|
|
| (2 intermediate revisions by 2 users not shown) |
| Line 1: |
Line 1: |
| − | ==Problem==
| + | #REDIRECT [[2016_AMC_10A_Problems/Problem_23]] |
| − | | |
| − | 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 the dot <math>\ \cdot</math> represents the usual multiplication operation.) The solution to the equation <math>2016\ \diamondsuit\ (6\ \diamondsuit\ x) = 100</math> can be written as <math>\frac{p}{q},</math> where <math>p</math> and <math>q</math> are relativelt 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>
| |
| − | | |
| − | ==Solution==
| |
| − | | |
| − | 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 second identity yields <math>( a\ \diamondsuit\ b) \cdot b = a\ \diamondsuit\ (b\ \diamondsuit\ b) = a\ \diamondsuit\ 1 = a\ \diamondsuit\ ( a\ \diamondsuit\ a) = ( a\ \diamondsuit\ a) \cdot a = a</math>. Hence, <math>( a\ \diamondsuit\ b) \cdot b = a,</math> or, dividing both sides of the equation by <math>a,</math> <math>( a\ \diamondsuit\ b) = \frac{a}{b}.</math>
| |
| − | | |
| − | Hence, the given equation becomes <math>\frac{2016}{\frac{60}{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>
| |
| − | | |
| − | ==See Also==
| |
| − | {{AMC12 box|year=2016|ab=A|num-b=19|num-a=21}}
| |
| − | {{MAA Notice}}
| |
