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Revision as of 20:30, 15 November 2023
Contents
Problem
Suppose 𝑎, 𝑏, and 𝑐 are positive integers such that
.
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.
Solution 1 (Guess and check + Contrapositive)
Try
which makes
false.
At this point, we can rule out answer A,B,C.
A => B or C. equiv. ~B AND ~C => ~A.
Let a = 14, b=15 (statisfying ~B and ~C). => C = 2*210. which is ~A.
is true.
So the answer is E.
~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
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. Therefore the answer is
~SpencerD.
See Also
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.