Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2016 AMC 10A Problems/Problem 23"
(→‎Solution)
Line 4: Line 4:
  
 
We see that <math>a \diamond a = 1</math>, and think of division. Testing, we see that the first condition <math>a \diamond (b \diamond c) = (a \diamond b) \cdot c</math> is satisfied, because <math>\frac{a}{\frac{b}{c}} = \frac{a}{b} \cdot c</math>. Therefore, division is the operation <math>\diamond</math>. Solving the equation, <math>\frac{2016}{\frac{6}{x}} = \frac{2016}{6} \cdot x = 336x = 100</math>, so <math>x=\frac{100}{336} = \frac{25}{84}</math>, so the answer is <math>25 + 84 = \boxed{109}</math> (A)
 
We see that <math>a \diamond a = 1</math>, and think of division. Testing, we see that the first condition <math>a \diamond (b \diamond c) = (a \diamond b) \cdot c</math> is satisfied, because <math>\frac{a}{\frac{b}{c}} = \frac{a}{b} \cdot c</math>. Therefore, division is the operation <math>\diamond</math>. Solving the equation, <math>\frac{2016}{\frac{6}{x}} = \frac{2016}{6} \cdot x = 336x = 100</math>, so <math>x=\frac{100}{336} = \frac{25}{84}</math>, so the answer is <math>25 + 84 = \boxed{109}</math> (A)
 +
 +
==See Also==
 +
{{AMC10 box|year=2016|ab=A|before=First Problem|num-a=2}}
 +
{{MAA Notice}}

Revision as of 18:13, 3 February 2016

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 $p+q?$

Solution

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, $\frac{2016}{\frac{6}{x}} = \frac{2016}{6} \cdot x = 336x = 100$, so $x=\frac{100}{336} = \frac{25}{84}$, so the answer is $25 + 84 = \boxed{109}$ (A)

See Also

2016 AMC 10A (Problems • Answer Key • Resources)
Preceded by
First Problem 		Followed by
Problem 2
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

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

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2016_AMC_10A_Problems/Problem_23&oldid=75125"