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Quiz about Those Odd Odd Integers 2
Quiz about Those Odd Odd Integers 2

Those Odd Odd Integers #2 Trivia Quiz


These questions involve the positive odd integers 1, 3, 5, ... . Most of these questions require more mathematics than my first quiz, but you'll learn some interesting results along the way! You will need pencil and paper for some of these. Good Luck!

A multiple-choice quiz by rodney_indy. Estimated time: 6 mins.
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Author
rodney_indy
Time
6 mins
Type
Multiple Choice
Quiz #
267,083
Updated
Dec 03 21
# Qns
10
Difficulty
Difficult
Avg Score
4 / 10
Plays
352
Question 1 of 10
1. It is well known that there are infinitely many primes. Since 2 is the only even prime, there are infinitely many primes among the positive odd integers. Which of the following arithmetic progressions of odd integers contains infinitely many primes, in addition to composite numbers? Hint


Question 2 of 10
2. How many odd integers are greater than 207 and less than 2007? Hint


Question 3 of 10
3. How many positive odd integers less than or equal to 2007 are not divisible by 3? Hint


Question 4 of 10
4. Let's take the list of the first 2007 positive odd integers and concatenate them to form one very large number: 135791113151719... . How many digits does this number have? Hint


Question 5 of 10
5. Let S be the set containing the first eight positive integers. How many subsets of S contain only odd integers? Hint


Question 6 of 10
6. Let f(x) = sin(x). Let N be a positive odd integer. What is the value of the Nth derivative of f evaluated at 0? Hint


Question 7 of 10
7. Let f(x) = cos(x). Let N be a positive odd integer. What is the value of the Nth derivative of f evaluated at 0? Hint


Question 8 of 10
8. Consider the following series- The alternating sum of the reciprocals of all positive odd integers: 1 - 1/3 + 1/5 - 1/7 + ... . Which of the following is the sum of this series? Hint


Question 9 of 10
9. Which of the following is true about the sum of the reciprocals of all positive odd integers? Hint


Question 10 of 10
10. It can be shown that the sum of the reciprocals of the squares of all positive integers is pi^2/6. What is the sum of the reciprocals of the squares of all positive odd integers? Hint



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Quiz Answer Key and Fun Facts
1. It is well known that there are infinitely many primes. Since 2 is the only even prime, there are infinitely many primes among the positive odd integers. Which of the following arithmetic progressions of odd integers contains infinitely many primes, in addition to composite numbers?

Answer: All of these

This is the result of Dirichlet's famous theorem on primes in arithmetic progressions. It says that if a and b are relatively prime, then there are infinitely many primes in an arithmetic progression whose nth term is an + b. The progressions above have nth terms 4n + 1, 10n + 1, and 1000n + 1. So by Dirichlet's Theorem, they all contain infinitely many primes.

This great (and difficult!) result belongs to analytic number theory.
2. How many odd integers are greater than 207 and less than 2007?

Answer: 899

Recall that the nth positive odd integer is 2n - 1. 207 = 2n - 1 implies n = 208/2 = 104, so 207 is the 104th positive odd integer. 2005 = 2n - 1 implies n = 2006/2 = 1003, so 2005 is the 1003rd positive odd integer. The odd integers being described here are: 209, 211, ..., 2005.

The list 1, 3, 5, ..., 2005 contains 1003 odd integers. The list 1, 3, 5, ..., 207 contains 104 odd integers. Therefore our list 209, 211, ..., 2005 contains 1003 - 104 = 899 odd integers.
3. How many positive odd integers less than or equal to 2007 are not divisible by 3?

Answer: 671

Keep in mind we're talking about the positive integers less than or equal to 2007, not all the integers less than or equal to 2007. These are precisely the positive integers less than or equal to 2007 that are not divisible by 2 and not divisible by 3. So let A be the set of all positive integers less than or equal to 2007 that are divisible by 2. Let B be the set of all positive integers less than or equal to 2007 that are divisible by 3.

Then A has 1003 elements and B has 667 elements. The intersection of A and B consists of all positive integers less than or equal to 2007 that are divisible by 6. So the intersection of A and B has 334 elements.

By the principle of inclusion/exclusion, the number of elements in the union of A and B is 1003 + 667 - 334 = 1336. We want the number of elements in the intersection of the complements of A and B, which by DeMorgan's law is just the number of elements in the complement of the union, which is 2007 - 1336 = 671. Of course, Venn diagrams can also be used to solve this problem.
4. Let's take the list of the first 2007 positive odd integers and concatenate them to form one very large number: 135791113151719... . How many digits does this number have?

Answer: 7473

The 2007th positive odd integer is 2*2007 - 1 = 4013. So this number is 13579...40114013. This number is built by concatenating one digit, two digit, three digit, and four digit numbers. We need to count the number of each type:

#one digit: 5

#two digit: 50 - 5 = 45

#three digit: 500 - 50 = 450

#four digit: Here we're counting the number of odd integers in the sequence 1001, 1003, ..., 4013. That is 2007 - 500 = 1507.

So the total number of digits of this number is

1 * 5 + 2 * 45 + 3 * 450 + 4 * 1507 = 7473.
5. Let S be the set containing the first eight positive integers. How many subsets of S contain only odd integers?

Answer: 15

The only subsets of S that contain all positive integers are subsets of the set containing 1, 3, 5, and 7. This set has 4 elements, thus it has 2^4 = 16 subsets. One of these is the empty set, hence there are 15 subsets total containing only odd integers.
6. Let f(x) = sin(x). Let N be a positive odd integer. What is the value of the Nth derivative of f evaluated at 0?

Answer: (-1)^( (N - 1)/2 )

f(x) = sin(x) has MacLaurin series sin(x) = x - x^3/3! + x^5/5! - ... . But this series is just Sum(j = 0, inf, (jth derivative of f evaluated at 0) * x^j/j!). Equating coefficients gives f^(2j - 1)(0)/(2j - 1)! = (-1)^(j + 1)/(2j - 1)!. Thus f^(2j - 1)(0) = (-1)^(j - 1). Now N = 2j - 1 which implies j = (N + 1)/2, thus j - 1 = (N - 1)/2. In other words, f^(N)(0) = (-1)^((N - 1)/2).
7. Let f(x) = cos(x). Let N be a positive odd integer. What is the value of the Nth derivative of f evaluated at 0?

Answer: 0

f(x) = cos(x) has MacLaurin series cos(x) = 1 - x^2/2! + x^4/4! - ... . But this series is just Sum(j = 0, inf, (jth derivative of f evaluated at 0) * x^j/j!). When we equate coefficients, we see that any odd derivative of f evaluated at 0 will be 0.
8. Consider the following series- The alternating sum of the reciprocals of all positive odd integers: 1 - 1/3 + 1/5 - 1/7 + ... . Which of the following is the sum of this series?

Answer: pi/4

Let f(x) = 1/(1 + x^2). f(x) represents the sum of the geometric series 1 - x^2 + x^4 - x^6 + ... . This series converges for |x| less than 1. Integrating this series gives x - x^3/3 + x^5/5 - x^7/7 + ... + C. The integral of f(x) is Arctan(x), so putting x = 0 we see the constant of integration C = 0. For x = 1, the series converges by the alternating series test. Now by Abel's Theorem, its sum is Arctan(1), which is pi/4.
9. Which of the following is true about the sum of the reciprocals of all positive odd integers?

Answer: It diverges.

The sum of the reciprocals of all positive odd integers is Sum(1,inf,1/(2n - 1)). We know that the harmonic series Sum(1, inf, 1/n) diverges. Since both series have positive terms, use the limit comparison test to show that the series of reciprocals of positive integers diverges.
10. It can be shown that the sum of the reciprocals of the squares of all positive integers is pi^2/6. What is the sum of the reciprocals of the squares of all positive odd integers?

Answer: pi^2/8

By use of the limit comparison test with Sum(1, inf, 1/n^2), we see that Sum(1, inf, 1/(2n - 1)^2) converges. In fact, both series converge absolutely since they have positive terms. So this means the computations below make sense:

1^2 + 1/3^2 + 1/5^2 + ...

= (1 + 1/2^2 + 1/3^2 + 1/4^2 + ...) - (1/2^2 + 1/4^2 + 1/6^2 + ...)

= (1 + 1/2^2 + 1/3^2 + 1/4^2 + ...) - 1/2^2 * (1 + 1/2^2 + 1/3^2 + ...)

= pi^2/6 - 1/4 * pi^2/6

= 3/4 * pi^2/6

= pi^2/8

I hope you enjoyed my quiz! Thanks for playing!
Source: Author rodney_indy

This quiz was reviewed by FunTrivia editor crisw before going online.
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