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Difference between revisions of "2016 AMC 10A Problems/Problem 23"

(→‎Solution 1)
(Added third solution; does not involve finding the identity of the operator.)
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Hence, the given equation becomes <math>\frac{2016}{\frac{6}{x}} = 100</math>.  Solving yields <math>x=\frac{100}{336} = \frac{25}{84},</math> so the answer is <math>25 + 84 = \boxed{\textbf{(A) }109.}</math>
 
Hence, the given equation becomes <math>\frac{2016}{\frac{6}{x}} = 100</math>.  Solving yields <math>x=\frac{100}{336} = \frac{25}{84},</math> so the answer is <math>25 + 84 = \boxed{\textbf{(A) }109.}</math>
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 +
==Solution 3==
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 +
One way to eliminate the <math>\diamondsuit</math> in this equation is to make <math>a = b</math> so that  <math>a\,\diamondsuit\, (b\,\diamondsuit \,c) = c</math>. In this case, we can make <math>b = 2016</math>.
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<math>2016 \,\diamondsuit\, (6\,\diamondsuit\, x)=100\vspace{1mm}\\
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(2016\, \diamondsuit\, 6) \cdot x = 100</math>
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By multiplying both sides by <math>\frac{6}{x}</math>, we get:
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 +
<math>(2016\, \diamondsuit\, 6) \cdot 6 = \frac{600}{x}\vspace{2mm}\\
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2016 \, \diamondsuit\, (6\, \diamondsuit\, 6) = \frac{600}{x}</math>
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 +
Because <math>6\, \diamondsuit\, 6 = 2016\, \diamondsuit\, 2016 = 1</math>:
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<math>2016 \, \diamondsuit\, (2016\, \diamondsuit\, 2016) = \frac{600}{x}\vspace{2mm}\\
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(2016\, \diamondsuit\, 2016) \cdot 2016 = \frac{600}{x}\vspace{1mm}\\
 +
2016 = \frac{600}{x}</math>
 +
 +
Therefore, <math>x = \frac{600}{2016} = \frac{25}{84}</math>, so the answer is <math>25 + 84 = \boxed{\textbf{(A) }109.}</math>
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==See Also==
 
==See Also==

Revision as of 16:18, 7 February 2016

Problem

A binary operation $\diamondsuit$ has the properties that $a\,\diamondsuit\, (b\,\diamondsuit \,c) = (a\,\diamondsuit \,b)\cdot c$ and that $a\,\diamondsuit \,a=1$ for all nonzero real numbers $a, b,$ and $c$. (Here $\cdot$ represents multiplication). The solution to the equation $2016 \,\diamondsuit\, (6\,\diamondsuit\, x)=100$ can be written as $\tfrac{p}{q}$, where $p$ and $q$ are relatively prime positive integers. What is $p+q?$

$\textbf{(A) }109\qquad\textbf{(B) }201\qquad\textbf{(C) }301\qquad\textbf{(D) }3049\qquad\textbf{(E) }33,601$

Solution 1

We see that $a \diamond a = 1$, and think of division. Testing, we see that the first condition $a \diamond (b \diamond c) = (a \diamond b) \cdot c$ is satisfied, because $\frac{a}{\frac{b}{c}} = \frac{a}{b} \cdot c$. Therefore, division is the operation $\diamond$. Solving the equation, \[\frac{2016}{\frac{6}{x}} = \frac{2016}{6} \cdot x = 336x = 100\implies x=\frac{100}{336} = \frac{25}{84},\] so the answer is $25 + 84 = \boxed{\textbf{(A) }109.}$

Solution 2

We can manipulate the given identities to arrive at a conclusion about the binary operator $\diamondsuit$. Substituting $b = c$ into the second identity yields $( a\ \diamondsuit\ b) \cdot b = a\ \diamondsuit\ (b\ \diamondsuit\ b) = a\ \diamondsuit\ 1 = a\ \diamondsuit\ ( a\ \diamondsuit\ a) = ( a\ \diamondsuit\ a) \cdot a = a$. Hence, $( a\ \diamondsuit\ b) \cdot b = a,$ or, dividing both sides of the equation by $b,$ $( a\ \diamondsuit\ b) = \frac{a}{b}.$

Hence, the given equation becomes $\frac{2016}{\frac{6}{x}} = 100$. Solving yields $x=\frac{100}{336} = \frac{25}{84},$ so the answer is $25 + 84 = \boxed{\textbf{(A) }109.}$

Solution 3

One way to eliminate the $\diamondsuit$ in this equation is to make $a = b$ so that $a\,\diamondsuit\, (b\,\diamondsuit \,c) = c$. In this case, we can make $b = 2016$.

$2016 \,\diamondsuit\, (6\,\diamondsuit\, x)=100\vspace{1mm}\\ (2016\, \diamondsuit\, 6) \cdot x = 100$

By multiplying both sides by $\frac{6}{x}$, we get:

$(2016\, \diamondsuit\, 6) \cdot 6 = \frac{600}{x}\vspace{2mm}\\ 2016 \, \diamondsuit\, (6\, \diamondsuit\, 6) = \frac{600}{x}$

Because $6\, \diamondsuit\, 6 = 2016\, \diamondsuit\, 2016 = 1$:

$2016 \, \diamondsuit\, (2016\, \diamondsuit\, 2016) = \frac{600}{x}\vspace{2mm}\\ (2016\, \diamondsuit\, 2016) \cdot 2016 = \frac{600}{x}\vspace{1mm}\\ 2016 = \frac{600}{x}$

Therefore, $x = \frac{600}{2016} = \frac{25}{84}$, so the answer is $25 + 84 = \boxed{\textbf{(A) }109.}$


See Also

2016 AMC 10A (Problems • Answer Key • Resources)
Preceded by
Problem 22 		Followed by
Problem 24
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
2016 AMC 12A (Problems • Answer Key • Resources)
Preceded by
Problem 19 		Followed by
Problem 21
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

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