Math 210 Answers and Comments

MATH 210 j ANSWERS & COMMENTS When comparing, note: Your problems may have different numbers, or be in different order. '9 1 ft = 12 in 2 ft = 24 in 10 ft = 129 in m ft = .... 1. Write an algebraic expression for the following: a. the number of inches in m feet 12m (inches) b. the cost in dollars of buying x CDs from a music club, if it costs $10 to join the club, then $15 for each CD purchased. 1 CD will cost $10 + $15 10 + 15x (dollars) 2 CDs cost $10 + $30 10 CDs cost $10 + $150.... c. the total value, in cents, of d dimes, n nickels and p “pennies”. The value of 4 dim es, 3 nickels and 2 pennies is 4@10 + 3@5 + 2 cents 10d + 5n + p (cents) '8 2. Multiply each of the following, expressing your final result without parentheses. a. ( a + b )2 = (a+b)(a+b) = (a+b)a + (a+b)b = a@a + b@a = a2 + 2ab = (a+b)a –(a+b)b b. ( a + b )( a – b ) = '8 + a@b + b@b + b a 2 + ab 2 – ( ab + b 2 ) = a2 – b2 3. Solve this problem algebraically. Give a “teacher’s solution”, identifying the variables, showing the work, and stating the conclusion. Problem: Sunny had three times as much money as Oscar. After Sunny spent $350 and Oscar earned $1000, Sunny had twice a much as Oscar. How much did they each have at the beginning? In the beginning: Oscar had x Sunny had 3x After: Oscar has x + $1000 Sunny has 3x – $350 AFTER that (8), Sunny had twice as much as Oscar.... 3x – $350 We solve: 3x – 350 x = 2 ( x + $1000 ) = 2x + 2000 = 2350 Oscar’s original amount of money was $2350. Sunny had three times as much, so that was $7050. Check: 7050/3 = 2350. So Oscar had 1/3 as much as Sunny. (U) After Sunny spent $350, she had 7050 – 350 = 6700 . After Oscar earned $1000, he had 2350 + 1000 = 3350. 2(3350) = 6700. (U) It all checks out to match the information given. U MATH 210 j ANSWERS & COMMENTS '6 4. Express 1015 ·1520 as a product of prime numbers (each prime may appear only once). 507 ·96 1015 ·1520 ))))) = 507 ·96 ))))))))) 2 7 2 6 (2@5)15 ·(3@5)20 (2@5 ) ·(3 ) 215 @515 ·320 @520 = ))))))))) 27 @ 514 ·312 = 27 @ 312 @514 )))))))) 1013 ·1520 50 · 9 ))))))))) 2 7 2 6 (2@5)13 ·(3@5)20 (2@5 ) ·(3 ) 213 @513 ·320 @520 = ))))))))) 27 @ 514 ·312 = 27 @ 312 @514 )))))))) )))))) = 7 6 '4 '16 15 20 35 2 @3 ·5 13 20 2 @3 ·5 33 215 320 535 = )) )) )) = 28 @ 38 @ 521 7 12 14 2 3 5 213 320 533 = )) )) )) = 26 @ 38 @ 519 27 312 514 5. List all the digits that can replace “d” (that can be placed in the blank) so that: a. 4 is a divisor of 123571d (123571_) 4*12 and 4*16, so d may be 2 or 6. b. 11 is a factor of 8070d2 (8070_2) Alternating sum is 8–0+7–0+d–2 = 13 + d , which is a multiple of 11 only if the digit d is 9. 6. In the statements below a, b and c represent whole numbers. Circle T if the statement is true, F if it is false. F Suppose a is 6 or 12 or 36 ! 4*6 & 6*6 but 24' *6 ! b. F T F If a is divisible by 2 & by 3, then a is divisible by 6. OK because 2 & 3 are relatively prime. c. F T F If a is any integer, then 0 is a multiple of a . 0 is on every list of multiples. @@@, –2, 0, 2,@@@ d. F T F The numbers 32 and 15 are relatively prime. GCF(32, 15) = 1 e. F T F 13 is a divisor of 2 @5 @7@13 @17 @3 + 26 . 13 clearly divides both terms. f. TF F 2@3 @5 @11@17 @43 + 24 is a multiple of 11. 11 divides the first term, but not 24. g. F T F If 5 | a and 5 | b then 5 | a+b . If a & b are both multiples of 5, then a+b is too. h. TF F If 5 '| a and 5 '| b then 5 '| a+b . See problem #9a. a. T F If 4* a and 6* a then 24* a . 4 2 '10 4 4 2 5 5 7. To test whether the numbers 211 & 221 are prime... a. we need to test for divisibility by certain primes. List all the primes we need to test. 2 3 5 7 11 13 There’s no need to test further, because 17 2 > N. If N is composite, at least one factor must be < %N & . b. Is 221 prime or composite? Explain or show how you know. 2' * 221 ( 2' *1 ) 3' * 221 ( 2+2+1=5) 5' * 221 ( 5' * 1) 7' * 221 ( 210 + 11 ) 11' * 221 ( 220 + 1 ) 13*221 ! 221 = 13@17 ; therefore 221 is composite. c. Is 211 prime or composite? Explain or show how you know. 2' * 211 ( 2' * 1, the last digit, so 2' * 210 + 1 ) 3' * 211 ( sum of digits = 2+1+1 = 4 and 3' *4 ) 5' * 211 ( 5' * 1, the last digit, so 5' * 210+1) 7' * 211 ( 7' * 210 + 1 ) 11' * 211 ( 11' * 220 – 9 ) 13' * 211 ( 13' * 260 – 49 ) None of the smaller prime factors divides 211; therefore 211 is prime. As stated in part a, if N is composite, it must have a prime factor less than %N & . So if 211 were composite, 211 would have a prime factor less than 14.something. But none of the prime numbers less than 17 divide 211, so 211 cannot be composite, therefore must be prime. MATH 210 j ANSWERS & COMMENTS ALL ANSWERS ON THIS PAGE MAY BE LEFT IN FACTORED FORM. '12 90 8. a. Find the Greatest Common Factor (GCF) of 90 and 84. 9 90 = 2 @ 32 @ 5 So GCF(90, 84) = 2 @ 3 84 = 2 2 @ 3 @ 7 A 3 2 2 2 3 2 3 2 2 84 10 3 2 4 5 2 21 2 3 7 2 b. Find GCF (2 @ 3 @ 7 @ 13 , 2 @ 5 @ 7 @13 , 2 @ 3 @ 5@13 ) 2 2 @30 @50 @70 @131 = 22 @13 or 52 Picking up the minimum power of each prime listed: 3 2 2 2 3 2 2 2 2 c. Find LCM (2 @ 3 @ 7 @ 13 , 2 @ 5 @ 7 @13 , 2 @ 3 @ 5@13 ) 23 @32 @53 @72 @132 or 74,529,000 Picking up the maximum power of each prime listed: '8 9. In each of the following, show the statement is false with a fully demonstrated counterexample. a. If 3' | a & 3' | b then 3 ' | (a + b) . The above claims: 3' | 4 & 3 ' | 5 so Let a = 4 b = 5 3 ' | (4 + 5) But 4+5 = 9 and 3*9. So the conclusion of this statement is false; thus the statement is false. b. If the sum of the digits of a number is divisible by 2, then the number is divisible by 2. The above statement says that since the sum of the digits in the number 121 is 4, which is divisible by 2, therefore the number 121 must be divisible by 2. This is nonsense, since 121 = 112 (prime factorization), and is clearly not a multiple of 2. '10 10. a. Use the Euclidean algorithm to find the GCF of 3150 and 306; b. then use the result from part a to find the LCM of 3150 and 306. Show the w ork (of course) ! 3150 – 3060 3150 = 10 @ 306 + 90 306 – 270 306 = 3 @ 90 + 36 90 = 2 @ 36 + 18 90 – 72 36 = 2 @ 18 + 0 !!!! So GCF ( 3150, 306 ) = 18 Since GCF ( a,b ) @ LCM ( a, b ) = a @ b for all whole numbers a and b, we know: a@b LCM ( a, b ) = ))))))))) GCF ( a, b ) 3150 @ 306 GCF(3150, 306) ))))))))))) LCM ( 3150, 306 ) = LCM ( 3150, 306 ) = 3150 @ 306 )))))))) 18 3150 @18 @17 18 = )))))))) = 53550 MATH 210 j ANSWERS & COMMENTS '6 11. Marie checks in at the office at 9:00 AM and every 72 minutes thereafter. Jean, from Qw ik Copy Service, makes his first pickup at 9:00 AM and returns every 45 minutes. Bot h businesses close at 5 PM. Is there any other time during the day (after 9:00 AM) w hen Marie and Jean w ill be at the office simultaneously? (If so, name the time.) Marie will be in the office at Jean will be in the office at 9:00 10:12 11:24 12:36 13:48 etc. 9:00 9:45 10:30 11:15 12:30 etc. To find when they will be in the office again simultaneously, we need the LCM(72, 45) 72 45 LCM = 23 @32 = 32 @5 3 = 2 @32 @5 = 4@9@10 = 360 Well that’s nice: 360 minutes = 6 hours. Marie & Jean will be in the office simultaneously again at 9:00 + 6 hours = 15:00 or 3 PM '3 12. How many factors has 8000? Make a table listing them all. 8000 6 8000 = 2 @5 3 So 8000 has (6+1) (3+1) = 7@4 = 28 factors. They are: '3 20 21 22 23 24 25 26 1 2 4 8 16 32 64 5 0 5 1 5 10 20 40 80 160 320 52 25 50 100 200 400 800 1600 53 125 250 500 1000 23 103 23 53 ² Factors of 8000 1000 2000 4000 8000 13. Write two composite numbers that are relatively prime. Answers will vary. Here are some possibilities: '3 8 14. If x and y and p and z are all w hole numbers, and 2 x 3 3 5 y 7 8 Find x and y and z. 4 12 12 24 and and and and 9 25 35 25 = 6 @1 5 2 7 z , By the Fundamental Theorem of Arithmetic, the whole numbers on each side, being equal, must have the same prime factorization. So we look at this through prime factors. 2X 33 5y 78 = 6@152 7Z break the 6 and 15 into prime factors... 2X 33 5y 78 = 2@3@32 52 7Z then equate the powers on each side 2X = 2 requires that x = 1 5y = 52 says y = 2 78 = 7Z says z = 8