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> | ||
| − | <math> \textbf{(A)}\ (1)\text{ only}\qquad\textbf{(B)}\ (2)\text{ only}\qquad\textbf{(C)}\ \text{(1) and (2) only} | + | |
| + | |||
| + | <math> \textbf{(A)}\ (1)\text{ only}\qquad\textbf{(B)}\ (2)\text{ only}\qquad\textbf{(C)}\ \text{(1) and (2) only}</math> | ||
| + | |||
| + | <math>\textbf{(D)}\ \text{(1) and (3) only}\qquad\textbf{(E)}\ \text{all three} </math> | ||
==Solution== | ==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 <math>\boxed{\textbf{E}}</math> | ||
| + | |||
| + | (Solution by someonenumber011) | ||
Latest revision as of 21:25, 17 April 2023
Problem
Let 
represent the operation on two numbers, 
and 
, which selects the larger of the two numbers, with 
. Let 
represent the operator which selects the smaller of the two numbers, with 
. Which of the following three rules is (are) correct?
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 
(Solution by someonenumber011)
