Difference between revisions of "1984 AIME Problems/Problem 10"
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<math>s=30+4c-w=30+4(c-1)-(w-4)=30+4(c+1)-(w+4)</math>. | <math>s=30+4c-w=30+4(c-1)-(w-4)=30+4(c+1)-(w+4)</math>. | ||
| − | Therefore, Mary could not have left at least five blank; otherwise, | + | 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 119. To see that no result other than 23 right/3 wrong produces 119, 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 31 questions, a contradiction. This makes the minimum score <math>\boxed{119}</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 23 right/3 wrong produces 119, 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 31 questions, a contradiction. This makes the minimum score <math>\boxed{119}</math>. |
== Solution 2 == | == Solution 2 == | ||
Revision as of 13:32, 4 September 2020
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
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 23 right/3 wrong produces 119, note that 
so 
. But if 
, then 
, which was the result given; otherwise 
and 
, but this implies at least 31 questions, a contradiction. This makes the minimum score 
.
Solution 2
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 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*5 or 4*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 4s 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*5 = 5*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 = 
Note that if we have the number 25, that can be solved for 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 =
P.S. didn't think the solution would be this complicated when I first wrote it but it's quite complicated. Look to sol 1 if you want a concise method using inequalities that's probably better than this solution
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 | ||
