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Math 373 — Practice Midterm 2
Multiple Choice Section
(1) Which of the following represents the prime factorization of 160?
(A) 16 × 10
(B) 20 × 8
(C) 24 × 10
(D) 25 × 5
(E) 26 × 5
(F) none of these.
Solution. D.
(2) d(160) =
(A) 8
(B) 10
(C) 12
(D) 14
(E) 16
(F) none of these.
(D) 400
(E) 450
(F) none of these.
(E) 3050
(F) none of these.
Solution. C.
d(160) = (5 + 1) × (1 + 1) = 12.
(3) φ(525) =
(A) 280
(B) 416
(C) 360
Solution. F.
525 = 3 × 52 × 7. So φ(525) = 3 × 52 × 7 × 2/3 × 4/5 × 6/7 = 240.
(4) σ(840) =
(A) 2280
(B) 2480
Solution. D.
840 = 23 × 3 × 5 × 7. So σ(840) =
(C) 2680
24 −1
2−1
(D) 2880
× (3 + 1) × (5 + 1) × (7 + 1) = 2880.
(5) The number of primitive roots modulo 100 is
(A) 0
(B) 5
(C) 16
(D) 40
(E) 1
(F) none of these.
Solution. A.
Because 100 = 22 × 52 , by our theorem there are no primitive roots modulo 100.
(6) The number of primitive roots modulo 101 is
(A) 0
(B) 5
(C) 16
(D) 40
(E) 1
(F) none of these.
Solution. D.
Our theorem says that there are primitive roots modulo 101 and the number is φ(φ(101)) = φ(100) = 100 ×
1/2 × 4/5 = 40.
(7) Which of the following statement is a known theorem? (i.e. not a conjecture)
(A) There are infinitely many Mersenne primes.
(B) There are infinitely many perfect numbers.
(C) There does not exist odd perfect number.
(D) There are infinitely many Carmichael numbers.
(E) There are infinitely many pairs of twin primes.
(F) There are infinitely many Fermat primes.
Solution. D.
(8) If an−1 ≡ 1 (mod n), then we say n passes Fermat’s primality test for a. Which one of the following statements
is TRUE
(A) If a number n passes Fermat’s test for some a, then it must be a prime number.
(B) If a number n passes Fermat’s test for every a coprime with n, then n must be a prime number.
(C) If a number n fails Fermat’s test for some a, then n must be composite.
(D) If a number n fails Fermat’s test for some a, then n still may be a prime number.
(E) none of these.
Solution. C.
Composite numbers satisfying (B) are known as Carmichael numbers. We have known that there are infinitely
many such numbers. (D) is false because of Fermat’s Little Theorem.
“Show work” for full credit problems
1 Use the Chinese Remainder Theorem to solve the linear diophantine system

 3x ≡ 2 (mod 5)
4x ≡ 5 (mod 7)

5x ≡ 7 (mod 11)
Solution. Observe that
c1 = 4, c2 = 3, c3 = −3
n1 = 77, n2 = 55, n3 = 35
n¯1 = 3, n¯2 = −1, n¯3 = 6.
So the solution to the system, by the Chinese Remainder Theorem, can be written as
x ≡ 4 × 77 × 3 + 3 × 55 × (−1) + (−3) × 35 × 6 ≡ 129
Here 385 = 5 × 7 × 11.
(mod 385).
2 Find all the primitive roots modulo 17.
Solution. φ(17) = 16. So the possible exponents are 1, 2, 4, 8, 16.
(mod 17). So 2 is not a primitive root. Also check that 34 ≡ 13 (mod
primitive root modulo 17 since it must belong to exponent 16. Recall
residue class modulo 16. Now our theory tells that all the primitive
modulo 17 are 3, 10, 5, 11, 14, 7, 12, 6.
Check that 24 ≡ −1 (mod 17) and 28 ≡ 1
17) and 38 ≡ 16 (mod 17). Therefore 3 is a
that {1, 3, 5, 7, 9, 11, 13, 15} forms a reduced
roots are 3, 33 , 35 , 37 , 39 , 311 , 313 , 315 , which
3 Find the prime factorization for 31!.
Solution.
b
31
31
31
31
c + b c + b c + b c = 15 + 7 + 3 + 1 = 26
2
4
8
16
31
31
31
b c + b c + b c = 10 + 3 + 1 = 14
3
9
27
31
31
b c+b c=6+1=7
5
25
31
b c=4
7
31
b c=2
11
31
b c=2
13
31
b c=1
17
31
b c=1
19
31
b c=1
23
31
b c=1
29
31
b c = 1.
31
So
31! = 226 × 314 × 57 × 74 × 112 × 132 × 17 × 19 × 23 × 29 × 31
is the prime factorization for 31!.
4 Let p = 5 and q = 11. So n = pq = 55 and φ(n) = (p − 1)(q − 1) = 40. Bob chooses his public exponent e = 27
which is coprime with 40. Then he publishes his public key (55, 27) openly.
(a) Compute Bob’s private key.
Solution. We need to solve
27d ≡ 1 (mod 40).
Clearly we may choose d = 3 (or one may see this by Euclid’s algorithm). So (55, 3) is Bob’s private key.
(b) Suppose that Alice wants to send an integer 15 to Bob. Compute the cyphertext c she will send in the open
channel.
Solution.
c ≡ 1527 ≡ (152 )13 × 15 ≡ 513 × 15 ≡ (53 )4 × 75 ≡ 154 × 20 ≡ 52 × 20 ≡ 5
(mod 55)
So Alice will send out 5.
(c) After receiving the cyphertext c, check that Bob is able to recover the original integer 15 using his private key.
Solution. Bob will recover the original integer by computing
m ≡ cd ≡ 53 ≡ 125 ≡ 15
(mod 55).