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

Revision as of 04:08, 5 May 2020

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 $p+q?$

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

Solutions

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 can be 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 first 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 $b = 2016$ .

$2016 \,\diamondsuit\, (6\,\diamondsuit\, x)=100\implies (2016\, \diamondsuit\, 6) \cdot x = 100$

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

$(2016\, \diamondsuit\, 6) \cdot 6 = \frac{600}{x}\implies 2016 \, \diamondsuit\, (6\, \diamondsuit\, 6) = \frac{600}{x}$

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

$2016 \, \diamondsuit\, (2016\, \diamondsuit\, 2016) = \frac{600}{x}\implies (2016\, \diamondsuit\, 2016) \cdot 2016 = \frac{600}{x}\implies 2016 = \frac{600}{x}$

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

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

@@ Line 37: / Line 37: @@
 ==See Also==
+Video Solution:
+https://www.youtube.com/watch?v=8GULAMwu5oE
 {{AMC10 box|year=2016|ab=A|num-b=22|num-a=24}}
 {{AMC12 box|year=2016|ab=A|num-b=19|num-a=21}}
 {{MAA Notice}}

Page

Toolbox

Search