Those Odd Odd Integers #1 Trivia Quiz

Answer: 4013

The nth positive odd integer is 2*n - 1. For example, the fifth positive integer is 9: 1, 3, 5, 7, 9 and 9 = 2*5 - 1. So the 2007th positive odd integer is 2*2007 - 1 = 4014 - 1 = 4013.

Answer: x + 2

Odd integers differ by 2. For example, 17 = 15 + 2 is the next consecutive odd integer after 15. So add 2 to x. Note that x + 1 and 2x will both be even if x is odd. x + 4 will be odd, but it is not the next consecutive odd integer.

Answer: All of these

Let n be a positive odd integer. Consider x = n^2 - 1. n^2 - 1 factors as a difference of squares:

x = n^2 - 1 = (n - 1)(n + 1)

Since n is odd, both n - 1 and n + 1 are even. Hence x is a product of two consecutive even integers.

Note by the same result, x is also divisible by 4: Both n - 1 and n + 1 are divisible by 2, which means n - 1 = 2r and n + 1 = 2s for some integers r and s. Thus

x = (n - 1)(n + 1) = (2r)(2s) = 4rs

Hence x is divisible by 4.

Now let's see why it must end in 0 or 4 or 8. Modulo 10, n is congruent to 1, 3, 5, 7, or 9. [This means that n leaves those remanders when divided by 10].
So n^2 is congruent to 1, 9, 25, 49, or 81 (mod 10), hence n^2 is congruent to 1, 5, or 9 (mod 10). Thus x = n^2 - 1 is congruent to 0, 4, or 8 (mod 10).

To learn more about modular arithmetic, check out books on "Number Theory".

Answer: No such triangle exists

Suppose there exists a right triangle all of whose sides have lengths that are positive odd integers. Let a, b be the lengths of the legs of the right triangle and let c be the length of the hypotenuse. Then by the Pythagorean Theorem, c^2 = a^2 + b^2. But a and b are both odd, so their squares are both odd. Therefore, a^2 + b^2 is even. But c is also odd, so c^2 is odd. This is a contradiction. Hence no such triangle exists.

Answer: 15

Recall that if a and b are the lengths of two sides of a triangle, then the length of the third side, which I'll call c, must satisfy

|a - b| is less than c is less than a + b

It's not possible for a triangle to have sides of lengths 1, 3, and 5 since 1 + 3 is less than 5. But there is no problem for a triangle with sides of lengths 3, 5, and 7. Its perimeter is 3 + 5 + 7 = 15.

Answer: 40000

The sum of the first n positive odd integers is n squared. This can be proven by a method called mathematical induction. I will denote the statement to be proved by P(n).

P(n): 1 + 3 + ... + (2n - 1) = n^2

You begin by proving the case where n = 1, but this is obvious here since the first odd integer is 1 which equals 1 squared.

P(1): 1 = 1^2 So P(1) is true.

Now we assume the result is true for n = k, that is,

P(k): 1 + 3 + ... + (2k - 1) = k^2.

[Note from question 1 I'm using the fact that the kth positive odd integer is 2k - 1]. Now add 2k + 1 to both sides of this equation to get

1 + 3 + ... + (2k - 1) + (2k + 1) = k^2 + 2k + 1.

The right hand side is a perfect square, so we now have

1 + 3 + ... + (2k - 1) + (2k + 1) = (k + 1)^2

This statement is just P(k + 1). Hence the statement is proved by mathematical induction. The principle of mathematical induction states the following:

If P(n) is a statement involving the natural numbers (n a natural number) for which

1. P(1) is true
2. P(k) true implies P(k + 1) true for all natural numbers k >= 1

then P(n) is true for every natural number n.

The principle of mathematical induction can be proved by using the fact that any subset of the natural numbers has a least element (the well-ordering principle). Mathematical induction is an important method of proof in mathematics and is fundamental in higher mathematics courses.

Now using the above result, the sum of the first 200 positive odd integers is just 200^2 = 40000.

Answer: 200

By question 7, the sum of the first n positive odd integers is n^2. Thus their average is n^2/n = n. So the answer is 200.

Answer: 9

The number 2007 can be factored the following ways: 1 * 2007, 3 * 669, and 9 * 223. The average of a list of an odd number of consecutive odd integers is the middle number in the list. So with these factorizations, we have 2007, 669, and 223 are the middle numbers of lists of consecutive numbers that have lengths of 1, 3, and 9 respectively. So here are the three ways we can write 2007 as a sum of consecutive positive integers:

2007 (itself)

667 + 669 + 671

215 + 217 + 219 + 221 + 223 + 225 + 227 + 229 + 231

Note that the middle numbers cannot be 1, 3, or 9, since then the lists would contain negative integers.

Answer: n(2n - 1)(2n + 1)/3

Here's the computation:

1^2 + 3^2 + ... + (2n - 1)^2

= 1^2 + 2^2 + 3^2 + 4^2 + ... + (2n - 1)^2 - (2^2 + 4 ^2 + ... + (2n - 2)^2)

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

=(2n - 1)(2n)(2(2n - 1) + 1)/6 - 4 * (n - 1)n(2(n - 1) + 1)/6

[Here I used the given result twice that 1^2 + 2^2 + ... + k^2 = k(k + 1)(2k + 1)/6. I used it for k = 2n - 1 and for k = n - 1].

= 2n(2n - 1)(4n - 1)/6 - 4n(n - 1)(2n - 1)/6

= (2n(2n - 1)/6) * (4n - 1 - 2(n - 1)

[I factored out the common factors]

= (n(2n - 1)/3) * (4n - 1 - 2n + 2)

= n(2n - 1)(2n + 1)/3

Answer: (2n - 1)! / (2^(n - 1) * (n - 1)! )

1 * 3 * 5 * ... * (2n - 1) =

1 * 2 * 3 * 4 * ... * (2n - 1)
--------------------------------- =
2 * 4 * ... * (2n - 2)


1 * 2 * 3 * 4 * ... * (2n - 1)
----------------------------------- =
2^(n - 1)* 1 * 2 * ... * (n - 1)


(2n - 1)!
------------------
2^(n-1) * (n - 1)!


I hope you enjoyed my first quiz! Thank you for playing!