Math 210, Home Work Section 23

HW Set 23 Section 5.5 P130 a) Find a common multiple of 15 and 12 The multiples of 15 are 15, 30, 45, 60,... 60 is also a multiple of 12 So 60 is a common multiple of 15 and 12 b) (4A, Practice IB, p27) Problem I FACTORS of 18: 1, 2, 3, 6, 9, 18 a) factors of 8: 1, 2, 4,8 e) factors of 75: Problem 4 75 ,25 ,15 ,5 ,3 را b) factors of 15: 1,3,5,15 f) factors of 98: c) factors of 20: 1, 2, 4, 5, 10, 20 d) factors of 50: 1, 2, 5, 10, 25, 50 1, 2, 7, 14, 49, 98 Problem 5 (find a common factor different from 1) a) 3 is a common factor of 15 and 6 b) 2 and 4 are common factors of 12 and 16 c) 3 is a common factor of 15 and 18 Problem 6 a) multiples of 2: 2, 4, 6, 8,... b) multiples of 6: 6, 12, 18, 24,... c) multiples of 8: 8, 16, 24, 32,... Problem 7 a) The multiples of 4 are 4, 8, 12,... 12 is also a multiple of 3 so 12 is a common multiple of 3 and 4 b) The multiples of 5 are 5, 10, 15, 20, ... 20 is also a multiple of 4 So 20 is a common multiple of 4 and 5 c) 12 is a common multiple of 4 and 6 3) Prove that if p is prime, then GCF(p,a) = 1 unless a is a multiple of p The only divisors of a prime p are 1 and p If pla (that is, a is a multiple of p), then p is the greatest divisor p and a have in common So GCF(p,a) = P in this case. If pta, then the only divisor p and a have in common is I So GCF(p,a) = 1 in this case. 4) a) 28 = 22.7 633.7 So GCF(28,63)= b) 104 = 23-13 132 = 22-3-11 So GCF (104, 132)=2² = 4 24 = 23.3 56 = 23.7 180 = 22.32.5 GCF (24,56, 180)=2²=4 5) EUCLID'S ALGORITHM a) GCF (91, 52) = GCF (52,39) = GCF (39, 13) = ECF (13,0) = 13 3 52/91 39)52 13/39 52 39 (13) 39 39 0 LAST NONZERO REMAINDER turns out to be the GCF b) GCF (812,336) = GCF (336, 140) = GCF (140,56) = GCF (56,28) 2 2 2 2 6) 812 140) 336 56) 140 28/56 336) 672 280 56 112 28 56 = GCF (28,0) = 28 140 C) GCF(2389485, 59675) = 6CF (59675, 2485) = GCF (2485, 35) = GCF (35, 0) = (35) 6) a) 32 = 25 So LCM (32, 1024) = 2 10 2=1024 1024=210 (This makes sense since 1024 is a multiple of namely 1024=32-32 32; b) 24 = 23.3 120=23-3.5 1056=25.3.11 } 25.3.5.11 So LCM (24, 120, 1056) = 2.3. = 5280 a) (5A, p38-39) Common multiples are being used as "common" denominators when adding / subtracting fractions. b) 84 = 2^237 147 = 37^2 } So LCM (84, 147) = 2^23*7^2=588 Now 14/84 + 3/147 + 14/588 + 20/588 = 34/588 (which reduces to 17/244) mult top/bottom by 7 to write as a fraction with denom 588. mult top/bottom by 4 to write as a fraction with denom 588 Find LCM (192, 320) 192 = 2^53 320 = 2^65 } LCM (192, 320) = 2^635 = 960 We need 960÷192=5 revolutions of the first gear to realign the mark (and 960÷320=3 revolutions of the second gear) Gear1 192 teeth Gear2 320 teeth we must find a number < 1000 that leaves remainder 1 upon division by 2,3,4,5,6,7 or 8 HINT: The number must follow one that is divisible by 2,3,4,5,6,7,8; that is, to find the number, ADD 1 to LCM (2,3,4,5,6,7,8) 2=2' 3=3' 4=2^2 5=5' 6=2^1*3^1 7=7' 8=2^3   LCM = 2^335*7! = 840 840 is The smallest multiple these numbers have in common; 840 is The smallest number divisible by EVERY number given The number we need is 840+1 or 841