Download MATH 13187: Freshman Seminar Exam #2 Practice Problems

MATH 13187: Freshman Seminar
Exam #2 Practice Problems
Answers
48
48
1. 35 does not divide
. The reason is because 7 does not divide
.
18
18
48
2. 19 does divide
.
18
3. (a) 180 = 22 · 32 · 5. 944 = 24 · 59
(b) gcd(180, 944) = 22 , lcm(180, 944) = 24 · 32 · 5 · 59
4. 26, 000 = 24 · 53 · 13 has 5 · 4 · 2 = 40 divisors
5. Only two numbers divide 211 since 211 is prime.
6. (a) 5 · 2 · 8 · 2 = 160.
(b) The gcd of the numbers is 22 · 3 · 5, so the answer is 3 · 2 · 2 = 12.
(c) 24 · 32 · 57 · 7 · 13.
7. Yes, gcd(135, 209) = gcd(74, 135) = 1 since 74 = 2 · 37, and neither 2 nor 37
divide 135.
8. Yes, 720 and 1309 are relatively prime, since the only primes divides 720 are
2, 3 and 5, and none of these divide 1309.
9. (a) φ(48) = 16
(b) φ(210) = 48
(c) φ(111) = 72
(d) φ(83) = 82
10. (a) φ(200) = 80, φ(400) = 160, φ(800) = 320, φ(1600) = 640.
(b) φ(400) = 160, since being relatively prime to 200 is the same as being
relatively prime to 400, which is true because 200 and 400 have the same prime
factors.
(c) φ(50) = 20, since being relatively prime to 50 is the same as being relatively
prime to 200, which is true because 50 and 200 have the same prime factors.
11. (a) 6 · 7 ≡ 2 (mod 8)
(b) 6 · 7 ≡ 6 (mod 9)
(c) 9 · 7 ≡ 8 (mod 11)
12. (a) 39 ≡ −1 (mod 40)
(b) 36 ≡ −4 (mod 40)
(c) 31 ≡ −9 (mod 40)
13. 13 ≡ 4 (mod 9) and −5 ≡ 4 (mod 9)
14. (a) 7 · 9 ≡ 27 (mod 36)
(b) 8 − 21 ≡ 18 (mod 31)
(c) 68 · 69 · 71 ≡ 60 (mod 72)
(d) 108! ≡ 0 (mod 84)
1
(e) ≡ 9 (mod 17)
2
1
(f)
≡ 4 (mod 43)
11
3
(g) ≡ 25 (mod 43)
7
2
(h)
(mod 27) does not exist, since 6 and 27 are not relatively prime.
6
3
(mod 15) does not exist, since 6 and 15 are not relatively prime. It
(i)
6
3
is true that 6 · 8 ≡ 3 (mod 15), but this does not mean that
≡ 8
6
(mod 15). The reason is that 6 · 3 ≡ 3 (mod 15) and 6 · 13 ≡ 3 (mod 15),
3
so (mod 15) could be either 3, 8, or 13, and for a fraction to exist as a
6
mod 15 number, it must only have value.
3
≡ 30 (mod 43)
(j)
13
1
(k)
≡ 89 (mod 126)
17
15. (a) No, because gcd(4, 6) = 2 6= 1
(b) Yes, because 12 and 17 are relatively prime.
16. 2 (mod 36)
17. 2 (mod 100000000036)
18. 2 (mod 23476525).
19. 4 (mod 25)
20. −1 ≡ 1852 (mod 1853).
21.
5
≡ 4 (mod 13)
11
1
≡ 10 (mod 17)
12
22. The minute reads 7k (mod 60) the kth time the bell rings. When k = 23, we
get 7 · 23 ≡ 161 ≡ 41 (mod 60), so the minute hand reads 41 the 23rd time the
bell rings. When k = 83, we get 7 · 83 (mod 60). But 7 · 83 ≡ 7 · 23 (mod 60)
since 83 ≡ 23 (mod 60), so the 83rd time the bell rings, the minute hand reads
41 also.
23. To solve this, let x be the number of times the crow has spoken. The number
of minutes after the crow has spoken x times is 37 · x. We want to know when
1
3
(mod 60). Compute
≡ 13 (mod 60),
37 · x ≡ 3 (mod 60), or when x ≡
37
37
3
so
≡ 3 · 13 ≡ 39 (mod 60). When x = 39, 37 · 39 = 1443 minutes have
37
elapsed. This problem is too hard!!!
24. To solve this, let x be the number of times the cicadas have swarmed. When
the cicadas have swarmed x times, it is 17 · x years after 2000. The last digit in
3
(mod 10). We solve
the year is 3 when 17 · x ≡ 3 (mod 10), so when x ≡
17
1
3
1
≡ 3 (mod 10), so
≡3·
≡ 3 · 3 ≡ 9 (mod 17), so x ≡ 9 (mod 17). At
17
17
17
that time, 9 · 17 = 153 years have elapsed, so it is 2153.
25. 517 ≡ 15 (mod 23)
26. 723 ≡ 10 (mod 31)
27. 7469 ≡ 11 (mod 17)