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

Revision as of 12:14, 8 October 2022

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 $a \, \diamondsuit \, a = 1$ , and think of division. Testing, we see that the first condition $a \, \diamondsuit \, (b \, \diamondsuit \, c) = (a \, \diamondsuit \, b) \cdot c$ is satisfied, because $\frac{a}{\frac{b}{c}} = \frac{a}{b} \cdot c$ . Therefore, division can be the operation $\diamondsuit$ . 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 $25 + 84 = \boxed{\textbf{(A) }109.}$

Solution 4 (Proving that $\diamondsuit$ is division)

Since given conditions hold for all nonzero numbers $a, b,$ and $c$ , they're valid even when the variables aren't distinct.

Let $a=b=c.$ From the first two givens, this implies that

$a\diamondsuit\, (a\diamondsuit\, {a})=(a\diamondsuit\, a)\cdot{a}.$

From $a\diamondsuit\,{a}=1,$ this equation simply becomes $a\diamondsuit\,{1}=a.$

Let $c=b.$ Substituting this into the first two conditions, we see that

$a\diamondsuit\, (b\diamondsuit\, {c})=(a\diamondsuit\, {b})\cdot{c} \implies a\diamondsuit\, (b\diamondsuit\, {b})=(a\diamondsuit\, {b})\cdot{b}.$

Substituting $b\diamondsuit\, {b} =1$ , the second equation becomes

$a\diamondsuit\, {1}=(a\diamondsuit\, {b})\cdot{b} \implies a=(a\diamondsuit\,{b})\cdot{b}.$

Since $a, b$ and $c$ are nonzero, we can divide by $b$ which yields,

$\frac{a}{b}=(a\diamondsuit\, {b}).$

Now we can find the value of $x$ straightforwardly:

$\frac{2016}{(\frac{6}{x})}=100 \implies 2016=\frac{600}{x} \implies x=\frac{600}{2016} = \frac{25}{84}.$

Therefore, $a+b=25+84=\boxed{\textbf{A)} 109}$

-Benedict T (countmath1)

Note: We only really cared about what $a\diamondsuit\,{b}$ was, so we used the existence of $c$ to get an expression in terms of just $a$ and $b$ .

Video Solution 1

https://www.youtube.com/watch?v=8GULAMwu5oE

Video Solution 2(Meta-Solving Technique)

https://youtu.be/GmUWIXXf_uk?t=1632

~ pi_is_3.14

@@ Line 65: / Line 65: @@
 -Benedict T (countmath1)
-Note: We only really cared about what <math>a\diamondsuit\,{b}</math> was, so we only used the existence of <math>c</math> to get an expression in terms of just <math>a</math> and <math>b</math>.
+Note: We only really cared about what <math>a\diamondsuit\,{b}</math> was, so we used the existence of <math>c</math> to get an expression in terms of just <math>a</math> and <math>b</math>.
 == Video Solution 1==

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

Revision as of 12:14, 8 October 2022

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