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Difference between revisions of "2024 AMC 12B Problems"
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==Problem 2==
 
==Problem 2==
 +
What is <math>10! - 7! \cdot 6!</math>?
 +
 +
<math>\textbf{(A) }-120 \qquad\textbf{(B) }0 \qquad\textbf{(C) }120 \qquad\textbf{(D) }600 \qquad\textbf{(E) }720 \qquad</math>
 +
 
[[2024 AMC 12B Problems/Problem 2|Solution]]
 
[[2024 AMC 12B Problems/Problem 2|Solution]]
  
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==Problem 4==
 
==Problem 4==
 +
 +
Balls numbered <math>1,2,3,\ldots</math> are deposited in <math>5</math> bins, labeled <math>A,B,C,D,</math> and <math>E</math>, using the following procedure. Ball <math>1</math> is deposited in bin <math>A</math>, and balls <math>2</math> and <math>3</math> are deposited in <math>B</math>. The next three balls are deposited in bin <math>C</math>, the next <math>4</math> in bin <math>D</math>, and so on, cycling back to bin <math>A</math> after balls are deposited in bin <math>E</math>. (For example, <math>22,23,\ldots,28</math> are deposited in bin <math>B</math> at step 7 of this process.) In which bin is ball <math>2024</math> deposited?
 +
 +
<math>\textbf{(A) }A\qquad\textbf{(B) }B\qquad\textbf{(C) }C\qquad\textbf{(D) }D\qquad\textbf{(E) }E</math>
 +
 
[[2024 AMC 12B Problems/Problem 4|Solution]]
 
[[2024 AMC 12B Problems/Problem 4|Solution]]
  
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==Problem 10==
 
==Problem 10==
 +
 +
A list of 9 real numbers consists of <math>1</math>, <math>2.2 </math>, <math>3.2 </math>, <math>5.2 </math>, <math>6.2 </math>, <math>7</math>, as well as <math>x, y,z</math> with <math>x\leq y\leq z</math>. The range of the list is <math>7</math>, and the mean and median are both positive integers. How many ordered triples <math>(x,y,z)</math> are possible?
 +
 +
<math>\textbf{(A) }1 \qquad\textbf{(B) }2 \qquad\textbf{(C) }3 \qquad\textbf{(D) }4 \qquad\textbf{(E) \text{infinitely many}}\qquad</math>
 +
 
[[2024 AMC 12B Problems/Problem 10|Solution]]
 
[[2024 AMC 12B Problems/Problem 10|Solution]]
  
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==Problem 14==
 
==Problem 14==
 +
How many different remainders can result when the <math>100</math>th power of an integer is divided by <math>125</math>?
 +
 +
<math>\textbf{(A) }1 \qquad\textbf{(B) }2 \qquad\textbf{(C) }5 \qquad\textbf{(D) }25 \qquad\textbf{(E) }125 \qquad</math>
 +
 
[[2024 AMC 12B Problems/Problem 14|Solution]]
 
[[2024 AMC 12B Problems/Problem 14|Solution]]
  
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==Problem 25==
 
==Problem 25==
 +
 +
Pablo will decorate each of <math>6</math> identical white balls with either a striped or a dotted pattern, using either red or blue paint. He will decide on the color and pattern for each ball by flipping a fair coin for each of the <math>12</math> decisions he must make. After the paint dries, he will place the <math>6</math> balls in an urn. Frida will randomly select one ball from the urn and note its color and pattern. The events "the ball Frida selects is red" and "the ball Frida selects is striped" may or may not be independent, depending on the outcome of Pablo's coin flips. The probability that these two events are independent can be written as <math>\frac mn,</math> where <math>m</math> and <math>n</math> are relatively prime positive integers. What is <math>m?</math> (Recall that two events <math>A</math> and <math>B</math> are independent if <math>P(A \text{ and }B) = P(A) \cdot P(B).</math>)
 +
 +
<math>\textbf{(A) } 243 \qquad \textbf{(B) } 245 \qquad \textbf{(C) } 247 \qquad \textbf{(D) } 249\qquad \textbf{(E) } 251</math>
 +
 
[[2024 AMC 12B Problems/Problem 25|Solution]]
 
[[2024 AMC 12B Problems/Problem 25|Solution]]
  
 
==See also==
 
==See also==
−
{{AMC12 box|year=2024|ab=B|before=[[2023 AMC 12B Problems]]|after=[[2025 AMC 12A Problems]]}}
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{{AMC12 box|year=2024|ab=B|before=[[2024 AMC 12A Problems]]|after=[[2025 AMC 12A Problems]]}}
−
* [[AMC 12]]
+
 
−
* [[AMC 12 Problems and Solutions]]
+
[[AMC 12]]
−
* [[Mathematics competitions]]
+
 
−
* [[Mathematics competition resources]]
+
[[AMC 12 Problems and Solutions]]
 +
 
 +
[[Mathematics competitions]]
 +
 
 +
[[Mathematics competition resources]]

Revision as of 00:18, 14 November 2024

2024 AMC 12B (Answer Key)
Printable versions: Wiki • AoPS Resources • PDF

Instructions

  1. This is a 25-question, multiple choice test. Each question is followed by answers marked A, B, C, D and E. Only one of these is correct.
  2. You will receive 6 points for each correct answer, 2.5 points for each problem left unanswered if the year is before 2006, 1.5 points for each problem left unanswered if the year is after 2006, and 0 points for each incorrect answer.
  3. No aids are permitted other than scratch paper, graph paper, ruler, compass, protractor and erasers (and calculators that are accepted for use on the test if before 2006. No problems on the test will require the use of a calculator).
  4. Figures are not necessarily drawn to scale.
  5. You will have 75 minutes working time to complete the test.
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

Problem 1

Solution

Problem 2

What is $10! - 7! \cdot 6!$?

$\textbf{(A) }-120 \qquad\textbf{(B) }0 \qquad\textbf{(C) }120 \qquad\textbf{(D) }600 \qquad\textbf{(E) }720 \qquad$

Solution

Problem 3

Solution

Problem 4

Balls numbered $1,2,3,\ldots$ are deposited in $5$ bins, labeled $A,B,C,D,$ and $E$, using the following procedure. Ball $1$ is deposited in bin $A$, and balls $2$ and $3$ are deposited in $B$. The next three balls are deposited in bin $C$, the next $4$ in bin $D$, and so on, cycling back to bin $A$ after balls are deposited in bin $E$. (For example, $22,23,\ldots,28$ are deposited in bin $B$ at step 7 of this process.) In which bin is ball $2024$ deposited?

$\textbf{(A) }A\qquad\textbf{(B) }B\qquad\textbf{(C) }C\qquad\textbf{(D) }D\qquad\textbf{(E) }E$

Solution

Problem 5

Solution

Problem 6

Solution

Problem 7

Solution

Problem 8

Solution

Problem 9

Solution

Problem 10

A list of 9 real numbers consists of $1$, $2.2$, $3.2$, $5.2$, $6.2$, $7$, as well as $x, y,z$ with $x\leq y\leq z$. The range of the list is $7$, and the mean and median are both positive integers. How many ordered triples $(x,y,z)$ are possible?

$\textbf{(A) }1 \qquad\textbf{(B) }2 \qquad\textbf{(C) }3 \qquad\textbf{(D) }4 \qquad\textbf{(E) \text{infinitely many}}\qquad$

Solution

Problem 11

Solution

Problem 12

Solution

Problem 13

Solution

Problem 14

How many different remainders can result when the $100$th power of an integer is divided by $125$?

$\textbf{(A) }1 \qquad\textbf{(B) }2 \qquad\textbf{(C) }5 \qquad\textbf{(D) }25 \qquad\textbf{(E) }125 \qquad$

Solution

Problem 15

Solution

Problem 16

Solution

Problem 17

Solution

Problem 18

Solution

Problem 19

Solution

Problem 20

Solution

Problem 21

Solution

Problem 22

Solution

Problem 23

Solution

Problem 24

Solution

Problem 25

Pablo will decorate each of $6$ identical white balls with either a striped or a dotted pattern, using either red or blue paint. He will decide on the color and pattern for each ball by flipping a fair coin for each of the $12$ decisions he must make. After the paint dries, he will place the $6$ balls in an urn. Frida will randomly select one ball from the urn and note its color and pattern. The events "the ball Frida selects is red" and "the ball Frida selects is striped" may or may not be independent, depending on the outcome of Pablo's coin flips. The probability that these two events are independent can be written as $\frac mn,$ where $m$ and $n$ are relatively prime positive integers. What is $m?$ (Recall that two events $A$ and $B$ are independent if $P(A \text{ and }B) = P(A) \cdot P(B).$)

$\textbf{(A) } 243 \qquad \textbf{(B) } 245 \qquad \textbf{(C) } 247 \qquad \textbf{(D) } 249\qquad \textbf{(E) } 251$

Solution

See also

2024 AMC 12B (Problems • Answer Key • Resources)
Preceded by
2024 AMC 12A Problems 		Followed by
2025 AMC 12A Problems
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

AMC 12

AMC 12 Problems and Solutions

Mathematics competitions

Mathematics competition resources

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