12$] #solutionbox[ #emph[x] = 11 ] #math.equation(block: true, alt: "376 equals n plus 265")[$376 = n + 265$] #strong[Solve Equations using the Addition Property of Equality] In the following exercises, solve each equation using the addition property of equality. #math.equation(block: true, alt: "y minus 7 equals 16")[$y − 7 = 16$] #solutionbox[ #emph[y] = 23 ] #math.equation(block: true, alt: "k minus 42 equals 113")[$k − 42 = 113$] #math.equation(block: true, alt: "19 equals p minus 15")[$19 = p − 15$] #solutionbox[ #emph[p] = 34 ] #math.equation(block: true, alt: "501 equals u minus 399")[$501 = u − 399$] #strong[Translate English Sentences to Algebraic Equations] In the following exercises, translate each English sentence into an algebraic equation. The sum of #math.equation(block: false, alt: "7")[$7$] and #math.equation(block: false, alt: "33")[$33$] is equal to #math.equation(block: false, alt: "40 .")[$40 .$] #solutionbox[ 7 + 33 = 40 ] The difference of #math.equation(block: false, alt: "15")[$15$] and #math.equation(block: false, alt: "3")[$3$] is equal to #math.equation(block: false, alt: "12 .")[$12 .$] The product of #math.equation(block: false, alt: "4")[$4$] and #math.equation(block: false, alt: "8")[$8$] is equal to #math.equation(block: false, alt: "32 .")[$32 .$] #solutionbox[ 4 ⋅ 8 = 32 ] The quotient of #math.equation(block: false, alt: "63")[$63$] and #math.equation(block: false, alt: "9")[$9$] is equal to #math.equation(block: false, alt: "7 .")[$7 .$] Twice the difference of #math.equation(block: false, alt: "n")[$n$] and #math.equation(block: false, alt: "3")[$3$] gives #math.equation(block: false, alt: "76 .")[$76 .$] #solutionbox[ 2(#emph[n] − 3) = 76 ] The sum of five times #math.equation(block: false, alt: "y")[$y$] and #math.equation(block: false, alt: "4")[$4$] is #math.equation(block: false, alt: "89 .")[$89 .$] #strong[Translate to an Equation and Solve] In the following exercises, translate each English sentence into an algebraic equation and then solve it. Eight more than #math.equation(block: false, alt: "x")[$x$] is equal to #math.equation(block: false, alt: "35 .")[$35 .$] #solutionbox[ #emph[x] + 8 = 35; #emph[x] = 27 ] #math.equation(block: false, alt: "21")[$21$] less than #math.equation(block: false, alt: "a")[$a$] is #math.equation(block: false, alt: "11 .")[$11 .$] The difference of #math.equation(block: false, alt: "q")[$q$] and #math.equation(block: false, alt: "18")[$18$] is #math.equation(block: false, alt: "57 .")[$57 .$] #solutionbox[ #emph[q] − 18 = 57; #emph[q] = 75 ] The sum of #math.equation(block: false, alt: "m")[$m$] and #math.equation(block: false, alt: "125")[$125$] is #math.equation(block: false, alt: "240 .")[$240 .$] #strong[Mixed Practice] In the following exercises, solve each equation. #math.equation(block: true, alt: "h minus 15 equals 27")[$h − 15 = 27$] #solutionbox[ #emph[h] = 42 ] #math.equation(block: true, alt: "k minus 11 equals 34")[$k − 11 = 34$] #math.equation(block: true, alt: "z plus 52 equals 85")[$z + 52 = 85$] #solutionbox[ #emph[z] = 33 ] #math.equation(block: true, alt: "x plus 93 equals 114")[$x + 93 = 114$] #math.equation(block: true, alt: "27 equals q plus 19")[$27 = q + 19$] #solutionbox[ #emph[q] = 8 ] #math.equation(block: true, alt: "38 equals p plus 19")[$38 = p + 19$] #math.equation(block: true, alt: "31 equals v minus 25")[$31 = v − 25$] #solutionbox[ #emph[v] = 56 ] #math.equation(block: true, alt: "38 equals u minus 16")[$38 = u − 16$] ==== Find Multiples and Factors #strong[Identify Multiples of Numbers] In the following exercises, list all the multiples less than #math.equation(block: false, alt: "50")[$50$] for each of the following. #math.equation(block: true, alt: "3")[$3$] #solutionbox[ 3, 6, 9, 12, 15, 18, 21, 24, 27, 30, 33, 36, 39, 42, 45, 48 ] #math.equation(block: true, alt: "2")[$2$] #math.equation(block: true, alt: "8")[$8$] #solutionbox[ 8, 16, 24, 32, 40, 48 ] #math.equation(block: true, alt: "10")[$10$] #strong[Use Common Divisibility Tests] In the following exercises, using the divisibility tests, determine whether each number is divisible by #math.equation(block: false, alt: "2 , by 3 , by 5 , by 6 , and by 10 .")[$2 , "by" #h(0.2em) 3 , "by" #h(0.2em) 5 , "by" #h(0.2em) 6 , "and by" #h(0.2em) 10 .$] #math.equation(block: true, alt: "96")[$96$] #solutionbox[ 2, 3, 6 ] #math.equation(block: true, alt: "250")[$250$] #math.equation(block: true, alt: "420")[$420$] #solutionbox[ 2, 3, 5, 6, 10 ] #math.equation(block: true, alt: "625")[$625$] #strong[Find All the Factors of a Number] In the following exercises, find all the factors of each number. #math.equation(block: true, alt: "30")[$30$] #solutionbox[ 1, 2, 3, 5, 6, 10, 15, 30 ] #math.equation(block: true, alt: "70")[$70$] #math.equation(block: true, alt: "180")[$180$] #solutionbox[ 1, 2, 3, 4, 5, 6, 9, 10, 12, 15, 18, 20, 30, 36, 45, 60, 90, 180 ] #math.equation(block: true, alt: "378")[$378$] #strong[Identify Prime and Composite Numbers] In the following exercises, identify each number as prime or composite. #math.equation(block: true, alt: "19")[$19$] #solutionbox[ prime ] #math.equation(block: true, alt: "51")[$51$] #math.equation(block: true, alt: "121")[$121$] #solutionbox[ composite ] #math.equation(block: true, alt: "219")[$219$] ==== Prime Factorization and the Least Common Multiple #strong[Find the Prime Factorization of a Composite Number] In the following exercises, find the prime factorization of each number. #math.equation(block: true, alt: "84")[$84$] #solutionbox[ 2 ⋅ 2 ⋅ 3 ⋅ 7 ] #math.equation(block: true, alt: "165")[$165$] #math.equation(block: true, alt: "350")[$350$] #solutionbox[ 2 ⋅ 5 ⋅ 5 ⋅ 7 ] #math.equation(block: true, alt: "572")[$572$] #strong[Find the Least Common Multiple of Two Numbers] In the following exercises, find the least common multiple of each pair of numbers. #math.equation(block: true, alt: "9 , 15")[$9 , 15$] #solutionbox[ 45 ] #math.equation(block: true, alt: "12 , 20")[$12 , 20$] #math.equation(block: true, alt: "25 , 35")[$25 , 35$] #solutionbox[ 175 ] #math.equation(block: true, alt: "18 , 40")[$18 , 40$] ==== Everyday Math Describe how you have used two topics from The Language of Algebra chapter in your life outside of your math class during the past month. #solutionbox[ Answers will vary ] === Chapter Practice Test In the following exercises, translate from an algebraic equation to English phrases. #math.equation(block: true, alt: "6 times 4")[$6 ⋅ 4$] #math.equation(block: true, alt: "15 minus x")[$15 − x$] #solutionbox[ 15 minus #emph[x], the difference of fifteen and #emph[x]. ] In the following exercises, identify each as an expression or equation. #math.equation(block: true, alt: "5 times 8 plus 10")[$5 ⋅ 8 + 10$] #math.equation(block: true, alt: "x plus 6 equals 9")[$x + 6 = 9$] #solutionbox[ equation ] #math.equation(block: true, alt: "3 times 11 equals 33")[$3 ⋅ 11 = 33$] + ⓐ Write #math.equation(block: false, alt: "n times n times n times n times n times n")[$n ⋅ n ⋅ n ⋅ n ⋅ n ⋅ n$] in exponential form. + ⓑ Write #math.equation(block: false, alt: "3 to the power 5")[$3^(5)$] in expanded form and then simplify. #solutionbox[ + ⓐ #emph[n]#super[6] + ⓑ 3 ⋅ 3 ⋅ 3 ⋅ 3 ⋅ 3 = 243 ] In the following exercises, simplify, using the order of operations. #math.equation(block: true, alt: "4 plus 3 times 5")[$4 + 3 ⋅ 5$] #math.equation(block: true, alt: "open parenthesis 8 plus 1 close parenthesis times 4")[$( 8 + 1 ) ⋅ 4$] #solutionbox[ 36 ] #math.equation(block: true, alt: "1 plus 6 open parenthesis 3 minus 1 close parenthesis")[$1 + 6 ( 3 − 1 )$] #math.equation(block: true, alt: "open parenthesis 8 plus 4 close parenthesis divided by 3 plus 1")[$( 8 + 4 ) ÷ 3 + 1$] #solutionbox[ 5 ] #math.equation(block: true, alt: "open parenthesis 1 plus 4 close parenthesis squared")[$attach(( 1 + 4 ), t: 2)$] #math.equation(block: true, alt: "5 [ 2 plus 7 open parenthesis 9 minus 8 close parenthesis ]")[$5 [ 2 + 7 ( 9 − 8 ) ]$] #solutionbox[ 45 ] In the following exercises, evaluate each expression. #math.equation(block: true, alt: "8 x minus 3 when x equals 4")[$8 x − 3 #h(0.2em) "when" #h(0.2em) x = 4$] #math.equation(block: true, alt: "y cubed when y equals 5")[$y^(3) #h(0.2em) "when" #h(0.2em) y = 5$] #solutionbox[ 125 ] #math.equation(block: true, alt: "6 a minus 2 b when a equals 5 , b equals 7")[$6 a − 2 b #h(0.2em) "when" #h(0.2em) a = 5 , b = 7$] #math.equation(block: true, alt: "h w when h equals 12 , w equals 3")[$h w #h(0.2em) "when" #h(0.2em) h = 12 , w = 3$] #solutionbox[ 36 ] Simplify by combining like terms. + ⓐ #math.equation(block: false, alt: "6 x plus 8 x")[$6 x + 8 x$] + ⓑ #math.equation(block: false, alt: "9 m plus 10 plus m plus 3")[$9 m + 10 + m + 3$] In the following exercises, translate each phrase into an algebraic expression. #math.equation(block: false, alt: "5")[$5$] more than #math.equation(block: false, alt: "x")[$x$] #solutionbox[ #emph[x] + 5 ] the quotient of #math.equation(block: false, alt: "12")[$12$] and #math.equation(block: false, alt: "y")[$y$] three times the difference of #math.equation(block: false, alt: "a and b")[$a #h(0.2em) "and" #h(0.2em) b$] #solutionbox[ 3(#emph[a] − #emph[b]) ] Caroline has #math.equation(block: false, alt: "3")[$3$] fewer earrings on her left ear than on her right ear. Call the number of earrings on her right ear, #math.equation(block: false, alt: "r .")[$r .$] Write an expression for the number of earrings on her left ear. In the following exercises, solve each equation. #math.equation(block: true, alt: "n minus 6 equals 25")[$n − 6 = 25$] #solutionbox[ #emph[n] = 31 ] #math.equation(block: true, alt: "x plus 58 equals 71")[$x + 58 = 71$] In the following exercises, translate each English sentence into an algebraic equation and then solve it. #math.equation(block: false, alt: "15")[$15$] less than #math.equation(block: false, alt: "y")[$y$] is #math.equation(block: false, alt: "32 .")[$32 .$] #solutionbox[ #emph[y] − 15 = 32; #emph[y] = 47 ] the sum of #math.equation(block: false, alt: "a")[$a$] and #math.equation(block: false, alt: "129")[$129$] is #math.equation(block: false, alt: "164 .")[$164 .$] List all the multiples of #math.equation(block: false, alt: "4")[$4$] that are less than #math.equation(block: false, alt: "50 .")[$50 .$] #solutionbox[ 4, 8, 12, 16, 20, 24, 28, 32, 36, 40, 44, 48 ] Find all the factors of #math.equation(block: false, alt: "90 .")[$90 .$] Find the prime factorization of #math.equation(block: false, alt: "1080 .")[$1080 .$] #solutionbox[ 2#super[3] ⋅ 3#super[3] ⋅ 5 ] Find the LCM (Least Common Multiple) of #math.equation(block: false, alt: "24")[$24$] and #math.equation(block: false, alt: "40 .")[$40 .$]