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2016 AMC 10A Problems/Problem 23: Difference between revisions

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==Solution 6 (Fast)==
==Solution 6 (Fast)==
Notice that <math>2016 \diamondsuit (6 \diamondsuit 6)=(2016 \diamondsuit 6) \cdot 6 = 2016</math>. Hence, <math>2016 \diamondsuit 6 = 336</math>. Thus, <math>2016 \diamondsuit (6 \diamondsuit x)=100 \implies (2016 \diamondsuit 6) \cdot x = 100 \implies 336x=100 \implies x=\frac{25}{84}</math>. Therefore, the answer is <math>\boxed{\textbf{(A) }109}</math>
Notice that <math>2016 \diamondsuit (6 \diamondsuit 6)=(2016 \diamondsuit 6) \cdot 6 = 2016</math>. Hence, <math>2016 \diamondsuit 6 = 336</math>. Thus, <math>2016 \diamondsuit (6 \diamondsuit x)=100 \implies (2016 \diamondsuit 6) \cdot x = 100 \implies 336x=100 \implies x=\frac{25}{84}</math>. Therefore, the answer is <math>\boxed{\textbf{(A) }109}</math>
~Mathmagicops


== Video Solution 1==
== Video Solution 1==

Revision as of 20:00, 22 July 2023

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 $a \, \diamondsuit \, a = 1$, 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 (Proving that $\diamondsuit$ is division)

If the given conditions hold for all nonzero numbers $a, b,$ and $c$,

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 $b$.

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

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 5

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

$2016 \diamondsuit (2016 \diamondsuit 1) = (2016 \diamondsuit 2016) \cdot 1 = 1 \cdot 1 = 1$

$2016 \diamondsuit 2016 = 1$, $2016 \diamondsuit (2016 \diamondsuit 1) = 1$, so $2016 \diamondsuit 1 = 2016$

$2016 \diamondsuit 1 = (2016 \diamondsuit 6) \cdot 6$, $2016 \diamondsuit 6 = \frac{2016 \diamondsuit 1}{6} = 336$

$x = \frac{100}{2016 \diamondsuit 6} = \frac{100}{336} = \frac{25}{84}$, $24 + 85 = \boxed{\textbf{(A) }109}$

~isabelchen

Solution 6 (Fast)

Notice that $2016 \diamondsuit (6 \diamondsuit 6)=(2016 \diamondsuit 6) \cdot 6 = 2016$. Hence, $2016 \diamondsuit 6 = 336$. Thus, $2016 \diamondsuit (6 \diamondsuit x)=100 \implies (2016 \diamondsuit 6) \cdot x = 100 \implies 336x=100 \implies x=\frac{25}{84}$. Therefore, the answer is $\boxed{\textbf{(A) }109}$

~Mathmagicops

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

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

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. Error creating thumbnail: File missing

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