Difference between revisions of "1984 AIME Problems/Problem 10"

Latest revision as of 22:53, 28 March 2022

Problem

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

Solution 1 (Inequalities)

Let Mary's score, number correct, and number wrong be $s,c,w$ respectively. Then $\begin{align*} s&=30+4c-w \\ &=30+4(c-1)-(w-4) \\ &=30+4(c+1)-(w+4). \end{align*}$ Therefore, Mary could not have left at least five blank; otherwise, one more correct and four 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 a 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 $w=3$ , then $c=23$ , 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 $\boxed{119}$ .

Solution 2 (Arithmetic)

A less technical approach that still gets the job done:

Pretend that the question is instead a game, where we are trying to get certain numbers by either adding $4$ or $5.$ The maximum number we can get is $70.$ The goal of the game is to find out what number we can achieve in only ONE method, while all other numbers above that can be achieved with TWO or MORE methods. (Note: This is actually the exact same problem as the original, just reworded differently and now we are adding the score. If this is already confusing, I suggest not looking further.)

For example, the number $"21"$ can be achieved with only $1$ method $(4+4+4+4+5).$ However, $25$ , which is a larger number than $21$ , can be achieved with multiple methods (e.g. $5 \cdot 5$ or $4 \cdot 5 + 5$ ), hence $21$ is not the number we are trying to find.

If we make a table of adding $4$ or adding $5$ , we will see we get $4, 8, 12, 16, 20,$ etc. if we add only $4$ s and if we add $5$ to those numbers then we will get $9, 13, 17, 21, 25,$ etc. Now a key observation to getting this problem correct is that if we can add one of those previous base numbers to $20$ , then there will be multiple methods (because $20 = 4 \cdot 5 = 5 \cdot 4$ ).

Hence, the number we are looking for cannot be $20$ plus one of those base numbers. Instead, it must be $10$ plus that base number, because that results in the same last digit while maintaining only one method to solve. For example, if we start with $4$ , the number $14$ would have only $1$ method to solve, but the number $24$ would have multiple (because $4 + 20 = 24$ and we are trying to avoid adding $20$ ). The largest number we see that is in our base numbers is $21.$ Hence, our maximum number is $21 + 10 = 31.$

Note that if we have the number $25$ , that can be solved via multiple methods, and if we keep repeating our cycle of base numbers, we are basically adding $20$ to a previous base number, which we don't want.

And since the maximum number of this game is $31$ , that is the number we subtract from the maximum score of $150$ , so we get $150 - 31 = \boxed{119}.$

P.S. didn't think the solution would be this complicated when I first wrote it but it's quite complicated. Look to solution $1$ if you want a concise method using inequalities that's probably better than this solution.

Solution 3 (Table)

Based on the value of $c,$ we construct the following table: $\begin{array}{c||c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c} &\hspace{5.5mm}&\hspace{5.5mm}&\hspace{5.5mm}&\hspace{5.5mm}&\hspace{5.5mm}&\hspace{5.5mm}&\hspace{5.5mm}&&&&&&&&&&&&& \\ [-2.5ex] \boldsymbol{c} &\boldsymbol{\cdots}&\boldsymbol{12}&\boldsymbol{13}&\boldsymbol{14}&\boldsymbol{15}&\boldsymbol{16}&\boldsymbol{17}&\boldsymbol{18}&\boldsymbol{19}&\boldsymbol{20}&\boldsymbol{21}&\boldsymbol{22}&\boldsymbol{23}&\boldsymbol{24}&\boldsymbol{25}&\boldsymbol{26}&\boldsymbol{27}&\boldsymbol{28}&\boldsymbol{29}&\boldsymbol{30} \\ \hline \hline &&&&&&&&&&&&&&&&&&& \\ [-2.25ex] \boldsymbol{s_{\min}} &\cdots&60&65&70&75&80&85&90&95&100&105&110&115&120&125&130&135&140&145&150 \\ \hline &&&&&&&&&&&&&&&&&&& \\ [-2.25ex] \boldsymbol{s_{\max}} &\cdots&78&82&86&90&94&98&102&106&110&114&118&122&126&130&134&138&142&146&150 \end{array}$ For a fixed value of $c,$ note that $s_{\min}$ occurs at $w=30-c,$ and $s_{\max}$ occurs at $w=0.$ Moreover, all integers from $s_{\min}$ through $s_{\max}$ are attainable. To find Mary's score, we look for the lowest score $\boldsymbol{s}$ such that $\boldsymbol{s\geq80}$ and $\boldsymbol{s}$ is contained in exactly one interval.

Let $S(c)$ denote the interval of all possible scores $s$ with $c$ correct answers. We need:

$S(c)\not\subset S(c-1)\cup S(c+1).$

$s\in S(c)$ but $s\not\in S(c-1)\cup S(c+1).$

It follows that the least such value of $c$ is $23,$ from which the lowest such score $s$ is $\boxed{119}.$

~MRENTHUSIASM

Solution 4 (Less Rigorous Version of Solution 1)

Given that Mary's score is $30+4c-w$ , two other ways to get that score are $30+4(c+1)-(w+4)$ and $30+4(c-1)-(w-4)$ . Since it is clear that $c>1$ , we must have $w<4$ . In order to minimize the score, assume that $w=3$ . The number of problems left blank must be less than $5$ because of the $30+4(c+1)-(w+4)$ case. In order to minimize the score, assume that the number of problems left blank is $4$ , making the number of correct problems $23$ . Substituting, we get that $s=30+23{\,\cdot\,}4-3$ , so $s=\boxed{119}$ .

~JeffersonJ

@@ Line 52: / Line 52: @@
 ~MRENTHUSIASM
-== Solution 4 (Less rigorous) ==
+== Solution 4 (Less Rigorous Version of Solution 1) ==
-Given that Mary's score is <math>30+4c-w</math>, 2 other ways to get that score are <math>30+4(c+1)-(w+4)</math> and <math>30+4(c-1)-(w-4)</math>.
+Given that Mary's score is <math>30+4c-w</math>, two other ways to get that score are <math>30+4(c+1)-(w+4)</math> and <math>30+4(c-1)-(w-4)</math>.
-Since it is obvious that <math>c>1</math>, <math>w</math> must be less than <math>4</math>. In order to minimize the score, assume that <math>w=3</math>.
+Since it is clear that <math>c>1</math>, we must have <math>w<4</math>. In order to minimize the score, assume that <math>w=3</math>.
-The number of problems left blank must be less than <math>5</math>, because of the <math>30+4(c+1)-(w+4)</math> case. In order to minimize the score, assume that the number of problems left blank is <math>4</math>, making the number of correct problems <math>23</math>.
+The number of problems left blank must be less than <math>5</math> because of the <math>30+4(c+1)-(w+4)</math> case. In order to minimize the score, assume that the number of problems left blank is <math>4</math>, making the number of correct problems <math>23</math>.
-Substituting, we get that <math>s=30+23 {\cdot} 4-3</math>, so <math>s</math> is <math>\boxed{119}</math>.
+Substituting, we get that <math>s=30+23{\,\cdot\,}4-3</math>, so <math>s=\boxed{119}</math>.
+~JeffersonJ
 == See also ==

1984 AIME (Problems • Answer Key • Resources)
Preceded by Problem 9	Followed by Problem 11
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions

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