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| == Problem ==
| | #redirect[[2023 AMC 12B Problems/Problem 15]] |
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| Suppose 𝑎, 𝑏, and 𝑐 are positive integers such that
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| <math>\dfrac{a}{14}+\dfrac{b}{15}=\dfrac{c}{210}</math>.
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| Which of the following statements are necessarily true?
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| I. If gcd(𝑎, 14) = 1 or gcd(𝑏, 15) = 1 or both, then gcd(𝑐, 210) = 1.
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| II. If gcd(𝑐, 210) = 1, then gcd(𝑎, 14) = 1 or gcd(𝑏, 15) = 1 or both.
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| III. gcd(𝑐, 210) = 1 if and only if gcd(𝑎, 14) = gcd(𝑏, 15) = 1.
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| == Solution 1 (Guess and check + Contrapositive)==
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| [[being revised]]
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| ~Technodoggo
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| ==Solution 2==
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| The equation given in the problem can be written as
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| <cmath>
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| \[
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| 15 a + 14 b = c. \hspace{1cm} (1)
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| \]
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| </cmath>
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| <math>\textbf{First, we prove that Statement I is not correct.}</math>
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| A counter example is <math>a = 1</math> and <math>b = 3</math>.
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| Thus, <math>{\rm gcd} (c, 210) = 3 \neq 1</math>.
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| <math>\textbf{Second, we prove that Statement III is correct.}</math>
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| First, we prove the ``if'' part.
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| Suppose <math>{\rm gcd}(a , 14) = 1</math> and <math>{\rm gcd}(b, 15) = 1</math>. However, <math>{\rm gcd} (c, 210) \neq 1</math>.
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| Thus, <math>c</math> must be divisible by at least one factor of 210. W.L.O.G, we assume <math>c</math> is divisible by 2.
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| Modulo 2 on Equation (1), we get that <math>2 | a</math>.
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| This is a contradiction with the condition that <math>{\rm gcd}(a , 14) = 1</math>.
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| Therefore, the ``if'' part in Statement III is correct.
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| Second, we prove the ``only if'' part.
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| Suppose <math>{\rm gcd} (c, 210) \neq 1</math>. Because <math>210 = 14 \cdot 15</math>, there must be one factor of 14 or 15 that divides <math>c</math>.
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| W.L.O.G, we assume there is a factor <math>q > 1</math> of 14 that divides <math>c</math>.
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| Because <math>{\rm gcd} (14, 15) = 1</math>, we have <math>{\rm gcd} (q, 15) = 1</math>.
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| Modulo <math>q</math> on Equation (1), we have <math>q | a</math>.
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| Because <math>q | 14</math>, we have <math>{\rm gcd}(a , 14) \geq q > 1</math>.
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| Analogously, we can prove that <math>{\rm gcd}(b , 15) > 1</math>.
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| <math>\textbf{Third, we prove that Statement II is correct.}</math>
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| This is simply a special case of the ``only if'' part of Statement III. So we omit the proof.
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| All analysis above imply
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| <math>\boxed{\textbf{(E) II and III only}}.</math>
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| ~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
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