Difference between revisions of "2023 AMC 12B Problems/Problem 15"

Revision as of 20:50, 15 November 2023

The following problem is from both the 2023 AMC 10B #18 and 2023 AMC 12B #15, so both problems redirect to this page.

Problem

Suppose $a$ , $b$ , and $c$ are positive integers such that $\frac{a}{14}+\frac{b}{15}=\frac{c}{210}.$ Which of the following statements are necessarily true?

I. If $\gcd(a,14)=1$ or $\gcd(b,15)=1$ or both, then $\gcd(c,210)=1$ .

II. If $\gcd(c,210)=1$ , then $\gcd(a,14)=1$ or $\gcd(b,15)=1$ or both.

III. $\gcd(c,210)=1$ if and only if $\gcd(a,14)=\gcd(b,15)=1$ .

$\textbf{(A)}~\text{I, II, and III}\qquad\textbf{(B)}~\text{I only}\qquad\textbf{(C)}~\text{I and II only}\qquad\textbf{(D)}~\text{III only}\qquad\textbf{(E)}~\text{II and III only}$

Solution 1 (Guess and check + Contrapositive)

$I.$ Try $a=3,b=5 => c = 17\cdot15$ which makes $\textbf{I}$ false. At this point, we can rule out answer A,B,C.

$II.$ A => B or C. equiv. ~B AND ~C => ~A. Let a = 14, b=15 (statisfying ~B and ~C). => C = 2*210. which is ~A.

$II$ is true.

So the answer is E. $\boxed{\textbf{(E) } II \text{ and } III \text{only}.}$ ~Technodoggo

Solution 2

The equation given in the problem can be written as $15 a + 14 b = c. \hspace{1cm} (1)$

$\textbf{First, we prove that Statement I is not correct.}$

A counter example is $a = 1$ and $b = 3$ . Thus, ${\rm gcd} (c, 210) = 3 \neq 1$ .

$\textbf{Second, we prove that Statement III is correct.}$

First, we prove the ``if part.

Suppose ${\rm gcd}(a , 14) = 1$ and ${\rm gcd}(b, 15) = 1$ . However, ${\rm gcd} (c, 210) \neq 1$ .

Thus, $c$ must be divisible by at least one factor of 210. W.L.O.G, we assume $c$ is divisible by 2.

Modulo 2 on Equation (1), we get that $2 | a$ . This is a contradiction with the condition that ${\rm gcd}(a , 14) = 1$ . Therefore, the ``if part in Statement III is correct.

Second, we prove the ``only if part.

Suppose ${\rm gcd} (c, 210) \neq 1$ . Because $210 = 14 \cdot 15$ , there must be one factor of 14 or 15 that divides $c$ . W.L.O.G, we assume there is a factor $q > 1$ of 14 that divides $c$ . Because ${\rm gcd} (14, 15) = 1$ , we have ${\rm gcd} (q, 15) = 1$ . Modulo $q$ on Equation (1), we have $q | a$ .

Because $q | 14$ , we have ${\rm gcd}(a , 14) \geq q > 1$ .

Analogously, we can prove that ${\rm gcd}(b , 15) > 1$ .

$\textbf{Third, we prove that Statement II is correct.}$

This is simply a special case of the ``only if part of Statement III. So we omit the proof.

All analyses above imply $\boxed{\textbf{(E) II and III only}}.$

~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 $a=1, b=5, c=85$ ), 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 $\boxed{\textbf{(E) II and III only}}.$

~SpencerD. ~edited by A_MatheMagician

@@ Line 72: / Line 72: @@
 ~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
-==Solution 3==
+==Solution 3 (Answer Choices)==
-It can easily be shown that statement I is false (a counterexample would be <math>a=1, b=5, c=85</math>), 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. Therefore the answer is
+It can easily be shown that statement I is false (a counterexample would be <math>a=1, b=5, c=85</math>), 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
 <math>\boxed{\textbf{(E) II and III only}}.</math>
-~SpencerD.
+~SpencerD. ~edited by A_MatheMagician
 ==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

Page

Toolbox

Search