Those Odd Odd Integers #2 Trivia Quiz

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.

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.

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.

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.

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.

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).

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.

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.

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.

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!