Mathematical Ideas 13th Edition By Miller - Test Bank

Mathematical Ideas 13th Edition By Miller – Test Bank

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Sample Questions Are 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 n.
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 B) 1, 2, 3, 7, 6, 14, 21, 42
C) 7, 6, 14, 42 D) 1, 7, 42
Answer: B
11) 125
A) 1, 5, 25, 125 B) 5, 62, 125 C) 5, 25, 125 D) 1, 5, 25
Answer: A
1
12) 110
A) 2, 5, 10, 11, 55, 110 B) 1, 2, 5, 10, 11, 22, 55, 110
C) 1, 2, 5, 10, 11, 22, 110 D) 1, 2, 4, 5, 10, 11, 22, 55, 110
Answer: B
13) 676
A) 1, 2, 4, 13, 26, 39, 52, 169, 338, 676 B) 1, 2, 4, 13, 26, 52, 169, 338, 676
C) 1, 2, 3, 4, 13, 26, 169, 338, 676 D) 1, 2, 13, 169, 676
Answer: B
14) 13
A) 1, 2, 3, 4, 6, 13 B) 1, 2, 6, 13 C) 1, 2, 4, 4, 6, 13 D) 1, 13
Answer: D
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
A) Yes B) No
Answer: B
17) 793,542; 3
A) Yes B) No
Answer: A
18) 788,913; 9
A) Yes B) No
Answer: A
19) 516,379; 9
A) Yes B) No
Answer: B
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
A) Yes B) No
Answer: B
2
22) 203,588; 7
A) Yes B) No
Answer: A
23) 324,841; 10
A) Yes B) No
Answer: B
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 B) 4 ∙ 3 C) 32 D) 4 ∙ 2
Answer: A
26) 252
A) 23 ∙ 32 ∙ 7 B) 22 ∙ 32 ∙ 7 C) 34 ∙ 7 D) 24 ∙ 7
Answer: B
27) 504
A) 24 ∙ 3 ∙ 7 B) 23 ∙ 33 ∙ 7 C) 23 ∙ 32 ∙ 7 D) 2 ∙ 34 ∙ 7
Answer: C
28) 30
A) 32 ∙ 2 B) 2 ∙ 3 ∙ 5 C) 6 ∙ 5 D) 22 ∙ 5
Answer: B
29) 126
A) 2 ∙ 3 ∙ 7 B) 22 ∙ 32 ∙ 7 C) 2 ∙ 32 ∙ 7 D) 14 ∙ 32
Answer: C
30) 69
A) 32 B) 3 ∙ 21 C) 32 ∙ 23 D) 3 ∙ 23
Answer: D
31) 2013
A) 33 ∙ 61 B) 112 ∙ 61 C) 3 ∙ 11 ∙ 61 D) 32 ∙ 61
Answer: C
32) 8775
A) 33 ∙ 52 ∙ 13 B) 54 ∙ 13 C) 32 ∙ 53 ∙ 13 D) 34 ∙ 13
Answer: A
3
SHORT 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
4
46) 400
A) 12 B) 15 C) 8 D) 16
Answer: B
47) 25 ∙ 36 ∙ 52
A) 60 B) 120 C) 90 D) 126
Answer: D
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 – 1 is also a composite number.
A) True B) False
Answer: A
50) The number 213 – 1 is an example of a Mersenne prime.
A) True B) False
Answer: A
51) Euler’s formula n2 – n + 41 generates primes for n up to 46 and fails at n = 47.
A) True B) False
Answer: B
52) Escott’s formula n2 – 79n + 1601 first fails at n = 80.
A) True B) False
Answer: A
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
5
57) p = n2 – n + 41; n = 12
A) 2693 B) 173 C) 115 D) 197
Answer: B
58) p = n2 – n + 41; n = 28
A) 797 B) 771 C) 1395 D) 853
Answer: A
59) p = n2 – n + 41; n = 37
A) 1365 B) 4251 C) 1373 D) 2691
Answer: C
60) p = n2 – 79n + 1,601; n = 1
A) 41 B) 1523 C) 1681 D) -1521
Answer: B
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 – 1.”
Answer: Two factors of M14. are M2 = 3 and M7 = 127.
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.
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MULTIPLE 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 – 1) is perfect.
A) True B) False
Answer: B
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
7
77) 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 B) 6 + 6 C) 3 + 9 D) 2 + 10
Answer: A
83) 28
A) 14 + 14 B) 5 + 23, 11 + 17
C) 3 + 25, 5 + 23, 7 + 21 D) 5 + 23, 13 + 15
Answer: B
84) 34
A) 21 + 13, 31 + 3, 29 + 5 B) 27 + 7, 29 + 5, 23 + 11
C) 29 + 5, 31 + 3, 23 + 11, 17 + 17 D) 15 + 19, 29 + 5, 23 + 11
Answer: C
85) 18
A) 9 + 9 B) 5 + 13, 7 + 11
C) 3 + 15, 5 + 13 , 7 + 11 D) 3 + 15, 7 + 11
Answer: B
86) 44
A) 3 + 41, 7 + 37, 11 + 33 B) 3 + 41, 5 + 39, 13 + 31
C) 22 + 22 D) 3 + 41, 7 + 37, 13 + 31
Answer: D
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 B) No C) 17, 19 D) 21, 23
Answer: C
8
88) 4 and 14
A) No B) 5, 7 and 11, 13 C) 5, 7 D) 5, 7 and 9, 11
Answer: B
89) 32 and 51
A) No B) 43, 47 C) 41, 43 D) 41, 42 and 47, 49
Answer: C
90) 100 and 106
A) 103, 105 B) No C) 101, 103, 105 D) 101, 103
Answer: D
91) 33 and 41
A) 37, 39 B) 35, 37 C) No D) 31, 33
Answer: C
92) 81 and 99
A) No B) 91, 93 C) 89, 91 D) 83, 85
Answer: A
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 B) one; happy C) neither; not happy D) one; not happy
Answer: D
95) 35,390,008 and 39,259,592
A) one; happy B) one; not happy C) neither; not happy D) both; happy
Answer: D
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
9
99) 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
10
108) 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
A) 6 B) 12 C) 1 D) 2
Answer: B
114) 168, 42
A) 6 B) 21 C) 42 D) 14
Answer: C
115) 240, 900
A) 30 B) 60 C) 240 D) 180
Answer: B
116) 9, 27, 30
A) 6 B) 1 C) 3 D) 9
Answer: C
117) 54, 72, 126
A) 18 B) 2 C) 9 D) 36
Answer: A
118) 120, 1000, 2250
A) 30 B) 10 C) 20 D) 50
Answer: B
119) 1562, 450, 6750
A) 12 B) 4 C) 2 D) 5
Answer: C
11
120) 12, 15, 24, 30
A) 3 B) 4 C) 12 D) 1
Answer: A
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
A) 34,375 B) 1375 C) 55 D) 275
Answer: B
123) 48, 162, 27
A) 324 B) 1296 C) 432 D) 648
Answer: B
124) 56, 96
A) 1344 B) 672 C) 168 D) 224
Answer: B
125) 60, 40, 70
A) 120 B) 420 C) 280 D) 840
Answer: D
126) 45, 56, 150
A) 1,050 B) 1260 C) 12,600 D) 2520
Answer: C
127) 43,378, 4715
A) 10 B) 943 C) 9430 D) 216,890
Answer: D
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
12
131) 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 C) F27 = 46,368 D) F27 = 196,418
Answer: D
138) F28 = 317,811, F30 = 832,040
Find F29.
A) F29 = 196,418 B) F29 = 1,149,851 C) F29 = 514,229 D) F29 = 1,346,269
Answer: C
13
139) 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 C) 1, 1, 3, 4, 7, 11, 18 D) 1, 1, 2, 3, 5, 8, 13
Answer: D
142) According to the Binet form, the nth Fibonacci number is given by
1 + 5
2
n
– 1 – 5
2
n
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 C) 13 – 8 = 5 D) 11 – 3 = 8
Answer: C
144) 1 + 1 = 2
3 + 2 = 5
8 + 5 = 13
21 + 13 = 34
A) 55 + 34 = 89 B) 21 + 34 = 55 C) 34 + 53 = 87 D) 13 + 21 = 34
Answer: A
145) 8 – 5 + 3 = 2 ∙ 3
13 – 8 + 5 = 2 ∙ 5
21 – 13 + 8 = 2 ∙ 8
A) 34 + 21 – 13 = 2 ∙ 21 B) 34 – 21 + 8 = 2 ∙ 8
C) 34 – 21 – 13 = 0 D) 34 – 21 + 13 = 2 ∙ 13
Answer: D
14
146) 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 B) 1 + 2 + 3 + 5 + 8 = 20
C) 1 + 1 + 2 + 3 + 5 + 8 + 13 = 33 D) 1 + 1 + 2 + 3 + 5 + 8 = 20
Answer: D
147) 22 – 12 = 3
32 – 12 = 8
52 – 22 = 21
A) 82 – 32 = 55 B) 82 – 22 = 60 C) 82 – 52 = 39 D) 132 – 82 = 55
Answer: A
148) 12 + 12 = 2
12 + 22 = 5
22 + 32 = 13
A) 32 + 52 = 35 B) 32 + 52 = 34 C) 32 + 82 = 73 D) 22 + 52 = 29
Answer: B
149) 1 = 2 – 1
1 + 1 = 3 – 1
1 + 1 + 2 = 5 – 1
A) 1 + 1 + 2 + 2 = 7 – 1 B) 1 + 1 + 2 + 3 = 13 – 1
C) 1 + 1 + 2 + 3 + 5 = 13 – 1 D) 1 + 1 + 2 + 3 = 8 – 1
Answer: D
150) 23 + 13 – 13 = 8
33 + 23 – 13 = 34
A) 53 + 33 – 23 = 144 B) 53 – 33 + 23 = 106
C) 53 + 23 – 13 = 132 D) 53 + 33 – 23 + 13 = 145
Answer: A
151) 12 = 1 ∙ 1
12 + 12 = 1 ∙ 2
12 + 12 + 22 = 2 ∙ 3
A) 12 + 22 + 32 = 2 ∙ 7 B) 12 + 12 + 22 + 22 = 2 ∙ 5
C) 12 + 12 + 22 + 32 = 3 ∙ 5 D) 12 + 12 + 22 + 32 + 52 = 5 ∙ 8
Answer: C
15
152) 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
16
161) Every natural number can be expressed as a sum of Fibonacci numbers, where no number is used more than
once.
A) True B) False
Answer: A
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
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