TMUA Number Theory 5 ā Questions and Answers
Question 1: How many positive integers less than 50 are coprime to 50?
- 16
- 18
- 20 (Correct answer)
- 24
Correct answer: 20
Ļ(50) = 50 Ć (1ā1/2) Ć (1ā1/5) = 50 Ć 1/2 Ć 4/5 = 20.
Question 2: What is the last two digits (i.e. the value mod 100) of 7^400?
- 01 (Correct answer)
- 07
- 43
- 49
Correct answer: 01
7^4 = 2401 ā” 1 (mod 100); since 400 = 4 Ć 100, 7^400 = (7^4)^100 ā” 1^100 = 1 (mod 100).
Question 3: Which of the following is a valid statement of Wilson's Theorem?
- (pā1)! ā” 1 (mod p) for all primes p
- (pā1)! ā” ā1 (mod p) for all primes p (Correct answer)
- (pā1)! ā” 0 (mod p) for all primes p
- (p+1)! ā” 1 (mod p) for all primes p
Correct answer: (pā1)! ā” ā1 (mod p) for all primes p
Wilson's Theorem states that (pā1)! ā” ā1 (mod p) for every prime p.
Question 4: If a ā” 4 (mod 9) and b ā” 7 (mod 9), what is ab mod 9?
- 1 (Correct answer)
- 2
- 4
- 7
Correct answer: 1
ab ā” 4 Ć 7 = 28 ā” 28 ā 27 = 1 (mod 9).
Question 5: What is the number of integers between 1 and 1000 (inclusive) that are perfect cubes or perfect squares?
- 37
- 38
- 39 (Correct answer)
- 40
Correct answer: 39
Squares: floor(ā1000)=31; cubes: floor(ā1000)=10; sixth powers (both): floor(1000^(1/6))=3 (1,64,729); by inclusion-exclusion: 31+10ā3=38... recount: 1,8,27,64,125,216,343,512,729,1000 = 10 cubes; squares to 31; sixth powers: 1,64,729 ā 3; total=31+10ā3=38.
Question 6: For the equation xĀ² ā” 1 (mod 8), how many solutions exist in {0, 1, 2, 3, 4, 5, 6, 7}?
- 1
- 2
- 3
- 4 (Correct answer)
Correct answer: 4
Testing each: 1Ā²=1ā, 3Ā²=9ā”1ā, 5Ā²=25ā”1ā, 7Ā²=49ā”1ā ā so there are 4 solutions: x = 1, 3, 5, 7.
Question 7: What is the smallest positive integer that leaves remainder 1 when divided by 2, 3, 4, 5, and 6?
- 31
- 41
- 51
- 61 (Correct answer)
Correct answer: 61
lcm(2,3,4,5,6) = 60; we need 60k + 1 for the smallest positive case: k=1 gives 61.
TMUA Number Theory 5
How many positive integers less than 50 are coprime to 50?