Difference between revisions of "1985 AJHSME Problem 25"
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== Solution 1== | == Solution 1== | ||
| − | To show Jane was wrong, Mary would have to show a card with a vowel. However, this is not possible. If Jane is correct, then the contrapositive ("If an odd number is on one side of the card, then a consonant is on the other side"). is true. So, Mary picked the card with a <math>\boxed{\text{A} 3}.</math> | + | To show Jane was wrong, Mary would have to show a card with a vowel. However, this is not possible. If Jane is correct, then the contrapositive ("If an odd number is on one side of the card, then a consonant is on the other side"). is true. So, Mary picked the card with a <math>\boxed{\text{(A)} 3}.</math> |
== Solution 2== | == Solution 2== | ||
Latest revision as of 22:18, 17 May 2024
Problem
Five cards are lying on a table as shown.
![\[\begin{matrix} & \qquad & \boxed{\tt{P}} & \qquad & \boxed{\tt{Q}} \\ \\ \boxed{\tt{3}} & \qquad & \boxed{\tt{4}} & \qquad & \boxed{\tt{6}} \end{matrix}\]](http://latex.artofproblemsolving.com/f/a/e/fae700ade12943277f8147039b058865f3fff819.png)
Each card has a letter on one side and a whole number on the other side. Jane said, "If a vowel is on one side of any card, then an even number is on the other side." Mary showed Jane was wrong by turning over one card. Which card did Mary turn over?
Solution 1
To show Jane was wrong, Mary would have to show a card with a vowel. However, this is not possible. If Jane is correct, then the contrapositive ("If an odd number is on one side of the card, then a consonant is on the other side"). is true. So, Mary picked the card with a 
Solution 2
Notice that if the answer was to flip a consonant (aka a non-vowel) or an even number, there would be two possible solutions. Since we know that there is only one solution, we conclude that Mary must have turned over the card with the
~Bradygho
