Difference between revisions of "1984 AIME Problems/Problem 10"
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== Problem == | == Problem == | ||
− | Mary told John her score on the American High School Mathematics Examination (AHSME), which was over <math> | + | Mary told John her score on the American High School Mathematics Examination (AHSME), which was over <math>80</math>. 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 <math>80</math>, John could not have determined this. What was Mary's score? (Recall that the AHSME consists of <math>30</math> multiple choice problems and that one's score, <math>s</math>, is computed by the formula <math>s=30+4c-w</math>, where <math>c</math> is the number of correct answers and <math>w</math> is the number of wrong answers. (Students are not penalized for problems left unanswered.) |
− | == Solution == | + | == Solution 1 (Inequalities) == |
− | Let Mary's score, number correct, and number wrong be <math> | + | Let Mary's score, number correct, and number wrong be <math>s,c,w</math> respectively. Then |
+ | <cmath>\begin{align*} | ||
+ | s&=30+4c-w \\ | ||
+ | &=30+4(c-1)-(w-4) \\ | ||
+ | &=30+4(c+1)-(w+4). | ||
+ | \end{align*}</cmath> | ||
+ | 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 <math>80</math>, or even <math>30</math>.) | ||
+ | It follows that <math>c+w\geq 26</math> and <math>w\leq 3</math>, so <math>c\geq 23</math> and <math>s=30+4c-w\geq 30+4(23)-3=119</math>. So Mary scored at least <math>119</math>. To see that no result other than <math>23</math> right/<math>3</math> wrong produces <math>119</math>, note that <math>s=119\Rightarrow 4c-w=89</math> so <math>w\equiv 3\pmod{4}</math>. But if <math>w=3</math>, then <math>c=23</math>, which was the result given; otherwise <math>w\geq 7</math> and <math>c\geq 24</math>, but this implies at least <math>31</math> questions, a contradiction. This makes the minimum score <math>\boxed{119}</math>. | ||
− | + | == 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 <math>4</math> or <math>5.</math> The maximum number we can get is <math>70.</math> 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 <math>"21"</math> can be achieved with only <math>1</math> method <math>(4+4+4+4+5).</math> However, <math>25</math>, which is a larger number than <math>21</math>, can be achieved with multiple methods (e.g. <math>5 \cdot 5</math> or <math>4 \cdot 5 + 5</math>), hence <math>21</math> is not the number we are trying to find. | |
+ | If we make a table of adding <math>4</math> or adding <math>5</math>, we will see we get <math>4, 8, 12, 16, 20,</math> etc. if we add only <math>4</math>s and if we add <math>5</math> to those numbers then we will get <math>9, 13, 17, 21, 25,</math> etc. Now a key observation to getting this problem correct is that if we can add one of those previous base numbers to <math>20</math>, then there will be multiple methods (because <math>20 = 4 \cdot 5 = 5 \cdot 4</math>). | ||
− | + | Hence, the number we are looking for cannot be <math>20</math> plus one of those base numbers. Instead, it must be <math>10</math> 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 <math>4</math>, the number <math>14</math> would have only <math>1</math> method to solve, but the number <math>24</math> would have multiple (because <math>4 + 20 = 24</math> and we are trying to avoid adding <math>20</math>). The largest number we see that is in our base numbers is <math>21.</math> Hence, our maximum number is <math>21 + 10 = 31.</math> | |
+ | |||
+ | Note that if we have the number <math>25</math>, that can be solved via multiple methods, and if we keep repeating our cycle of base numbers, we are basically adding <math>20</math> to a previous base number, which we don't want. | ||
+ | |||
+ | And since the maximum number of this game is <math>31</math>, that is the number we subtract from the maximum score of <math>150</math>, so we get <math>150 - 31 = \boxed{119}.</math> | ||
+ | |||
+ | P.S. didn't think the solution would be this complicated when I first wrote it but it's quite complicated. Look to solution <math>1</math> if you want a concise method using inequalities that's probably better than this solution. | ||
+ | |||
+ | == Solution 3 (Table) == | ||
+ | Based on the value of <math>c,</math> we construct the following table: | ||
+ | <cmath>\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}</cmath> | ||
+ | For a fixed value of <math>c,</math> note that <math>s_{\min}</math> occurs at <math>w=30-c,</math> and <math>s_{\max}</math> occurs at <math>w=0.</math> Moreover, all integers from <math>s_{\min}</math> through <math>s_{\max}</math> are attainable. <i><b>To find Mary's score, we look for the lowest score <math>\boldsymbol{s}</math> such that <math>\boldsymbol{s\geq80}</math> and <math>\boldsymbol{s}</math> is contained in exactly one interval.</b></i> | ||
+ | |||
+ | Let <math>S(c)</math> denote the interval of all possible scores <math>s</math> with <math>c</math> correct answers. We need: | ||
+ | <ol style="margin-left: 1.5em;"> | ||
+ | <li><math>S(c)\not\subset S(c-1)\cup S(c+1).</math></li><p> | ||
+ | <li><math>s\in S(c)</math> but <math>s\not\in S(c-1)\cup S(c+1).</math></li><p> | ||
+ | </ol> | ||
+ | It follows that the least such value of <math>c</math> is <math>23,</math> from which the lowest such score <math>s</math> is <math>\boxed{119}.</math> | ||
+ | |||
+ | ~MRENTHUSIASM | ||
+ | |||
+ | == Solution 4 (Less Rigorous Version of Solution 1) == | ||
+ | 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 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>. | ||
+ | Substituting, we get that <math>s=30+23{\,\cdot\,}4-3</math>, so <math>s=\boxed{119}</math>. | ||
+ | |||
+ | ~JeffersonJ | ||
− | |||
== See also == | == See also == | ||
− | * [[ | + | {{AIME box|year=1984|num-b=9|num-a=11}} |
− | * [[ | + | * [[AIME Problems and Solutions]] |
− | + | * [[American Invitational Mathematics Examination]] | |
+ | * [[Mathematics competition resources]] | ||
+ | |||
+ | [[Category:Intermediate Algebra Problems]] | ||
+ | |||
+ | [[Category:Intermediate Number Theory Problems]] |
Latest revision as of 23:53, 28 March 2022
Contents
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 answers and
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 respectively. Then
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
, or even
.)
It follows that and
, so
and
. So Mary scored at least
. To see that no result other than
right/
wrong produces
, note that
so
. But if
, then
, which was the result given; otherwise
and
, but this implies at least
questions, a contradiction. This makes the minimum score
.
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 or
The maximum number we can get is
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 can be achieved with only
method
However,
, which is a larger number than
, can be achieved with multiple methods (e.g.
or
), hence
is not the number we are trying to find.
If we make a table of adding or adding
, we will see we get
etc. if we add only
s and if we add
to those numbers then we will get
etc. Now a key observation to getting this problem correct is that if we can add one of those previous base numbers to
, then there will be multiple methods (because
).
Hence, the number we are looking for cannot be plus one of those base numbers. Instead, it must be
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
, the number
would have only
method to solve, but the number
would have multiple (because
and we are trying to avoid adding
). The largest number we see that is in our base numbers is
Hence, our maximum number is
Note that if we have the number , that can be solved via multiple methods, and if we keep repeating our cycle of base numbers, we are basically adding
to a previous base number, which we don't want.
And since the maximum number of this game is , that is the number we subtract from the maximum score of
, so we get
P.S. didn't think the solution would be this complicated when I first wrote it but it's quite complicated. Look to solution if you want a concise method using inequalities that's probably better than this solution.
Solution 3 (Table)
Based on the value of we construct the following table:
For a fixed value of
note that
occurs at
and
occurs at
Moreover, all integers from
through
are attainable. To find Mary's score, we look for the lowest score
such that
and
is contained in exactly one interval.
Let denote the interval of all possible scores
with
correct answers. We need:
but
It follows that the least such value of is
from which the lowest such score
is
~MRENTHUSIASM
Solution 4 (Less Rigorous Version of Solution 1)
Given that Mary's score is , two other ways to get that score are
and
.
Since it is clear that
, we must have
. In order to minimize the score, assume that
.
The number of problems left blank must be less than
because of the
case. In order to minimize the score, assume that the number of problems left blank is
, making the number of correct problems
.
Substituting, we get that
, so
.
~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 |