2023 AMC 12B Problems/Problem 15
- The following problem is from both the 2023 AMC 10B #18 and 2023 AMC 12B #15, so both problems redirect to this page.
Contents
Problem
Suppose , , and are positive integers such thatWhich of the following statements are necessarily true?
I. If or or both, then .
II. If , then or or both.
III. if and only if .
Video Solution by MegaMath
https://www.youtube.com/watch?v=le0KSx3Cy-g&t=28s
Solution 1 (Guess and check + Contrapositive)
We examine each of the conditions.
The first condition is false. A simple counterexample is and . The corresponding value of is . Clearly, and , so condition would imply that However, is clearly not (they share a common factor of ). Obviously, condition is false, so we can rule out choices and .
We are now deciding between the two answer choices and . What differs between them is the validity of condition , so it suffices to simply check .
We look at statement 's contrapositive to prove it. The contrapositive states that if and , then In other words, if shares some common factor that is not with and shares some common factor that is not with , then also shares a common factor with . Let's say that , where is a factor of not equal to . (So is the common factor.)
We can rewrite the given equation as We can express as , for some positive integer (this can be ). We can factor out to get
We know that all values in this equation are integers, so must be divisible by . Since is a factor of , must also be a factor of , a multiple of . Therefore, we know that shares a common factor with (which is ), so . This is what states, so therefore is true.
Thus, our answer is ~Technodoggo
Solution 2
The equation given in the problem can be written as
A counter example is and . Thus, .
First, we prove the ``if part.
Suppose and . However, .
Thus, must be divisible by at least one factor of 210. W.L.O.G, we assume is divisible by 2.
Modulo 2 on Equation (1), we get that . This is a contradiction with the condition that . Therefore, the ``if part in Statement III is correct.
Second, we prove the ``only if part.
Suppose . Because , there must be one factor of 14 or 15 that divides . W.L.O.G, we assume there is a factor of 14 that divides . Because , we have . Modulo on Equation (1), we have .
Because , we have .
Analogously, we can prove that .
This is simply a special case of the ``only if part of Statement III. So we omit the proof.
All analyses above imply
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
Solution 3 (Answer Choices)
It can easily be shown that statement I is false (a counterexample would be ), meaning the only viable answer choices are D and E. Since both of these answer choices include statement III, this means III is true. Since III is true, this actually implies that statement II is true, as III is just a stronger version of statement II (or it's contrapositive, to be precise). Therefore the answer is
~SpencerD. ~edited by A_MatheMagician
Solution 4
First, we will show that statement I is false. A simple counterexample is and . Then, is even, and it is not relatively prime with .
Now, we will focus on statement III. It's pretty easy to realize that if either isn't relatively prime to or isn't relatively prime to , it will always share factors with because (think ). Therefore, statement III must be true.
Since statement III is a stricter case of statement II, the last two statements are true, hence choose
~xHypotenuse
Video Solution
https://youtu.be/pg1pyRa6C24?si=bOM1mQkWzkYZzfv_
~MathProblemSolvingSkills.com
Video Solution 2 by OmegaLearn
Video Solution
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
See Also
2023 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 17 |
Followed by Problem 19 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
All AMC 10 Problems and Solutions |
2023 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 14 |
Followed by Problem 16 |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | |
All AMC 12 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.