Mathematical Ideas 13th Edition By Miller – Test Bank

Complete Test Bank With Answers

 

 

Sample Questions Posted Below

 

MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question.

Determine whether the statement is true or false.

1) If n is a natural number and 10|n, then 20|n.

A) True B) False

Answer: B

2) If a natural number is divisible by 3, then it must also be divisible by 15.

A) True B) False

Answer: B

3) If a natural number is divisible by 7 and 2, then it must also be divisible by 14.

A) True B) False

Answer: A

4) A prime number may have one or two different natural number factors but may not have more than two

different natural number factors.

A) True B) False

Answer: B

5) Every composite number is divisible by 2.

A) True B) False

Answer: B

6) All prime numbers are odd.

A) True B) False

Answer: B

7) Any composite number in the range 2 to n must be a multiple of some prime number less than or equal to A) True B) False

Answer: A

8) There are 35 prime numbers smaller than 150.

A) True B) False

Answer: A

9) If a number is divisible by both 3 and 9 then it is divisible by 27.

A) True B) False

Answer: B

Find all natural number factors of the number.

10) 42

A) 1, 2, 3, 7, 6, 14, 28, 42 C) 7, 6, 14, 42 Answer: B

B) 1, 2, 3, 7, 6, 14, 21, 42

D) 1, 7, 42

11) 125

A) 1, 5, 25, 125 Answer: A

B) 5, 62, 125 C) 5, 25, 125 D) 1, 5, 25

1

n.12) 110

A) 2, 5, 10, 11, 55, 110 C) 1, 2, 5, 10, 11, 22, 110 Answer: B

13) 676

A) 1, 2, 4, 13, 26, 39, 52, 169, 338, 676 C) 1, 2, 3, 4, 13, 26, 169, 338, 676 Answer: B

B) 1, 2, 5, 10, 11, 22, 55, 110

D) 1, 2, 4, 5, 10, 11, 22, 55, 110

B) 1, 2, 4, 13, 26, 52, 169, 338, 676

D) 1, 2, 13, 169, 676

14) 13

A) 1, 2, 3, 4, 6, 13 Answer: D

B) 1, 2, 6, 13 C) 1, 2, 4, 4, 6, 13 D) 1, 13

Use divisibility tests to decide whether the first number is divisible by the second.

15) 281,032; 4

A) Yes B) No

Answer: A

16) 772,047; 4

Answer: B

A) Yes B) No

17) 793,542; 3

Answer: A

A) Yes B) No

18) 788,913; 9

Answer: A

A) Yes B) No

19) 516,379; 9

Answer: B

A) Yes B) No

20) 6,955,200; 7

[Note that a divisibility test for 7 is as follows:

Double the last digit of the number and subtract this value from the original number with the last digit omitted.

Repeat this process as many times as necessary until the number obtained can easily be divided by 7. If the final

number obtained is divisible by 7, then so is the original number. If the final number obtained is not divisible by

7, then neither is the original number.]

A) Yes B) No

Answer: A

21) 968,461; 5

Answer: B

A) Yes B) No

222) 203,588; 7

Answer: A

A) Yes B) No

23) 324,841; 10

Answer: B

A) Yes B) No

24) 139,243,105; 11

[Note that a divisibility test for 11 is as follows:

Starting at the left of the number, add together every other digit. Add together the remaining digits. Subtract the

smaller of the two sums from the larger. If the final number obtained is divisible by 11, then so is the original

number. If the final number obtained is not divisible by 11, then neither is the original number.]

A) Yes B) No

Answer: B

Give the prime factorization of the number. Use exponents when possible.

25) 12

A) 22 ∙ 3 Answer: A

B) 4 ∙ 3 C) 32 D) 4 ∙ 2

26) 252

A) 23 ∙ 32 ∙ 7 Answer: B

B) 22 ∙ 32 ∙ 7 C) 34 ∙ 7 D) 24 ∙ 7

27) 504

A) 24 ∙ 3 ∙ 7 Answer: C

B) 23 ∙ 33 ∙ 7 C) 23 ∙ 32 ∙ 7 D) 2 ∙ 34 ∙ 7

28) 30

A) 32 ∙ 2 Answer: B

B) 2 ∙ 3 ∙ 5 C) 6 ∙ 5 D) 22 ∙ 5

29) 126

A) 2 ∙ 3 ∙ 7 Answer: C

B) 22 ∙ 32 ∙ 7 C) 2 ∙ 32 ∙ 7 D) 14 ∙ 32

30) 69

A) 32 Answer: D

B) 3 ∙ 21 C) 32 ∙ 23 D) 3 ∙ 23

31) 2013

A) 33 ∙ 61 Answer: C

B) 112 ∙ 61 C) 3 ∙ 11 ∙ 61 D) 32 ∙ 61

32) 8775

A) 33 ∙ 52 ∙ 13 B) 54 ∙ 13 Answer: A

C) 32 ∙ 53 ∙ 13 D) 34 ∙ 13

3SHORT ANSWER. Write the word or phrase that best completes each statement or answers the question.

Determine all values for the digit x that make the first number divisible by the second number. If none exist, so state.

33) 43×1 is divisible by 3.

Answer: 1, 4, 7

34) 43×2 is divisible by 9.

Answer: 0, 9

35) 43×6 is divisible by 4.

Answer: 1, 3, 5, 7, 9

36) 533x is divisible by 11.

Answer: 5

37) 214,21x is divisible by 11.

Answer: 4

38) 74,3×2 is divisible by 6.

Answer: 2, 5, 8

39) 4×21 is divisible by 4.

Answer: There are no values for x that make 4×21 divisible by 4.

40) 414,3×2 is divisible by 8 but not 16.

Answer: 1, 9

41) 417,30x is divisible by 8 but not 16.

Answer: There are no values for x that make 417,30x divisible by 8 but not 16.

42) 43×0 is divisible by 6.

Answer: 2, 5, 8

MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question.

Find the number of divisors of the number.

43) 60

A) 16 B) 14 C) 12 D) 10

Answer: C

44) 54

A) 9 B) 6 C) 8 D) 12

Answer: C

45) 360

A) 18 B) 24 C) 12 D) 36

Answer: B

446) 400

A) 12 B) 15 C) 8 D) 16

Answer: B

47) 25∙ 36∙ 52

Answer: D

A) 60 B) 120 C) 90 D) 126

Determine whether the statement is true or false.

48) If n is a prime number, then the Mersenne number 2n

– 1 is also a prime number.

A) True B) False

Answer: B

49) If n is a composite number, then the Mersenne number 2n

A) True B) False

Answer: A

– 1 is also a composite number.

50) The number 213

Answer: A

– 1 is an example of a Mersenne prime.

A) True B) False

51) Euler’s formula n2

Answer: B

– n + 41 generates primes for n up to 46 and fails at n = 47.

A) True B) False

52) Escott’s formula n2

Answer: A

– 79n + 1601 first fails at n = 80.

A) True B) False

53) Every natural number of the form 4k + 4 is prime.

A) True B) False

Answer: B

54) Every natural number of the form 4k + 5 is prime.

A) True B) False

Answer: B

Use the formula indicated to determine the prime number generated for the given value of n.

55) p = n2 – n + 41; n = 10

A) 131 B) 151 C) 49 D) 2491

Answer: A

56) p = n2 – 79n + 1,601; n = 52

A) 197 B) 2693 C) 5211 D) 8413

Answer: A

557) p = n2 – n + 41; n = 12

Answer: B

A) 2693 B) 173 C) 115 D) 197

58) p = n2 – n + 41; n = 28

Answer: A

A) 797 B) 771 C) 1395 D) 853

59) p = n2 – n + 41; n = 37

Answer: C

A) 1365 B) 4251 C) 1373 D) 2691

60) p = n2 – 79n + 1,601; n = 1

Answer: B

A) 41 B) 1523 C) 1681 D)–1521

61) p = n2 – 79n + 1,601; n = 39

A) 3001 B) 1523 C) 41 D) 6203

Answer: C

SHORT ANSWER. Write the word or phrase that best completes each statement or answers the question.

Solve the problem.

62) Given that the following statement is true, explain how you would use this fact to find two distinct factors of the

Mersenne number M14. Then calculate those two factors.

“If p is a prime factor of n, then 2p – 1 is a factor of the Mersenne number 2n

Answer: Two factors of M14. are M2 = 3 and M7 = 127.

– 1.”

63) Fermat conjectured that the formula 22n

+ 1 would always produce a prime for any whole number n. The sixth

Fermat number is 225

+ 1. Evaluate this number. In seeking possible prime factors of this Fermat number, what

is the largest potential prime factor that one would have to try?

Answer: Sixth Fermat number = 4,294,967,297.

Largest potential prime factor that one would have to try = 65,521.

64) Euler conjectured that the formula n2

– n + 41 would generate prime numbers for whole numbers n. The

formula generates prime numbers for n up to 40 and fails at n = 41.

Evaluate Euler’s formula for n = 42 and determine whether this is a prime number. If it is not a prime number,

give its prime factors.

Answer: 1763. Not a prime number. Factors are 41 and 43.

65) Escott used the formula n2

– 79n + 1601 to generate prime numbers for whole numbers n. The formula fails to

produce a prime for n = 80.

Evaluate Escott’s formula for n = 81 and determine whether this is a prime number. If it is not a prime number,

give its prime factors.

Answer: 1763. Not a prime number. Factors are 41 and 43.

6MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question.

Determine whether the statement is true or false.

66) The proper divisors of a natural number include all divisors of the number except 1.

A) True B) False

Answer: B

67) If 2n – 1 is prime, then 2n–1(2n – 1) is perfect.

A) True B) False

Answer: A

68) The number 27(28

Answer: B

– 1) is perfect.

A) True B) False

69) Not all perfect numbers end in 6 or 28.

A) True B) False

Answer: B

70) All prime numbers are also deficient numbers.

A) True B) False

Answer: A

71) All composite numbers are also abundant numbers.

A) True B) False

Answer: B

72) A natural number which is not deficient must be abundant.

A) True B) False

Answer: B

73) There is no largest prime number.

A) True B) False

Answer: A

74) Twin primes are prime numbers whose difference is a multiple of 3.

A) True B) False

Answer: B

75) 520 and 616 are amicable numbers.

A) True B) False

Answer: B

Determine whether the number is abundant or deficient.

76) 32

A) Abundant B) Deficient

Answer: B

777) 18

A) Deficient B) Abundant

Answer: B

78) 108

A) Deficient B) Abundant

Answer: B

79) 55

A) Deficient B) Abundant

Answer: A

80) 30

A) Abundant B) Deficient

Answer: A

81) 32

A) Deficient B) Abundant

Answer: A

Write the number as the sum of two primes. There may be more than one way to do this.

82) 12

A) 5 + 7 Answer: A

B) 6 + 6 C) 3 + 9 83) 28

A) 14 + 14 C) 3 + 25, 5 + 23, 7 + 21 Answer: B

B) 5 + 23, 11 + 17

D) 5 + 23, 13 + 15

84) 34

A) 21 + 13, 31 + 3, 29 + 5 C) 29 + 5, 31 + 3, 23 + 11, 17 + 17 Answer: C

B) 27 + 7, 29 + 5, 23 + 11

D) 15 + 19, 29 + 5, 23 + 11

85) 18

A) 9 + 9 C) 3 + 15, 5 + 13 , 7 + 11 Answer: B

B) 5 + 13, 7 + 11

D) 3 + 15, 7 + 11

86) 44

A) 3 + 41, 7 + 37, 11 + 33 C) 22 + 22 Answer: D

B) 3 + 41, 5 + 39, 13 + 31

D) 3 + 41, 7 + 37, 13 + 31

D) 2 + 10

Determine whether or not one or more pairs of twin primes exist between the pair of numbers given. If so, identify the

twin primes.

87) 16 and 27

A) 17, 19 and 21, 23 Answer: C

B) No C) 17, 19 D) 21, 23

888) 4 and 14

A) No Answer: B

B) 5, 7 and 11, 13 C) 5, 7 D) 5, 7 and 9, 11

89) 32 and 51

A) No Answer: C

B) 43, 47 C) 41, 43 D) 41, 42 and 47, 49

90) 100 and 106

Answer: D

A) 103, 105 B) No C) 101, 103, 105 D) 101, 103

91) 33 and 41

Answer: C

A) 37, 39 B) 35, 37 C) No D) 31, 33

92) 81 and 99

Answer: A

A) No B) 91, 93 C) 89, 91 D) 83, 85

For the following amicable pair, determine whether neither, one, or both of the members are happy, and whether the pair

is a happy amicable pair.

93) 66,928 and 66,992

A) one; not happy B) neither; not happy C) both; happy D) one; happy

Answer: B

94) 69,615 and 87,633

A) both; happy Answer: D

B) one; happy C) neither; not happy D) one; not happy

95) 35,390,008 and 39,259,592

A) one; happy Answer: D

B) one; not happy C) neither; not happy D) both; happy

SHORT ANSWER. Write the word or phrase that best completes each statement or answers the question.

Solve the problem.

96) Factorial primes are those of the form n! ± 1 for natural numbers n. Determine the two numbers that are

obtained when this formula is applied to the value n = 6, and state whether both, neither, or exactly one of them

is prime.

Answer: 6! – 1 = 719, 6! +1 = 721.

One of these numbers (719) is prime.

97) Primorial primes are those of the form p# ± 1.

Apply this formula to the value p = 11 to obtain two numbers and state whether both, neither, or exactly one of

these numbers is prime.

Answer: 11# – 1 = 2309 and 11# + 1 = 2311. Both are prime.

98) List all the abundant numbers between 20 and 30 (inclusive)

Answer: 20, 24, 30

999) Fermat proved that every odd prime number can be expressed as the difference of two squares in one and only

one way. Express each of the first 6 odd prime numbers as the difference of two squares.

Answer: 3 = 4 – 1

5 = 9 – 4

7 = 16 – 9

11 = 36 – 25

13 = 49 – 36

17 = 81 – 64

100) Show that the number 132 is not perfect.

Answer: The proper divisors of 132 are 1, 2, 3, 4, 6, 11, 12, 22, 33, 44, 66

1 + 2 + 3 + 4 + 6 + 11 + 12 + 22 + 33 + 44 + 66 = 204 ≠ 132.

Therefore 132 is not perfect

101) Show that the numbers 140 and 165 are not amicable.

Answer: The proper divisors of 140 are 1, 2, 4, 5, 7, 10, 14, 20, 28, 35, 70

1 + 2 + 4 + 5 + 7 + 10 + 14 + 20 + 28 + 35 + 70 = 196 ≠ 165.

Therefore 140 and 165 are not amicable.

MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question.

Determine whether the statement is true or false.

102) The greatest common factor of two relatively prime numbers is always 1.

A) True B) False

Answer: A

103) The greatest common factor of a group of natural numbers is the largest natural number that is a factor of all the

numbers in a group.

A) True B) False

Answer: A

104) Greatest common factors can not be found by using prime factorization, only by using the division method.

A) True B) False

Answer: B

105) Two natural, relatively prime numbers have at most one common factor.

A) True B) False

Answer: A

106) The set of all common multiples of two given whole numbers is finite.

A) True B) False

Answer: B

107) Two composite numbers can never be relatively prime.

A) True B) False

Answer: B

10108) The least common multiple of p and q cannot be larger than pq.

A) True B) False

Answer: A

109) If the least common multiple of p and q is smaller than pq, then p and q have a common factor other than 1.

A) True B) False

Answer: A

110) If a > b > c, then the greatest common factor of paqb

, pbqc, and pcqa is paqa

.

A) True B) False

Answer: B

111) If a > b > c, then the least common multiple of paqbrc and pcqarb is paqarb

.

A) True B) False

Answer: A

Find the greatest common factor of the numbers in the group.

112) 88, 567

A) 6 B) 1 C) 77 D) 44

Answer: B

113) 60, 72

Answer: B

A) 6 B) 12 C) 1 D) 2

114) 168, 42

Answer: C

A) 6 B) 21 C) 42 D) 14

115) 240, 900

Answer: B

A) 30 B) 60 C) 240 D) 180

116) 9, 27, 30

Answer: C

A) 6 B) 1 C) 3 D) 9

117) 54, 72, 126

Answer: A

A) 18 B) 2 C) 9 D) 36

118) 120, 1000, 2250

Answer: B

A) 30 B) 10 C) 20 D) 50

119) 1562, 450, 6750

Answer: C

A) 12 B) 4 C) 2 D) 5

11120) 12, 15, 24, 30

Answer: A

A) 3 B) 4 C) 12 D) 1

Find the least common multiple of the numbers in the group.

121) 168, 42

A) 42 B) 336 C) 504 D) 168

Answer: D

122) 125, 275

Answer: B

A) 34,375 B) 1375 C) 55 D) 275

123) 48, 162, 27

Answer: B

A) 324 B) 1296 C) 432 D) 648

124) 56, 96

Answer: B

A) 1344 B) 672 C) 168 D) 224

125) 60, 40, 70

Answer: D

A) 120 B) 420 C) 280 D) 840

126) 45, 56, 150

Answer: C

A) 1,050 B) 1260 C) 12,600 D) 2520

127) 43,378, 4715

Answer: D

A) 10 B) 943 C) 9430 D) 216,890

Answer the question.

128) Planets A, B, and C orbit a certain star once every 3, 7, and 18 months, respectively. If the three planets are now

in the same straight line, what is the smallest number of months that must pass before they line up again?

A) 54 months B) 378 months C) 126 months D) 28 months

Answer: C

129) Bob’s frog travels 7 inches per jump, Kim’s frog travels 8 inches and Jack’s frog travels 11 inches. If the three

frogs start off side–by–side, what is the smallest distance they must all travel before they are side–by–side

again?

A) 26 inches B) 616 inches C) 56 inches D) 88 inches

Answer: B

130) A brick layer is hired to build three walls of equal length. He has three lengths of brick, 4 inches, 12 inches, and

10 inches. He plans to build one wall out of each type. What is the shortest length of wall possible?

A) 120 inches B) 480 inches C) 26 inches D) 60 inches

Answer: D

12131) Three taxi cabs make a complete trip from downtown to the airport and back in 21, 39 and 91 minutes,

respectively. If all three cabs leave at the same time, what is the shortest time that must pass before they are all

together again?

A) 273 minutes B) 819 minutes C) 151 minutes D) 3549 minutes

Answer: A

132) Three clocks chime every 10 minutes, 22 minutes, and 55 minutes, respectively. If the three clocks chime

together, how much time must pass before they will chime together again?

A) 220 minutes B) 110 minutes C) 87 minutes D) 1210 minutes

Answer: B

133) Pete has 575 hot dogs and 475 hot dog buns. He wants to put the same number of hot dogs and hot dog buns on

each tray. What is the greatest number of trays Pete can use to accomplish this?

A) 5 B) 437 C) 25 D) 115

Answer: C

134) Two runners run around a circular track. The first runner completes each lap in 7 minutes. The second runner

completes each lap in 11 minutes. If they both start at the same place and the same time and go in the same

direction, after how many minutes will they meet again at the starting place?

A) 154 B) 18 C) 78 D) 77

Answer: D

135) Several different bus routes stop at the corner of Second St. and Lincoln Ave. A Wilkenson bus arrives every 21

minutes and a Harris Road bus arrives every 15 minutes. If both buses arrive at the stop at 5:07 AM, when will

they again arrive at the same time?

A) 10:22 AM B) 8:22 AM C) 6:52 AM D) 6:12 AM

Answer: C

136) At Northwest High School, there are 513 students in the Junior Class and 741 students in the Senior Class. To let

the juniors work with more experienced students, the teachers want to assign the students to committees with

the same number of juniors in each committee and the same number of seniors in each committee. (For example,

there might be 2 juniors and 3 seniors in every committee). What is the largest number of committees that can be

formed?

A) 19 committees B) 171 committees C) 3 committees D) 57 committees

Answer: D

Solve the problem relating to the Fibonacci sequence.

137) F25 = 75,025, F26 = 121,393

Find F27.

A) F27 = 167,761 B) F27 = 242,786 Answer: D

C) F27 = 46,368 D) F27 = 196,418

138) F28 = 317,811, F30 = 832,040

Find F29.

A) F29 = 196,418 Answer: C

B) F29 = 1,149,851 C) F29 = 514,229 D) F29 = 1,346,269

13139) If an 8–inch wide rectangle is to approach the golden ratio, what should its length be?

A) 10 in B) 5 in C) 12 in D) 13 in

Answer: D

140) If a 13–inch wide rectangle is to approach the golden ratio, what should its length be?

A) 21 in B) 8 in C) 34 in D) 20 in

Answer: A

141) List the first seven terms of the Fibonacci sequence.

A) 1, 2, 3, 5, 8, 13, 21 B) 1, 2, 4, 6, 10, 16, 26 Answer: D

C) 1, 1, 3, 4, 7, 11, 18 142) According to the Binet form, the nth Fibonacci number is given by

n

–

D) 1, 1, 2, 3, 5, 8, 13

1 + 5

1 – 5

n

2

2

5

Use this Binet form to find the 13th Fibonacci number.

A) 233 B) 235 C) 234 D) 232

Answer: A

Use inductive reasoning to find the next equation in the pattern.

143) 2 – 1 = 1

3 – 2 = 1

5 – 3 = 2

8 – 5 = 3

A) 10 – 5 = 5 B) 13 – 5 = 8 Answer: C

144) 1 + 1 = 2

3 + 2 = 5

8 + 5 = 13

21 + 13 = 34

A) 55 + 34 = 89 Answer: A

B) 21 + 34 = 55 145) 8 – 5 + 3 = 2 ∙ 3

13 – 8 + 5 = 2 ∙ 5

21 – 13 + 8 = 2 ∙ 8

A) 34 + 21 – 13 = 2 ∙ 21 C) 34 – 21 – 13 = 0 Answer: D

C) 13 – 8 = 5 C) 34 + 53 = 87 D) 11 – 3 = 8

D) 13 + 21 = 34

B) 34 – 21 + 8 = 2 ∙ 8

D) 34 – 21 + 13 = 2 ∙ 13

14146) 1 + 1 = 2

1 + 1 + 2 = 4

1 + 1 + 2 + 3 = 7

1 + 1 + 2 + 3 + 5 = 12

A) 1 + 1 + 2 + 3 + 5 + 8 + 13 = 20 C) 1 + 1 + 2 + 3 + 5 + 8 + 13 = 33 Answer: D

147) 22 – 12 = 3

32 – 12 = 8

52 – 22 = 21

A) 82 – 32 = 55 Answer: A

148) 12 + 12 = 2

12 + 22 = 5

22 + 32 = 13

A) 32 + 52 = 35 Answer: B

149) 1 = 2 – 1

1 + 1 = 3 – 1

1 + 1 + 2 = 5 – 1

A) 1 + 1 + 2 + 2 = 7 – 1 C) 1 + 1 + 2 + 3 + 5 = 13 – 1 Answer: D

150) 23 + 13 – 13 = 8

33 + 23 – 13 = 34

A) 53 + 33 – 23 = 144 C) 53 + 23 – 13 = 132 Answer: A

B) 82 – 22 = 60 B) 32 + 52 = 34 151) 12 = 1 ∙ 1

12 + 12 = 1 ∙ 2

12 + 12 + 22 = 2 ∙ 3

A) 12 + 22 + 32 = 2 ∙ 7 C) 12 + 12 + 22 + 32 = 3 ∙ 5 Answer: C

B) 1 + 2 + 3 + 5 + 8 = 20

D) 1 + 1 + 2 + 3 + 5 + 8 = 20

C) 82 – 52 = 39 D) 132 – 82 = 55

C) 32 + 82 = 73 D) 22 + 52 = 29

B) 1 + 1 + 2 + 3 = 13 – 1

D) 1 + 1 + 2 + 3 = 8 – 1

B) 53 – 33 + 23 = 106

D) 53 + 33 – 23 + 13 = 145

B) 12 + 12 + 22 + 22 = 2 ∙ 5

D) 12 + 12 + 22 + 32 + 52 = 5 ∙ 8

15152) 1

1 = 1

2

1 = 2

3

2 = 1.5

A) 5

3 = 1.666… B) 3

1 = 3 C) 8

5 = 1.6 D) 2

3 = 0.666…

Answer: A

Use the Lucas sequence to answer the question.

153) What is the 15th term of the Lucas sequence?

A) 843 B) 3,571 C) 2,207 D) 1,364

Answer: D

154) What is the 26th term of the Lucas sequence?

A) 271,443 B) 103,682 C) 167,761 D) 165,061

Answer: A

155) What is the 20th term of the Lucas sequence?

A) 15,217 B) 24,476 C) 15,127 D) 9,349

Answer: C

156) What is the 30th term of the Lucas sequence?

A) 710,647 B) 1,149,851 C) 3,010,349 D) 1,860,498

Answer: D

Determine whether the statement is true or false.

157) For any odd natural number n: If you square the nth term of the Lucas sequence and add 2 you will obtain the

(2n)th term of the Lucas sequence.

A) True B) False

Answer: A

158) The following quotient, where Fn represents the nth term of the Fibonacci sequence, approaches the golden

ratio as n gets larger:

Fn+1

.

Fn

A) True B) False

Answer: A

159) If you square any term of the Lucas sequence and subtract 5 you will obtain the product of the 2 terms either

side of the term squared.

A) True B) False

Answer: B

160) For any natural number n, if you multiply the nth term of the Fibonacci sequence and the nth term of the Lucas

sequence, you will obtain the (2n)th term of the Fibonacci sequence.

A) True B) False

Answer: A

16161) Every natural number can be expressed as a sum of Fibonacci numbers, where no number is used more than

once.

Answer: A

A) True B) False

162) If m divides n, then Fm is a factor of Fn. Fn represents the nth term of the Fibonacci sequence.

A) True B) False

Answer: A

163) For any prime number p except 2, either Fp+1 or Fp–1 is divisible by p. Fn represents the nth term of the

Fibonacci sequence.

A) True B) False

Answer: B

164) A Pythagorean triple can be obtained as follows:

Choose any four successive terms of the Fibonacci sequence. Multiply the first and fourth. Double the product of

the second and third. Add the squares of the second and third. The three numbers obtained will always form a

Pythagorean triple.

A) True B) False

Answer: A

17