Difference between revisions of "2002 AMC 12P Problems/Problem 22"
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Let <math>c</math> denote the amount of correct answers, <math>u</math> denote the amount of unanswered questions. Then <math>c+u \leq 25.</math> We want to find the LCM of <math>2.5</math> and <math>6</math> so we can convert between different ways, similar to solution <math>1.</math> The LCM is equal to <math>30</math>, since <math>6 \cdot \textbf{5}=30</math> and <math>2.5 \cdot \textbf{12}=30.</math> | Let <math>c</math> denote the amount of correct answers, <math>u</math> denote the amount of unanswered questions. Then <math>c+u \leq 25.</math> We want to find the LCM of <math>2.5</math> and <math>6</math> so we can convert between different ways, similar to solution <math>1.</math> The LCM is equal to <math>30</math>, since <math>6 \cdot \textbf{5}=30</math> and <math>2.5 \cdot \textbf{12}=30.</math> | ||
− | The first way will take the form <math>(c,u).</math> The second way will take the form <math>(c+5, u-12).</math> The third way will take the form <math>(c+10, u-24).</math> In particular, since we cannot have a negative amount of unanswered questions, <math>u \leq 24.</math> Combined with <math>c+u | + | The first way will take the form <math>(c,u).</math> The second way will take the form <math>(c+5, u-12).</math> The third way will take the form <math>(c+10, u-24).</math> In particular, since we cannot have a negative amount of unanswered questions, <math>u \leq 24.</math> Combined with <math>c+u \leq 25</math>, our three cases are <math>(0,24), (1,24),</math> and <math>(0,25).</math> |
== See also == | == See also == | ||
{{AMC12 box|year=2002|ab=P|num-b=21|num-a=23}} | {{AMC12 box|year=2002|ab=P|num-b=21|num-a=23}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 03:54, 14 July 2024
Contents
Problem
Under the new AMC scoring method,
points are given for each correct answer,
points are given for each unanswered question, and no points are given for an incorrect answer. Some of the possible scores between
and
can be obtained in only one way, for example, the only way to obtain a score of
is to have
correct answers and one unanswered question. Some scores can be obtained in exactly two ways, for example, a score of
can be obtained with
correct answers,
unanswered question, and
incorrect, and also with
correct answers and
unanswered questions. There are (three) scores that can be obtained in exactly three ways. What is their sum?
Solution 1
Let denote the amount of correct answers,
denote the amount of unanswered questions, and
denote the amount of wrong answers. The conversion rate is what differentiates between two ways to get the same score.
There are two ways to realize the conversion rate of wrong answers to unanswered questions and correct answers to unanswered questions.
Sub Solution 1.1 (Cheese)
Notice how the way to get points directly tells us the conversion rate.
correct questions,
unanswered question, and
wrong answers is equivalent to
correct answers,
unanswered questions, and
wrong answers. Since
and
, the conversion rate is
Sub Solution 1.2
The second way to find out the conversion rate is to actually do the math. If a wrong answer is converted to an unanswered question, the total value will be more than the total value. If a correct answer is converted to an unanswered question, the total value will be
less than the total value. Thus, we want the "more than" to be equal to the "less than".
Let denote the amount of correct answers converted to unanswered questions and
denote the amount of wrong answers converted to unanswered questions. Since we want "more than" to be equal to the "less than",
, where
and
are integers. Multiply by
on both sides,
implies
and
Again, the conversion rate is
Sub Solutions Combine
Either way, we find If there are three ways, this implies the first way
the second way is
and the third way is
Since
it follows that our three cases' first ways are as follows:
correct answers,
unanswered questions,
wrong answers
correct answers,
unanswered questions,
wrong answers
correct answers,
unanswered questions,
wrong answers.
Thus, our answer is
Solution 2
This solution is formulated in more formal language rather than cheesy methods, though the idea is the same. Additionally, we do not need to take into account the amount of wrong answers, which is easier to deal with.
Let denote the amount of correct answers,
denote the amount of unanswered questions. Then
We want to find the LCM of
and
so we can convert between different ways, similar to solution
The LCM is equal to
, since
and
The first way will take the form The second way will take the form
The third way will take the form
In particular, since we cannot have a negative amount of unanswered questions,
Combined with
, our three cases are
and
See also
2002 AMC 12P (Problems • Answer Key • Resources) | |
Preceded by Problem 21 |
Followed by Problem 23 |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | |
All AMC 12 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.