Difference between revisions of "2016 AMC 10A Problems/Problem 23"
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− | ==Problem== | + | == Problem == |
A binary operation <math>\diamondsuit</math> has the properties that <math>a\,\diamondsuit\, (b\,\diamondsuit \,c) = (a\,\diamondsuit \,b)\cdot c</math> and that <math>a\,\diamondsuit \,a=1</math> for all nonzero real numbers <math>a, b,</math> and <math>c</math>. (Here <math>\cdot</math> represents multiplication). The solution to the equation <math>2016 \,\diamondsuit\, (6\,\diamondsuit\, x)=100</math> can be written as <math>\tfrac{p}{q}</math>, where <math>p</math> and <math>q</math> are relatively prime positive integers. What is <math>p+q?</math> | A binary operation <math>\diamondsuit</math> has the properties that <math>a\,\diamondsuit\, (b\,\diamondsuit \,c) = (a\,\diamondsuit \,b)\cdot c</math> and that <math>a\,\diamondsuit \,a=1</math> for all nonzero real numbers <math>a, b,</math> and <math>c</math>. (Here <math>\cdot</math> represents multiplication). The solution to the equation <math>2016 \,\diamondsuit\, (6\,\diamondsuit\, x)=100</math> can be written as <math>\tfrac{p}{q}</math>, where <math>p</math> and <math>q</math> are relatively prime positive integers. What is <math>p+q?</math> | ||
<math>\textbf{(A) }109\qquad\textbf{(B) }201\qquad\textbf{(C) }301\qquad\textbf{(D) }3049\qquad\textbf{(E) }33,601</math> | <math>\textbf{(A) }109\qquad\textbf{(B) }201\qquad\textbf{(C) }301\qquad\textbf{(D) }3049\qquad\textbf{(E) }33,601</math> | ||
− | |||
− | + | == Solutions == | |
+ | === Solution 1 === | ||
− | ==See Also== | + | 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 can be the operation <math>\diamond</math>. Solving the equation, <cmath>\frac{2016}{\frac{6}{x}} = \frac{2016}{6} \cdot x = 336x = 100\implies x=\frac{100}{336} = \frac{25}{84},</cmath> so the answer is <math>25 + 84 = \boxed{\textbf{(A) }109}</math>. |
+ | |||
+ | === Solution 2 === | ||
+ | We can manipulate the given identities to arrive at a conclusion about the binary operator <math>\diamondsuit</math>. Substituting <math>b = c</math> into the first identity yields <cmath>( a\ \diamondsuit\ b) \cdot b = a\ \diamondsuit\ (b\ \diamondsuit\ b) = a\ \diamondsuit\ 1 = a\ \diamondsuit\ ( a\ \diamondsuit\ a) = ( a\ \diamondsuit\ a) \cdot a = a.</cmath> Hence, <math>( a\ \diamondsuit\ b) \cdot b = a,</math> or, dividing both sides of the equation by <math>b,</math> <math>( a\ \diamondsuit\ b) = \frac{a}{b}.</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> | ||
+ | |||
+ | === Solution 3 === | ||
+ | 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>. | ||
+ | |||
+ | <cmath>2016 \,\diamondsuit\, (6\,\diamondsuit\, x)=100\implies | ||
+ | (2016\, \diamondsuit\, 6) \cdot x = 100</cmath> | ||
+ | |||
+ | By multiplying both sides by <math>\frac{6}{x}</math>, we get: | ||
+ | |||
+ | <cmath>(2016\, \diamondsuit\, 6) \cdot 6 = \frac{600}{x}\implies | ||
+ | 2016 \, \diamondsuit\, (6\, \diamondsuit\, 6) = \frac{600}{x}</cmath> | ||
+ | |||
+ | Because <math>6\, \diamondsuit\, 6 = 2016\, \diamondsuit\, 2016 = 1:</math> | ||
+ | |||
+ | <cmath>2016 \, \diamondsuit\, (2016\, \diamondsuit\, 2016) = \frac{600}{x}\implies | ||
+ | (2016\, \diamondsuit\, 2016) \cdot 2016 = \frac{600}{x}\implies | ||
+ | 2016 = \frac{600}{x}</cmath> | ||
+ | |||
+ | Therefore, <math>x = \frac{600}{2016} = \frac{25}{84}</math>, so the answer is <math>25 + 84 = \boxed{\textbf{(A) }109.}</math> | ||
+ | |||
+ | == 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 == | ||
{{AMC10 box|year=2016|ab=A|num-b=22|num-a=24}} | {{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}} | {{MAA Notice}} |
Latest revision as of 22:12, 24 January 2021
Contents
Problem
A binary operation has the properties that and that for all nonzero real numbers and . (Here represents multiplication). The solution to the equation can be written as , where and are relatively prime positive integers. What is
Solutions
Solution 1
We see that , and think of division. Testing, we see that the first condition is satisfied, because . Therefore, division can be the operation . Solving the equation, so the answer is .
Solution 2
We can manipulate the given identities to arrive at a conclusion about the binary operator . Substituting into the first identity yields Hence, or, dividing both sides of the equation by
Hence, the given equation becomes . Solving yields so the answer is
Solution 3
One way to eliminate the in this equation is to make so that . In this case, we can make .
By multiplying both sides by , we get:
Because
Therefore, , so the answer is
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 |
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