Student’s Name [print]
Student Number
Mathematics 2156B Midterm Exam
6:30–9:30 p.m.
March 1, 2013
Instructions: Print your name on the SCANTRON answer sheet. Sign the SCANTRON
answer sheet, and use a PENCIL to mark your student number on the SCANTRON answer
sheet. Use a PENCIL to mark your answers to questions 1–20 on the SCANTRON answer
sheet, and as well, circle your answers on the question sheet.
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A1. When the Euclidean algorithm is applied to a pair of
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(unknown) positive integers a and b, the results are as
shown at right. Based on this information the gcd of a
and b can be written as
A: 7a − 22b
B: −7a + 24b
C: 7a − 52b
a
b
r1
r2
D: 7a + 29b
=
=
=
=
7b + r1
2r1 + r2
3r2 + r3
3r3
E: None of A, B, C, D
Solution: We have ( a, b ) = r3 = r1 − 3(b − 2r1 ) = −3b + 7(a − 7b) = 7a − 52b, so the answer is C.
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A2. Suppose that q ∈ Z and a = 3366q + 1824. What is the greatest common divisor of a and
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3366?
A: 32
B: 48
C: 149
D: 307
E: None of A, B, C, D
Solution: ( a, 3366 ) = ( 3366, 1824 ). We have 3366 = 1824 + 1542, 1824 = 1542 + 282, 1542 = 5(282)+ 132,
282 = 2(132) + 18, 132 = 7(18) + 6, and 18 = 3(6), so ( 3366, 1824 ) = 6. The answer is E.
Alternatively, we could write 3366 = 11(306) = (6)(11)(51) = (2)(3)(11)(3)(17) = 21 32 111 171 , and
1824 = (6)(304) = (6)(8)(38) = 25 31 191 , so ( 3366, 1824 ) = 21 31 = 6.
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A3. How many pairs (x, y) of positive integers x and y satisfy 20x + 8y = 244?
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A: 7
B: 5
C: 3
D: 6
E: 0
Solution: If (x, y) is a solution, then 8y ≡ 244 (mod 20), or 8y ≡ 4 (mod 20). This is equivalent to its
reduced form, 2y ≡ 1 (mod 5), which has solved form y ≡ 3 (mod 5). Thus y = 3 + 5k, k ∈ Z. Note
that for any k ∈ Z, setting y = 3 + 5k in the equation 20x + 8y = 244 gives 20x + 24 + 40k = 244, or
20(x + 2k) = 220; equivalently x + 2k = 11. Thus for any integer k, x = 11 − 2k and y = 3 + 5k give a
solution to the equation 20x + 8y = 244, and every solution is of this form. Thus we need only determine
the values of k ∈ Z such that 3 + 5k > 0 and 11 − 2k > 0, and these are 0 ≤ k ≤ 5. There are therefore 6
pairs (x, y) of positive integers x and y such that 20x + 8y = 244. The answer is D.
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A4. Let p > 2 and q > 2 be distinct prime numbers. Exactly which of the following statements
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are true?
(i)
(iii)
( pq , q p ) = 1.
(ii)
[ p + q, q ] = (p + q)q. (iv)
A: (i), (ii)
B: (ii)
( p2 q 2 , p3 q ) = p2 q.
[ 2p2 q, pq 2 ] = (pq)2 .
C: (i), (ii), (iv)
D: (ii),(iv)
E: None of A, B, C, D
March 1, 2013
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Mathematics 2156B
Midterm Exam
Solution:
(i) is true, since ( p, q ) = 1 implies ( pn , q ) = 1 for any positive integer n. Thus in particular, ( pq , q ) = 1,
and thus ( pq , q n ) = 1 for any positive integer n.
(ii) is true.
(iii) is true, since ( p + q, q ) = ( p, q ) = 1.
(iv) is false, since [ 2p2 q, pq 2 ] = 21 p2 q 2 = 2(pq)2 . Here is where we need to know that neither p nor q is
equal to 2.
Since (i), (ii), and (iii) are true, while (iv) is false, the answer is E.
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A5. Exactly which of the statements to the right (i)
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are true for all positive integers a and b?
(ii)
(iii)
(iv)
A: (i), (ii), (iii)
B: (i), (ii), (iv)
( a, b ) = 1
( a, b ) = 1
( a, b ) = 1
( a, b ) = 1
C: (i), (iii)
implies
implies
implies
implies
D: (ii), (iv)
( a, b2 ) = 1.
( a, a − b ) = 1.
( a, a + 2b ) = 1.
( a, 2a + b ) = 1.
E: All of them
Solution:
(i) is true by a result in the text.
(ii) is true, since ( a, a − b ) = ( a, b ).
(iii) is false. ( a, a + 2b ) = ( a, 2b ) = ( a, 2 ) since ( a, b ) = 1. Thus for any even integer a, ( a, a + 2b ) = 2.
(iv) is true, since ( a, 2a + b ) = ( a, b ).
Since (i), (ii), and (iv) are true, while (iii) is false, the answer is B.
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A6. Exactly which of the statements to the right
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are true for all positive integers a and b?
A: (iv)
B: (ii), (iii)
C: (ii), (iv)
(i)
(ii)
(iii)
(iv)
( a + b, a2 ) = ( a, b ).
( a − b, a + b ) = ( a, b ).
( a, b2 ) = ( a, b )2 .
( a2 , b2 ) = ( a, b )2 .
D: (i), (iv)
E: None of A, B, C, D
Solution:
(i) is false. For example, a = b even ( 2a, a2 ) = a( 2, a ) = 2a on the one hand, and ( a, a ) = a on the
other.
(ii) is false. For example, a = 3, b = 1 results in ( 2, 4 ) = 2 on the one hand, and ( 3, 1 ) = 1 on the other.
(iii) may be false if ( a, b ) > 1. For example, take a = 2 and b = 2.
(iv) is true. Let d = ( a, b ), so that a = a1 d and b = b1 d for some relatively prime integers a1 and b1 .
Then ( a2 , b2 ) = ( a21 d2 , b21 d2 ) = d2 ( a21 , b21 ) = d2 , as required.
Since only (iv) is true, the answer is A.
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A7. What is the exponent of the greatest power of 3 that divides ( 72900, 34326 )?
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A: 3
B: 2
C: 1
D: 0
E: None of A, B, C, D
Solution: We have 72900 = (100)(9)(81) = 22 52 36 , and 34326 = (6)(5721) = (6)(3)(1907) = (2)(32 )(1907),
and ( 3, 1907 ) = 1, so 32 divides 34326 but 33 does not. Since 32 also divides 72900, it follows that the
exponent of the greatest power of three that divides ( 72900, 34326 ) is 2.
The answer is B.
Mathematics 2156B
Midterm Exam
March 1, 2013
Page 3
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A8. Exactly which of [25]75 , [35]75 , [38]75 are units in Z75 ?
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A: [25]75
B: [35]75
C: [38]75
D: None of them
E: All of them
Solution: [a]75 is a unit of Z75 if and only if ( a, 75 ) = 1, and only ( 38, 75 ) = 1. The answer is C.
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A9. The number of integer values of x with 1 ≤ x ≤ 23 for which there is some integer y with
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xy ≡ 1 (mod 24) is:
A: 8
B: 20
C: 16
D: 12
E: None of A, B, C, D
Solution: The number is ϕ(24) = ϕ(23 31 ) = (4)(1)(2) = 8, so the answer is A.
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A10. Exactly
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(ii)
(iii)
A: (i)
which of the following statements are valid for any integer n > 1?
The product of any two zero divisors in Zn is always a zero divisor.
The product of any two nonunits in Zn is never a unit.
The product of a unit and a zero divisor in Zn is always a zero divisor.
B: (i), (iii)
C: (ii), (iii)
D: (i), (ii), (iii)
E: None of A, B, C, D
Solution:
(i) is false, since in Z6 for example, we have [2]6 and [3]6 are zero divisors but [2]6 [3]6 = [0]6 , which is not
a zero divisor.
(ii) is true, since a nonunit is a zero divisor or equal to 0, and the product of zero divisors is either a zero
divisor or else 0.
(iii) is true.
Since (ii) and (iii) are true, while (i) is false, the answer is C.
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A11. The number of integers x which satisfy 36x ≡ 30 (mod 69) and 0 ≤ x ≤ 68 is:
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A: 11
B: 0
C: 1
D: 3
E: 6
Solution: Since ( 36, 69 ) = 3 and 3 is a divisor of 30, there are ( 36, 69 ) = 3 solutions in the interval from
0 to 69-1=68 inclusive. The answer is D.
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A12. The number of integer solutions with 0 ≤ x ≤ 5774
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of the system of congruences shown to the right is:
A: 0
B: 1
C: 5
x ≡ 14
x ≡ 5
x ≡ 7
D: 10
(mod 21)
(mod 33)
(mod 35)
E: 21
Solution: Since ( 21, 33 ) = 3 and 3 divides 14 − 5 = 9, ( 21, 35 ) = 7 and 7 divides 14 − 7 = 7, and
( 33, 35 ) = 1 the system has a solution, unique modulo [ 21, 33, 35 ] = (3)(7)(11)(5) = (231)(5) = 1155.
Since 5775 = 5(1155), there are five integers in the interval 0..5774 that are solutions to the system of
congruences. The answer is C.
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A13. The number of integer solutions with 0 ≤ x ≤ 419
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of the system of congruences shown to the right is:
x ≡ 2
x ≡ 8
x ≡ 4
(mod 6)
(mod 10)
(mod 14)
March 1, 2013
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Mathematics 2156B
Midterm Exam
A: 0
B: 1
C: 10
D: 2
E: None of A, B, C, D
Solution: Since ( 6, 10 ) = 2 divides 2 − 8, ( 6, 14 ) = 2 divides 2 − 4, and ( 10, 14 ) = 2 divides 8 − 4, there is
a solution c, and the solved form is x ≡ c (mod [ 6, 10, 14 ]). Since [ 6, 10, 14 ] = 210, there is one solution
between 0 and 209, and one solution between 210 and 419, so there are two solutions in the interval from
0 to 419 inclusive. The answer is D.
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A14. Exactly which of the systems of congruences shown below have solutions?
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(i) x ≡ 6 (mod 8)
(ii) 2x ≡ 3 (mod 11) (iii) x ≡ 2 (mod 8)
x ≡ 9 (mod 20)
5x ≡ 8 (mod 12)
x ≡ 4 (mod 10)
x ≡ 4 (mod 12)
A: (i)
B: (ii)
C: (iii)
D: (i), (ii)
E: (ii), (iii)
Solution:
(i) has no solution since ( 8, 20 ) = 4 and 4 does not divide 6 − 9.
(ii) has a solution since each of the congruences has a solution and ( 11, 12 ) = 1.
(iii) has no solution since ( 8, 12 ) = 4 and 4 does not divide 2 − 4.
Since only (ii) has a solution, the answer is B.
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A15. ϕ(260) is:
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A: 128
B: 84
C: 210
D: 96
E: None of A, B, C, D
Solution: Since 260 = (26)(10) = (2)(13)(2)(5) = 22 51 131 , ϕ(260) = 21 (2 − 1)50 (5 − 1)130 (13 − 1) = 96.
The answer •is D.
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A16. Which of the following statements concerning the graphs drawn above is true?
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(i) No two of the graphs are isomorphic.
(ii) G3 and G4 are isomorphic, and no other isomorphisms exist between
pairs of the above graphs.
(iii) G3 and G6 are isomorphic, and no other isomorphisms exist between
pairs of the above graphs.
(iv) G4 and G6 are isomorphic, and no other isomorphisms exist between
pairs of the above graphs.
(v) G3 , G4 , and G6 are all isomorphic.
A: (i)
B: (ii)
C: (iii)
D: (iv)
E: (v)
Solution: If two graphs are isomorphic, then they have identical degree sequences. The degree sequences
of G1 , G2 , G3 , G4 , G5 , and G6 , respectively, are (4, 4, 3, 3, 2), (3, 3, 3, 3, 3), (4, 3, 3, 3, 3), (4, 3, 3, 3, 3),
(3, 3, 3, 3, 3, 3), and 4, 3, 3, 3, 3). The only possiblity would be that G3 , G4 , and G6 are isomorphic. In fact,
if we redraw each of G3 and G4 , placing the vertex of degree 4 inside the 4-cycle, we obtain the diagram
of G6 . Thus G3 , G4 , and G6 are isomorphic.
Mathematics 2156B
Midterm Exam
March 1, 2013
Page 5
The answer is E.
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A17. Exactly which of the graphs G1 , G2 , G3 , G4 , G5 and G6 are regular?
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A: G2 , G3 , G4
B: G2
C: G2 , G5
D: G5
E: G2 , G5 , G6
Solution: G1 has a vertex of degree 2 and a vertex of degree 3, so G1 is not regular. G2 = K5 is 4-regular.
Each of G3 and G4 has a vertex of degree 3 and a vertex of degree 4, so neither G3 nor G4 is regular. G5
is 3-regular. Finally, G6 has a vertex of degree 3 and a vertex of degree 4, so G6 is not regular.
Thus G2 and G5 are regular, and the others are not. The answer is C.
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A18. Exactly which of
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(i)
(ii)
(iii)
A: (i), (iii)
the following statements are true?
There exists a simple graph with 6 vertices and 14 edges.
There exists a simple 4-regular graph with 9 edges.
There exists a simple 3-regular graph with 15 edges.
B: (iii)
C: (ii)
D: (ii), (iii)
E: None of A, B, C, D
Solution: (i) true. K6 has 62 = 15 edges, so the graph obtained by removing a single edge from K6 is a
simple graph with 6 vertices and 14 edges.
(ii) false. If G were a simple 4-regular graph with n vertices and 9 edges, then by the handshake lemma,
we would have 4n = 18. Since 4 does not divide 18, there is no such graph.
(iii) true. The Petersen graph is simple, 3-regular, has 10 vertices and 15 edges.
Since (i) and (iii) are true while (ii) is false, the answer is A.
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A19. If G is a graph with degree sequence (4, 4, 3, 3, 2, 2, 1, 1), how many edges does G have?
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A: 8
B: 10
C: 12
D: 21
E: None of A, B, C, D
Solution: By the handshake lemma, such a graph would have m edges, where 2 m = 4 + 4 + 3 + 3 + 2 +
2 + 1 + 1 = 20, so m = 10. Here is a graph with this degree sequence:
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The answer is B.
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A20. If G is a graph with 26 edges and degG (v) ≥ 3 for each vertex v of G, then the largest
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possible value for νG is:
A: 18
B: 17
C: 15
D: 13
Solution: By the handshake lemma, we have 3 νG ≤
P
v∈V
G
E: None of A, B, C, D
degG (v) = 2 εG = 52, so it must be that
νG ≤ ⌊ 52/3 ⌋ = 17. Is there actually a graph with 17 vertices and 26 edges? Yes, consider either of
the following graphs. Each has 17 vertices, sixteen of degree 3 and one of degree 4. Thus each has
(3(16) + 4)/2 = 26 edges.
Mathematics 2156B
Midterm Exam
March 1, 2013
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The answer is B.
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PART B
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B1. Find (210, 116) and express it as a linear combination of 210 and 116.
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Solution: We have ( 210, 116 ) = 2( 105, 58 ), and 105 = 1(58) + 47, 58 = 1(47) + 11, 47 = 4(11) + 3,
11 = 3(3) + 2, 3 = 1(2) + 1, and 2 = 2(1) + 0, so ( 105, 58 ) = 1 and from
105
1
−38
1
21
4
−17
3
4
1
−1
2
1
58
0
we find that 1 = (−38)(58) + (21)(105), so 2 = (−38)(116) + (21)(210).
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B2. Find all solutions to the linear congruence 15x ≡ 20 (mod 115).
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Solution: First, compute ( 15, 115 ) = 5, and verify that 5 divides 20. It does, so form the reduced congruence by dividing by 5, obtaining 3x ≡ 4 (mod 23). Now ( 3, 23 ) = 1, so we must find the multiplicative
inverse of 3 modulo 23. Since we observe that (3)(8) = 24 ≡ 1 (mod 23), we multiply by 8 to get
x ≡ 32 ≡ 8 (mod 23). Thus x ≡ 9 (mod 23) is the solved form of the congruence. Note that there are
( 15, 115 ) = 5 solutions that lie in the interval from 0 to 115 − 1 = 114, and these are 9, 9 + 23 = 32,
32 + 23 = 55, 55 + 23 = 78, and 78 + 23 = 101.
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B3. Find all solutions to the following system of linear congruences:
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x≡4 (mod 16)
x≡8 (mod 18)
x≡6 (mod 22)
Solution: Set x = 8 + 18y and solve 8 + 18y ≡ 4 (mod 16) for y. We have 2y ≡ −4 (mod 16), which
reduced gives y ≡ −2 (mod 8). Take y = −2 to find that x = 8 + 18(−2) = −28 is a solution to the
first two congruences, and so x ≡ −28 (mod [ 16, 18 ]) is the solved form of the first two congruences.
Thus it remains to solve x ≡ −28 (mod (16)(9)) and x ≡ 6 (mod 22). Set x = −28 + (16)(9)y and solve
−28 + (16)(9)y ≡ 6 (mod 22) for y. We find that (−6)(9)y ≡ 34 (mod 22), or −10y ≡ 12 (mod 22).
Reduce to find that −5y ≡ 6 (mod 11). Evidently, y = 1 is a solution, so we have x = −28 + (16)(9)(1)
is a solution to the system of all three congruences, and thus x ≡ −28 + (16)(9) (mod [ (16)(9), 22 ]) is
the solved form of the system. The modulus is [ (16)(9), 22 ] = (16)(9)(22)/2 = (144)(11) = 1584, and the
constant is −28 + 144 = 116, so x ≡ 116 (mod 1584) is the canonical form of the solution to the system.
Check: 116 = 7(16)+4 ≡ 4 (mod 16), 116 = 6(18)+8 ≡ 8 (mod 18), and 116 = (5)(22)+6 ≡ 6 (mod 22).
March 1, 2013
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Mathematics 2156B
Midterm Exam
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B4. (a) List the units of Z24 .
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Solution: These are the congruence classes [a]24 where ( a, 24 ) = 1, and there will be ϕ(24) =
ϕ(23 31 ) = 22 (2) = 8 of them. We find [1]24 , [5]24 , [7]24 , [11]24 , [13]24 , [17]24 , [19]24 , and [23]24 .
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(b) List the zero divisors of Z24 .
Solution: There are 24 − 1 − 8 = 15 of them, given by
(Z24 − { [0]24 }) − U24 = { [2]24 , [3]24 , [4]24 , [6]24 , [8]24 , [9]24 , [10]24 , [12]24 ,
[14]24 , [15]24 , [16]24 , [18]24 , [20]24 , [21]24 , [22]24 }.
B5. In each case below, find b ∈ { 1, . . . , n − 1 } such that if [a]n is a unit in Zn , then [a]n [b]n =
[1]n , while if [a]n is a zero divisor in Zn , then [a]n [b]n = [0]n .
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(a) a = 22, n = 32.
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(b) a = 22, n = 31.
Solution: Since ( 22, 32 ) = 2 > 1, [22]32 is a nonzero, nonunit element of Z32 , and thus is a zero
divisor. We note that (2)(16) = 32, so (2)(16) ≡ 0 (mod 32). As a result, (11)(2)(16) ≡ 0 (mod 32),
or (22)(16) ≡ 0 (mod 32), and 16 ∈ { 1, 2, . . . , 31 }. Thus b = 16 is suitable.
Solution: Since ( 22, 31 ) = 1, [22]31 is a unit. To find b such that 22b ≡ 1 (mod 31), we calculate
31 = 22 + 9, 22 = 2(9) + 4, 9 = 2(4) + 1, so 1 = 9 − 2(22 − 2(9)) = −2(22) + 5(9) = −2(22) +
5(31 − 22) = 5(31) − 7(22). Thus 22(−7) ≡ 1 (mod 31) and since −7 ≡ 24 (mod 31), we find that
b = 24 ∈ { 1, 2, . . . , 30 } and 22b ≡ 1 (mod 31). Thus b = 24 is suitable.
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B6. Find the least common multiple of 9, 12 and 20.
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Solution: [ 9, 12, 20 ] = [ [ 9, 12 ], 20 ] = [ (3)(12), 20 ] = (3)(12)(20)/4 = 180.
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B7. Prove that if ( a, b ) = 2, then ( a2 + b2 , ab ) = 4.
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Solution: Suppose that ( a, b ) = 2, so that a = 2a1 and b = 2b1 with ( a1 , b1 ) = 1. Then ( a2 + b2 , ab ) =
( 22 a21 + 22 b21 , 4a1 b1 ) = 4( (a21 + b21 , a1 b1 ). Now ( a21 + b21 , a1 ) = ( a1 , b21 ) = 1 since ( a1 , b1 ) = 1, and
( a21 + b21 , b1 ) = ( a21 , b1 ) = 1 since ( a1 , b1 ) = 1. Thus ( a2 + b2 , ab ) = 4( (a21 + b21 , a1 b1 ) = 4.
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B8. Find integers R1 and R2 such that for any integers c1 and c2 , x = R1 c1 + R2 c2 is a solution
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to the system of linear congruences x ≡ c1 (mod 15) and x ≡ c2 (mod 28).
Solution: We take R1 = 28r1 where 28r1 ≡ 1 (mod 15), and we take R2 = 15r2 , where 15r2 ≡ 1 (mod 28).
Now, 13r1 ≡ 28r1 ≡ 1 (mod 15) and 13 ≡ −2 (mod 15), so 2r1 ≡ −1 ≡ 14 (mod 15). Since ( 2, 15 ) = 1, we
have r1 ≡ 7 (mod 15) and so we choose r1 = 7 to obtain R1 = 28(7) = 196. To solve 15r2 ≡ 1 (mod 28),
we can write 15r2 ≡ −27 (mod 28), and since ( 3, 28 ) = 1, we can cancel the common factor of 3 to obtain
5r2 ≡ −9 (mod 28). Then 28 = 5(5) + 3, 5 = 3 + 2 and 3 = 2 + 1, so 1 = 3 − (5 − 3) = −5 + 2(28 − 5(5)) =
2(28) − 11(5). Thus (−11)(5) ≡ 1 (mod 28) and so r2 ≡ 99 ≡ 15 (mod 28). We shall take r2 = 15 and get
R2 = (15)(15) = 225.
Check: 196 = 13(15) + 1 = 7(28), so 196 ≡ 1 (mod 15) and 196 ≡ 0 (mod 28). As well, 225 =
(8)(28) + 1 = (15)(15), so 225 ≡ 1 (mod 28) and 225 ≡ 0 (mod 15).
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B9. Is there a simple graph that has (4, 4, 3, 3, 2, 2, 1, 1) as its degree sequence? If so, draw one.
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If not, prove that there is no such graph.
Solution: There are many, and we list several below.
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March 1, 2013
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Mathematics 2156B
Midterm Exam
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There are none with three components, as part (b) establishes.
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B10. Draw two non-isomorphic simple graphs with 5 vertices and 7 edges. Verify that your
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graphs are non-isomorphic.
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Solution:
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one on the left has no such vertex. (There are many other examples.)
B11. Let p be any prime, and let n be a positive integer.
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(a) Prove that [p] ∈ Upn −1 .
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Solution: ( p, p − 1 ) = ( p, 1 ) = 1, so [p] ∈ Upn −1 .
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(b) Prove that [p]n = [1] in Upn −1 , while [p]k 6= [1] for any k with 1 ≤ k < n.
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(c) Explain how (a) and (b) can be used to conclude that n divides ϕ(pn − 1).
Solution: Since [pn − 1] = [0] in Zpn −1 , we have [p]n − [1] = [0] and thus [p]n = [1] in Zpn −1 ; hence in
Upn −1 .
Solution: (a) tells us that [p] is an element of the group of units Upn −1 , which has size ϕ(pn − 1), and
(b) tells us that the order of [p] in Upn −1 is n. By Lagrange’s theorem, the order of an element in a
finite group divides the size of the group, so n divides ϕ(pn − 1).
3 Bonus: Determine necessary and sufficient conditions on distinct primes p and q in order that
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there exists a positive integer n with ϕ(n) = pq.
Solution: First of all, if pq is odd, then there will be no solution, so one of the primes must be 2. Let q = 2
and p be a prime different from 2. Suppose first that there is n with ϕ(n) = 2p. Then either n is prime, in
which case n = 2p + 1, or else n is composite. If n is composite, it can’t be a power of 2, so it must have at
least one odd prime factor f , say, so that n = f i a and ( a, f ) = 1. But then 2p = ϕ(n) = f i−1 (f − 1)ϕ(a).
Since f is odd, f − 1 is even, say f − 1 = 2m and thus p = f i−1 mϕ(a). Let us check all possibilities.
Case 1: p = f i−1 . Then f = p, i = 2, m = 1 = ϕ(a), and thus a = 1 or 2 and f − 1 = 2, so p = f = 3.
We note that for n = 32 or (2)(32 ), ϕ(n) = 2p.
Case 2: p = m, so i = 1 and ϕ(a) = 1, giving a = 1 or 2 and f = 2p + 1. We have n = 2p + 1 or
(2p + 1)(2) and 2p + 1 is prime. We note that if 2p + 1 is prime, then ϕ(2p + 1) = ϕ(2(2p + 1)) = 2p.
Case 3: p = ϕ(a), which is not possible since p is odd.
Thus for an odd prime p, for there to exist n with ϕ(n) = 2p, it is necessary and sufficient that either
p = 3 or else 2p + 1 be prime; that is, 2p + 1 is prime.