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Quiz about Those Odd Odd Integers 1
Quiz about Those Odd Odd Integers 1

Those Odd Odd Integers #1 Trivia Quiz


All questions have to deal with the positive odd integers: 1, 3, 5, 7, 9, ... . Some questions require a little thought and a pencil and paper. None of the questions in this quiz require calculus. Good Luck!

A multiple-choice quiz by rodney_indy. Estimated time: 5 mins.
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Author
rodney_indy
Time
5 mins
Type
Multiple Choice
Quiz #
266,994
Updated
Dec 03 21
# Qns
10
Difficulty
Difficult
Avg Score
5 / 10
Plays
356
Last 3 plays: colbymanram (1/10), Guest 75 (4/10), stedman (3/10).
Question 1 of 10
1. Which of the following is the 2007th positive odd integer? Hint


Question 2 of 10
2. If x is a positive odd integer, then which of the following is the next consecutive odd integer? Hint


Question 3 of 10
3. Which of the following is true about a number that is 1 less than the square of a positive odd integer? Hint


Question 4 of 10
4. A right triangle has sides whose lengths are all positive odd integers. Which of the following could be the length of its hypotenuse? Hint


Question 5 of 10
5. A triangle has sides whose lengths are all consecutive odd integers. What is the smallest possible perimeter of such a triangle? Hint


Question 6 of 10
6. What is the sum of the first 200 positive odd integers? Hint


Question 7 of 10
7. What is the average of the first 200 positive odd integers? Hint


Question 8 of 10
8. The number 2007 can be written as a sum of consecutive positive odd integers in more than one way. What is the largest number of positive odd integers in such a sum? Hint


Question 9 of 10
9. The sum of the squares of the first n positive integers can be shown to be
n(n+1)(2n+1)/6. Which of the following is the sum of the squares of the first n positive odd integers?
Hint


Question 10 of 10
10. Final question. Which of the following is the product of the first n positive odd integers?

[Recall that k! is read "k factorial" and means k * (k - 1) * ... * 1].
Hint



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Most Recent Scores
Mar 10 2024 : colbymanram: 1/10
Mar 07 2024 : Guest 75: 4/10
Feb 09 2024 : stedman: 3/10

Score Distribution

quiz
Quiz Answer Key and Fun Facts
1. Which of the following is the 2007th positive odd integer?

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.
2. If x is a positive odd integer, then which of the following is the next consecutive odd integer?

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.
3. Which of the following is true about a number that is 1 less than the square of a positive 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".
4. A right triangle has sides whose lengths are all positive odd integers. Which of the following could be the length of its hypotenuse?

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.
5. A triangle has sides whose lengths are all consecutive odd integers. What is the smallest possible perimeter of such a triangle?

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.
6. What is the sum of the first 200 positive odd integers?

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.
7. What is the average of the first 200 positive odd integers?

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.
8. The number 2007 can be written as a sum of consecutive positive odd integers in more than one way. What is the largest number of positive odd integers in such a sum?

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.
9. The sum of the squares of the first n positive integers can be shown to be n(n+1)(2n+1)/6. Which of the following is the sum of the squares of the first n positive odd 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
10. Final question. Which of the following is the product of the first n positive odd integers? [Recall that k! is read "k factorial" and means k * (k - 1) * ... * 1].

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!
Source: Author rodney_indy

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