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| {{duplicate|[[2025 AMC 10A Problems/Problem 8|2025 AMC 10A #8]] and [[2025 AMC 12A Problems/Problem 4|2025 AMC 12A #4]]}}
| | #redirect [[2025 AMC 12A Problems/Problem 4]] |
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| ==Problem==
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| Agnes writes the following four statements on a blank piece of paper.
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| <imath>\bullet</imath> At least one of these statements is true.
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| <imath>\bullet</imath> At least two of these statements are true.
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| <imath>\bullet</imath> At least two of these statements are false.
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| <imath>\bullet</imath> At least one of these statements is false.
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| Each statement is either true or false. How many false statements did Agnes write on the paper?
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| <imath>\textbf{(A) } 0 \qquad\textbf{(B) } 1 \qquad\textbf{(C) } 2 \qquad\textbf{(D) } 3 \qquad\textbf{(E) } 4</imath>
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| ==Solution 1==
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| We first number all the statements:
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| 1) At least one of these statements is true.
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| 2) At least two of these statements are true.
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| 3) At least two of these statements are false.
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| 4) At least one of these statements is false.
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| We can immediately see that statement 4 must be true, as it would contradict itself if it were false. Similarly, statement 1 must be true, as all the other statements must be false if it were true, which is contradictory because statement 4 is true. Since both 1 and 4 are true, statement 2 has to be true. Therefore, statement 3 is the only false statement, making the answer <imath>\boxed{\text{(B) }1}</imath>.
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| -Rainjs
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| ==Solution 2==
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| Statements <imath>I,II,</imath> and <imath>IV</imath> are true, while statement <imath>III</imath> is false. Hence, there are <imath>3</imath> true statements and <imath>\boxed{\text{(B) }1}</imath> false statement. This result can be checked by examining the statements individually again.
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| Statements <imath>I</imath> and <imath>II</imath> will be true because there are <imath>3\ge2</imath> true statements. Statement <imath>IV</imath> is also true because there is <imath>1\ge1</imath> false statement. Finally, statement <imath>III</imath> is false because there are <imath>1\ngeq2</imath> false statements.
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| ~Tacos_are_yummy_1
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| ==Solution 3==
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| Let's say there are <imath>T</imath> true statements. We know that <imath>T</imath> can be any integer from <imath>0</imath> to <imath>4</imath>. We denote <imath>A</imath> as <imath>T \geq 1</imath>, Statement <imath>B</imath> as <imath>T \geq 2</imath>, Statement <imath>C</imath> as <imath>T \leq 2</imath>, and Statement <imath>D</imath> as <imath>T \leq 3</imath>.
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| If <imath>T=0</imath>, then <imath>C</imath> and <imath>D</imath> are met, so there are <imath>2</imath> true statements, which is a contradiction.
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| If <imath>T=1</imath>, then <imath>A,C,D</imath> are met, so there are <imath>3</imath> true statements, which is a contradiction.
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| If <imath>T=2</imath>, then <imath>A,B,C,D</imath> are met, so there are <imath>4</imath> true statements, which is a contradiction.
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| If <imath>T=3</imath>, then <imath>A,B,D</imath> are met, so there are <imath>3</imath> true statements, which is consistent with our assumption that <imath>T=3</imath>.
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| If <imath>T=4</imath>, then <imath>A,B</imath> are met, so there are <imath>2</imath> true statements, which is a contradiction.
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| Only <imath>T=3</imath> was consistent, so there are <imath>3</imath> true statements and <imath>4-3=\boxed{1}</imath> false statement. (In particular, Statement C is the false statement).
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| ~lprado
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| ==Solution 4==
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| Suppose Statement <imath>I</imath> is false, then none of the Statements are true, which contradicts the fact that a false Statement <imath>III</imath> or <imath>IV</imath> is telling the truth. Therefore, Statement <imath>I</imath> is true and assume Statement <imath>II</imath> is false.
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| Statement <imath>II</imath> thus implies that only Statement <imath>I</imath> was the truth, and the rest, false. But then, there are 3 false statements but then Statement <imath>III</imath> and Statement <imath>IV</imath> are telling the truth. So Statement <imath>II</imath> is also true.
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| Now, if Statement <imath>III</imath> is true, then both Statement <imath>III</imath> and Statement <imath>IV</imath> is false, contradicting the fact that it is true. Statement <imath>III</imath> is hence false and Statement <imath>IV</imath> tells the truth since Statement <imath>III</imath> lied so indeed, there are at least one lie. There are a total of 3 truths and 1 lie, making the answer <imath>\boxed{\text{(B) }1}</imath>. ~hxve
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| ==Solution 5 (Quick)==
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| I tried setting that the statements regarding the true ones (I and II) were true and
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| regarding the false ones (III and IV) were false, and it clearly doesn't work.
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| It doesn't work because statement III was true, not false as I set it to be originally.
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| What if we make that statement (Statement III) true too?
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| Well, then we have 3 T and 1 F statements, so statement IV is false,
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| statement III is true, and statements I and II are true too.
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| So there is <imath>\boxed{\text{(B) }1}</imath> false statement.
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| ~Aarav22
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| ==Chinese Video Solution==
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| https://www.bilibili.com/video/BV1t72uBREof/
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| ~metrixgo
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| ==Video Solution by SpreadTheMathLove==
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| https://www.youtube.com/watch?v=dAeyV60Hu5c
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| == Video Solution (In 1 Min) ==
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| https://youtu.be/uv3uIMwIkrg?si=XCbsXL7ikMawCGyM ~ Pi Academy
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| ==Video Solution by Daily Dose of Math==
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| https://youtu.be/gPh9w3X3QSw
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| ~Thesmartgreekmathdude
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| ==See Also==
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| {{AMC10 box|year=2025|ab=A|num-b=7|num-a=9}}
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| {{AMC12 box|year=2025|ab=A|num-b=3|num-a=5}}
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| * [[AMC 10]]
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| * [[AMC 10 Problems and Solutions]]
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| * [[Mathematics competitions]]
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| * [[Mathematics competition resources]]
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| {{MAA Notice}}
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