1984 AIME Problems/Problem 10

Revision as of 19:45, 26 March 2007 by Scorpius119 (talk | contribs) (solution added)


Mary told John her score on the American High School Mathematics Examination (AHSME), which was over $\displaystyle 80$. From this, John was able to determine the number of problems Mary solved correctly. If Mary's score had been any lower, but still over $\displaystyle 80$, John could not have determined this. What was Mary's score? (Recall that the AHSME consists of $\displaystyle 30$ multiple choice problems and that one's score, $\displaystyle s$, is computed by the formula $\displaystyle s=30+4c-w$, where $\displaystyle c$ is the number of correct answers and $\displaystyle w$ is the number of wrong answers. Students are not penalized for problems left unanswered.)


Let Mary's score, number correct, and number wrong be $\displaystyle s,c,w$ respectively. Then

$\displaystyle s=30+4c-w=30+4(c-1)-(w-4)=30+4(c+1)-(w+4)$

Therefore, Mary could not have left at least five blank; otherwise, 1 more correct and 4 more wrong would produce the same score. Similarly, Mary could not have answered at least four wrong (clearly Mary answered at least one right to have score above 80, or even 30.)

It follows that $c+w\geq 26$ and $w\leq 3$, so $c\geq 23$ and $s=30+4c-w\geq 30+4(23)-3=119$. So Mary scored at least 119. To see that no result other than 23 right/3 wrong produces 119, note that $s=119\Rightarrow 4c-w=89$ so $w\equiv 3\pmod{4}$. But if $\displaystyle w=3$, then $\displaystyle c=23$ which was the result given; otherwise $w\geq 7$ and $c\geq 24$, but this implies at least 31 questions, a contradiction. This makes the minimum score


See also

Invalid username
Login to AoPS