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

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==Solution==
 
==Solution==
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The first rule must be correct as both sides of the equation pick the larger out of a and b.
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The second rule must also be correct as both sides would end up picking the largest out of a, b, and c.
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WLOG, lets assume b < c.
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The third rule is a little more complex. To see if it works, let’s split the possibilities into three cases. a < b < c, b < a < c, and b < c < a. For the first case, the equation simplifies to a = a, which is correct. In the second case, the equation simplifies to a = a, which is correct. For the last case, the equation simplifies to c = c, which is also correct. Since all three cases work, this rule also works.
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Thus, our answer is \boxed{\bold{E}}

Revision as of 21:32, 19 November 2019

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

The first rule must be correct as both sides of the equation pick the larger out of a and b.

The second rule must also be correct as both sides would end up picking the largest out of a, b, and c.

WLOG, lets assume b < c.

The third rule is a little more complex. To see if it works, let’s split the possibilities into three cases. a < b < c, b < a < c, and b < c < a. For the first case, the equation simplifies to a = a, which is correct. In the second case, the equation simplifies to a = a, which is correct. For the last case, the equation simplifies to c = c, which is also correct. Since all three cases work, this rule also works.

Thus, our answer is \boxed{\bold{E}}

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