Difference between revisions of "1962 AHSME Problems/Problem 27"

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<math> \textbf{(1)}\ a@b = b@a\qquad\textbf{(2)}\ a@(b@c) = (a@b)@c\qquad\textbf{(3)}\ a ! (b@c) = (a ! b) @ (a ! c) </math>
 
<math> \textbf{(1)}\ a@b = b@a\qquad\textbf{(2)}\ a@(b@c) = (a@b)@c\qquad\textbf{(3)}\ a ! (b@c) = (a ! b) @ (a ! c) </math>
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<math> \textbf{(A)}\ (1)\text{ only}\qquad\textbf{(B)}\ (2)\text{ only}\qquad\textbf{(C)}\ \text{(1) and (2) only}\qquad\textbf{(D)}\ \text{(1) and (3) only}\qquad\textbf{(E)}\ \text{all three} </math>
 
<math> \textbf{(A)}\ (1)\text{ only}\qquad\textbf{(B)}\ (2)\text{ only}\qquad\textbf{(C)}\ \text{(1) and (2) only}\qquad\textbf{(D)}\ \text{(1) and (3) only}\qquad\textbf{(E)}\ \text{all three} </math>

Revision as of 22:56, 9 November 2013

Problem

Let $a @ b$ represent the operation on two numbers, $a$ and $b$, which selects the larger of the two numbers, with $a@a = a$. Let $a ! b$ represent the operator which selects the smaller of the two numbers, with $a ! a = a$. Which of the following three rules is (are) correct?

$\textbf{(1)}\ a@b = b@a\qquad\textbf{(2)}\ a@(b@c) = (a@b)@c\qquad\textbf{(3)}\ a ! (b@c) = (a ! b) @ (a ! c)$


$\textbf{(A)}\ (1)\text{ only}\qquad\textbf{(B)}\ (2)\text{ only}\qquad\textbf{(C)}\ \text{(1) and (2) only}\qquad\textbf{(D)}\ \text{(1) and (3) only}\qquad\textbf{(E)}\ \text{all three}$

Solution

"Unsolved"