Problem:
Mary told John her score on the American High School Mathematics Examination (AHSME), which was over . 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 , John could not have determined this. What was Mary's score? (Recall that the AHSME consists of multiple-choice problems and that one's score, , is computed by the formula , where is the number of correct and is the number of wrong answers; students are not penalized for problems left unanswered.)
Solution:
Given that , the problem calls for finding the lowest value of , for which the corresponding value of is unique. To this end, first observe that if , then by increasing the number of correct answers by and the number of wrong answers by , one attains the same score. (This is made possible by the equivalence of the above inequality to . Consequently, one must have
Next observe that
for otherwise one could reduce the number of wrong answers by and the number of correct answers by , and still attain the same score. (The latter is possible since clearly implies that .) Now to minimize , we must minimize and maximize , subject to Inequalities and above. This leads to and .
The problems on this page are the property of the MAA's American Mathematics Competitions