Math 201: Homework 7 Solutions


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1 Math 201: Homework 7 Solutions 1. ( 5.2 #4) (a) The factors of 72 are 1, 2, 3, 4, 6, 8, 9, 12, 18, 24, 36, and 72. (b) The factors of 81 are 1, 3, 9, 27, and 81. (c) The factors of 62 are 1, 2, 31, and 62. (d) The factors of 54 are 1, 2, 3, 6, 9, 18, 27, and 54. (e) The factors of 326 are 1, 2, 163, and 326. (f) The factors of 242 are 1, 2, 11, 22, 121, and ( 5.2 #10) We begin by drawing out the list of numbers as given. Since the square root of 150 is about 12, we know we only need to cross out multiples of primes through 12. So we cross out all multiples of the primes 2, 3, 5, 7, and 11, and we are left with all the prime numbers through 150: We found the new primes 127, 131, 137, 139, and ( 5.2 #12) (a) The prime factorization of 16 is = 2 4. (b) The prime factorization of 27 is = 3 3. (c) The prime factorization of 52 is = (d) The prime factorization of 75 is = (e) The prime factorization of 112 is =
2 (f) The prime factorization of 125 is = ( 5.2 #14) We can find the top number using only the given numbers. Since the Fundamental Theorem of Arithmetic guarantees that every number has a unique prime factorization, and the factor tree we are given tells us the prime factorization of the top number, then we need only multiply the numbers up the tree to find the omitted numbers ( 5.2 #16) (a) The square root of 16 is 4, since 4 2 = 16. (b) The square root of 81 is 9, since 9 2 = 81. (c) The square root of 100 is 10, since 10 2 = 100. (d) The square root of 144 is 12, since 12 2 = ( 5.2 #18) (a) 417 is composite since, for instance, 3 divides it (note that 3 ( )). (b) 729 is composite since, for instance, 3 divides it (note that 3 ( )). (c) 1571 is prime. Note that we need only check primes through 40, since (d) 4587 is composite since, for instance 3 divides it (note that 3 ( )). (e) 35, 721 is composite since, for instance, 3 divides it (note that 3 ( )). (f) 87, 451 is composite since, for instance, 7 divides it (you can do our little splitting trick a few times, or you can just solve 7 ) 87, 451). 7. ( 5.2 #22) (a) The prime factorization of 432 is = (b) The prime factorization of 1568 is = (c) The prime factorization of 2079 is = (d) The prime factorization of 6318 is = (e) The prime factorization of 6048 is = (f) The prime factorization of 8281 is =
3 8. ( 5.2 #24) (a) False. The number 2 is even, but it is also prime. (b) False. 6 is not prime. The prime factorization of 60 would be (c) False. Both 2 and 5 are prime, and 2 5 = 10 is even. (d) True. Since the square root of 1393 is about 37, then we need only check for prime factors less than or equal to ( 5.2 #26) This question is equivalent to asking for all the factors of 24, which are 1, 2, 3, 4, 6, 8, 12, and 24. That is, you could divide the class up into 24 groups of 1 person each, 12 groups of 2 people each, 8 groups of 3 people each, 6 groups of 4 people each, 4 groups of 6 people each, 3 groups of 8 people each, 2 groups of 12 people each, or 1 big group of 24 people. 10. ( 5.2 #40) (a) We first write each given base in its prime factorization, then combine using properties of exponents = (2 2 3) 9 ( ) 15 (13 2 ) 5 = (2 2 ) (2 2 ) 15 (3 2 ) 15 (13 2 ) 5 Power of a Product Property = Power of a Power Property = Product of Powers Property (b) We first write each given base in its prime factorization, then combine using properties of exponents. 11. ( 5.3 #4) = (2 4 ) 13 (2 7) 7 (3 4 ) 14 = (2 4 ) (3 4 ) 14 Power of a Product Property = Power of a Power Property = Product of Powers Property (a) The prime factorizations of 9 and of 15 are 9 = 3 2 and 15 = 3 5 The only common factor is a single 3. Hence, GCF(9, 15) = 3. (b) The prime factorizations of 13 and of 20 are 13 = 13 and 20 = These have no common prime factor. Hence, GCF(13, 20) = 1. 3
4 (c) The prime factorizations of 6 and of 8 are 6 = 2 3 and 8 = 2 3 The only common factor is a single 2. Hence, GCF(6, 8) = 2. (d) The prime factorizations of 21 and of 63 are 21 = 3 7 and 63 = The common factors are a 3 and a 7. Hence, GCF(21, 63) = 3 7 = 21. (e) The prime factorizations of 36 and of 54 are 36 = and 54 = The common factors are a single 2 and two 3s. Hence, GCF(36, 54) = = 18. (f) The prime factorizations of 100 and of 360 are 12. ( 5.3 #6) (a) (b) 100 = and 360 = The common factors are two 2s and a single 5. Hence, GCF(100, 360) = = 20. GCF(6, 26) = GCF(6, 26 6) = GCF(6, 20) = GCF(6, 20 6) = GCF(6, 14) = GCF(6, 14 6) = GCF(6, 8) = GCF(6, 8 6) = GCF(6, 2) = 2 GCF(8, 28) = GCF(8, 28 8) = GCF(8, 20) = GCF(8, 20 8) = GCF(8, 12) = GCF(8, 12 8) = GCF(8, 4) = 4 4
5 (c) GCF(40, 56) = GCF(40, 56 40) = GCF(40, 16) = GCF(40 16, 16) = GCF(24, 16) = GCF(24 16, 16) = GCF(8, 16) = 8 (d) GCF(35, 42) = GCF(35, 42 35) = GCF(35, 7) = 7 (e) GCF(34, 85) = GCF(34, 85 34) = GCF(34, 51) = GCF(34, 51 34) = GCF(34, 17) = 17 (f) GCF(32, 55) = GCF(32, 55 32) = GCF(32, 23) = GCF(32 23, 23) = GCF(9, 23) = GCF(9, 23 9) = GCF(9, 14) = GCF(9, 14 9) = GCF(9, 5) = GCF(9 5, 5) = GCF(4, 5) = GCF(4, 5 4) = GCF(4, 1) = 1 5
6 13. ( 5.3 #8) (a) 2R48 84 ) 216 1R36 48 ) 84 1R12 36 ) 48 3R0 12 ) 36 So GCF(84, 216) = 12. (b) 1R ) 192 1R48 72 ) 120 1R24 48 ) 72 2R0 24 ) 48 (c) (d) (e) (f) (g) (h) So GCF(120, 192) = 24. So GCF(63, 84) = 21. 1R42 98 ) 140 So GCF(98, 140) = 14. So GCF(45, 90) = 45. So GCF(160, 800) = 160. So GCF(56, 588) = 28. 1R ) 935 So GCF(544, 935) = 17. 1R21 63 ) 84 10R28 56 ) 588 1R68 85 ) R14 42 ) 98 2R0 45 ) 90 5R0 160 ) 800 1R ) 544 3R0 21 ) 63 2R0 28 ) 56 1R17 68 ) 85 3R0 14 ) 42 2R ) 391 4R0 17 ) 68
7 14. ( 5.3 #10) (a) The prime factorizations of 3 and of 8 are 3 = 3 and 8 = 2 3 There are no common factors, so the diagram looks like Factors of 3 Factors of Hence, LCM(3, 8) = 3 (2 2 2) = 24. (b) The prime factorizations of 9 and of 12 are 9 = 3 2 and 12 = These only share a factor of 3, so the diagram looks like Factors of 9 Factors of Hence, LCM(9, 12) = 3 3 (2 2) = 36. (c) The prime factorizations of 12 and of 36 are 12 = and 36 = These share two 2s and a 3, so the diagram looks like Factors of 12 Factors of Hence, LCM(12, 36) = (2 2 3) 3 = 36. (d) The prime factorizations of 2 and of 9 are 2 = 2 and 9 = 3 2 7
8 There are no common factors, so the diagram looks like Factors of 2 Factors of Hence, LCM(2, 9) = 2 (3 3) = 18. (e) The prime factorizations of 24 and of 45 are 24 = and 45 = These only share the factor 3, so the diagram looks like Factors of 24 Factors of Hence, LCM(24, 45) = (2 2 2) 3 (3 5) = 360. (f) The prime factorizations of 15 and of 25 are 15 = 3 5 and 25 = 5 2 These share only a factor of 5, so the diagram looks like Factors of 15 Factors of Hence, LCM(15, 25) = = ( 5.3 #12) (a) So LCM(3, 6) = 6. 8
9 (b) (c) So LCM(4, 5) = So LCM(2, 7) = ( 5.3 #14) (a) The prime factorizations of 15 and of 20 are 15 = 3 5 and 20 = These share only a factor of 5, so the diagram looks like Factors of 15 Factors of Hence, GCF(15, 20) = 5, and LCM(15, 20) = 3 5 (2 2) = 60. (b) The prime factorizations of 50 and of 100 are 50 = and 100 = These share one factor of 2 and two factors of 5, so the diagram looks like Factors of 50 Factors of Hence, GCF(50, 100) = = 50, and LCM(50, 100) = (2 5 5) 2 =
10 (c) The prime factorizations of 24 and of 30 are 24 = and 30 = These share a factor of 2 and a factor of 3, so the diagram looks like Factors of 24 Factors of Hence, GCF(24, 30) = 2 3 = 6, and LCM(24, 30) = (2 2) (2 3) 5 = ( 5.3 #20) If the gift bags are to be identical, then you must use the same number of candles and gift cards in each bag. This is equivalent to asking for GCF(18, 24) = 6. Hence, the largest number of gift bags you can make is 6, where each contains 3 candles and 4 gift cards. 18. ( 5.3 #22) (a) The prime factorizations of 12, 18, and 24 are 12 = and 18 = and 24 = The only factors that are common to all three numbers are a single 2 and a single 3. Hence, the greatest common factor of 12, 18, and 24 is 2 3 = 6. (b) The prime factorizations of 24, 36, and 60 are 24 = and 36 = and 60 = The only factors that are common to all three numbers are two 2s and a single 3. Hence, the greatest common factor of 24, 36, and 60 is = 12. (c) The prime factorizations of 26, 52, and 78 are 19. ( 5.3 #24) 26 = 2 13 and 52 = and 78 = The only factors that are common to all three numbers are a single 2 and a single 13. Hence, the greatest common factor of 26, 52, and 78 is 2 13 = 26. (a) We ll compute the least common multiple of 4, 6, and 12 by looking at their multiples and identifying the first common one. The multiples of 4 are 4, 8, 12, 16, 20, 24,.... The multiples of 6 are 6, 12, 18, 24, 30, 36,.... The multiples of 12 are 12, 24, 36, 48, 60, 72,.... The first common multiple among these is 12, so this is our answer. 10
11 (b) We ll compute the least common multiple of 5, 16, and 20 by looking at their multiples and identifying the first common one. The multiples of 5 are 5, 10, 15, 20, 25, 30, 35, 40, 45, 50, 55, 60, 65, 70, 75, 80, The multiples of 16 are 16, 32, 48, 64, 80, 96,.... The multiples of 20 are 20, 40, 60, 80, 100, 120,.... The first common multiple among these is 80, so this is our answer. (c) We ll compute the least common multiple of 24, 36, and 48 by looking at their multiples and identifying the first common one. The multiples of 24 are 24, 48, 72, 96, 120, 144,.... The multiples of 36 are 36, 72, 108, 144, 180,.... The multiples of 48 are 48, 96, 144, 192, 240,.... The first common multiple among these is 144, so this is our answer. 20. ( 5.3 #32) We would like to know how far each of the runners will have run when they pass the starting line at the same time. We first find LCM(75, 90), which will tell us how much time has elapsed when they pass the starting line at the same time. The prime factorizations of 75 and 90 are 75 = and 90 = These share a factor of 3 and a factor of 5, so the diagram looks like Factors of 75 Factors of So LCM(75, 90) = 5 (3 5) (2 3) = 450. That is, 450 seconds after starting, the runners will pass each other at the starting line. The first runner will have run = 6 laps around the track, and the second will have run = 5 laps around the track. 11
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