Difference between revisions of "2022 AMC 10A Problems/Problem 12"
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| − | The condition of the third question is extraneous. However, we know that <math>A=9</math> and <math>L=6,</math> | + | The condition of the third question is extraneous. However, we know that <math>A=9</math> and <math>L=6,</math> so there are <math>9</math> alternaters who responded lie-truth-lie, <math>6</math> liars, and <math>9</math> alternaters who responded truth-lie-truth from this condition. |
~sigma ~[[OrenSH|orenbad]] ~MRENTHUSIASM | ~sigma ~[[OrenSH|orenbad]] ~MRENTHUSIASM | ||
Latest revision as of 11:45, 31 October 2024
- The following problem is from both the 2022 AMC 10A #12 and 2022 AMC 12A #9, so both problems redirect to this page.
Contents
Problem
On Halloween 
children walked into the principal's office asking for candy. They
can be classified into three types: Some always lie; some always tell the truth; and
some alternately lie and tell the truth. The alternaters arbitrarily choose their first
response, either a lie or the truth, but each subsequent statement has the opposite
truth value from its predecessor. The principal asked everyone the same three
questions in this order.
"Are you a truth-teller?" The principal gave a piece of candy to each of the 
children who answered yes.
"Are you an alternater?" The principal gave a piece of candy to each of the 
children who answered yes.
"Are you a liar?" The principal gave a piece of candy to each of the 
children who
answered yes.
How many pieces of candy in all did the principal give to the children who always tell the truth?
Solution 1
Note that:
- Truth-tellers would answer yes-no-no to the three questions in this order.
- Liars would answer yes-yes-no to the three questions in this order.
- Alternaters who responded truth-lie-truth would answer no-no-no to the three questions in this order.
- Alternaters who responded lie-truth-lie would answer yes-yes-yes to the three questions in this order.
Suppose that there are 
truth-tellers, 
liars, and 
alternaters who responded lie-truth-lie.
The conditions of the first two questions imply that
Subtracting the second equation from the first, we have 
Remark
The condition of the third question is extraneous. However, we know that 
and 
so there are alternaters who responded lie-truth-lie, 
liars, and alternaters who responded truth-lie-truth from this condition.
~sigma ~orenbad ~MRENTHUSIASM
Solution 2
Consider when the principal asks "Are you a liar?": The truth tellers truthfully say no, and the liars lie and say no. This leaves only alternaters who lie on this question to answer yes. Thus, all children that answered yes are alternaters that falsely answer Questions 1 and 3, and truthfully answer Question 2. The rest of the alternaters, however many there are, have the opposite behavior.
Consider the second question, "Are you an alternater?": The truth tellers again answer no, the liars falsely answer yes, and alternaters that truthfully answer also say yes. From the previous part, we know that alternaters truthfully answer here. Because only liars and alternaters answer yes, we can deduce that there are 
liars.
Consider the first question, "Are you a truth teller?": Truth tellers say yes, liars also say yes, and alternaters that lie on this question also say yes. From the first part, we know that alternaters lie here. From the previous part, we know that there are liars. Because only the number of truth tellers is unknown here, we can deduce that there are 
truth tellers.
The final question is how many pieces of candy did the principal give to truth tellers. Because truth tellers answer yes on only the first question, we know that all 
of them said yes once, resulting in 
pieces of candy.
~phuang1024
Remark (Fake Solve)
This problem is broken in an interesting way that helps the test-taker. Since the true answers alternate for alternaters, you can still get the correct answer if you misinterpret the problem as "alternaters alternate their yes/no answers, but not their truth values". Suppose that alternaters answer yes-no-yes, and 
alternaters answer no-yes-no. You still get 
.
In addition, you can even get the correct answer with the misinterpretation "alternaters have two types: alternaters of the same type give the same arbitrary answers to the three questions, and alternaters of different types give the opposite arbitrary answers to the three questions".
It's also notable that the misinterpretation makes the problem harder, so that the solution actually relies on all the information. This suggests that the question-writer may have been mistaken but got lucky.
~oinava, based on demonstration of misinterpretation fakesolve by ~orenbad
Video Solution 1 (One Key Observation)
~Education, the Study of Everything
Video Solution 2 (Let's first understand the question)
https://youtu.be/7yAh4MtJ8a8?si=_jfMzsAiHgAGJ1uo&t=1552
~Math-X
Video Solution 3
Video Solution by TheBeautyofMath
https://youtu.be/0kkc4-y8TkU?t=254
~IceMatrix
See Also
| 2022 AMC 10A (Problems • Answer Key • Resources) | ||
| Preceded by Problem 11 |
Followed by Problem 13 | |
| 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 10 Problems and Solutions | ||
| 2022 AMC 12A (Problems • Answer Key • Resources) | |
| Preceded by Problem 8 |
Followed by Problem 10 |
| 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. 
