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Difference between revisions of "2024 AMC 10B Problems/Problem 24"

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Here is the English translation with selective LaTeX formatting:
 
Here is the English translation with selective LaTeX formatting:
  
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First, we know that <math>P(2022)</math> and <math>P(2024)</math> must be integers since they are both divisible by <math>2</math>. Next, let’s consider the remaining two numbers. Since they are not divisible by <math>2</math>, the result of the first term must be a certain number <math>+\frac{1}{2}</math>, and the result of the second term must be a certain number <math>+\frac{1}{4}</math>. Similarly, the remaining two terms must each be <math>\frac{1}{8}</math>.
+
First, we know that <math>P(2022)</math> and <math>P(2024)</math> must be integers since they are both divisible by <math>2</math>.  
  
−
Their sum is <math>\frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \frac{1}{8} = 1</math>, so <math>P(2023)</math> and <math>P(2024)</math> are also integers.
+
Then Let’s consider the remaining two numbers. Since they are not divisible by <math>2</math>, the result of the first term must be a certain number <math>+\frac{1}{2}</math>, and the result of the second term must be a certain number <math>+\frac{1}{4}</math>. Similarly, the remaining two terms must each be <math>\frac{1}{8}</math>.
 +
Their sum is <math>\frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \frac{1}{8} = 1</math>, so <math>P(2023)</math> and <math>P(2025)</math> are also integers.
  
 
Therefore, the answer is <math>\boxed{\textbf{(E) }4}</math>.
 
Therefore, the answer is <math>\boxed{\textbf{(E) }4}</math>.

Revision as of 06:08, 14 November 2024

Problem

Let \[P(m)=\frac{m}{2}+\frac{m^2}{4}+\frac{m^4}{8}+\frac{m^8}{8}\] How many of the values $P(2022)$, $P(2023)$, $P(2024)$, and $P(2025)$ are integers?

$\textbf{(A) } 0 \qquad\textbf{(B) } 1 \qquad\textbf{(C) } 2 \qquad\textbf{(D) } 3 \qquad\textbf{(E) } 4$

Solution (The simplest way)

Here is the English translation with selective LaTeX formatting:

First, we know that $P(2022)$ and $P(2024)$ must be integers since they are both divisible by $2$.

Then Let’s consider the remaining two numbers. Since they are not divisible by $2$, the result of the first term must be a certain number $+\frac{1}{2}$, and the result of the second term must be a certain number $+\frac{1}{4}$. Similarly, the remaining two terms must each be $\frac{1}{8}$. Their sum is $\frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \frac{1}{8} = 1$, so $P(2023)$ and $P(2025)$ are also integers.

Therefore, the answer is $\boxed{\textbf{(E) }4}$.

~Athmyx

See also

2024 AMC 10B (Problems • Answer Key • Resources)
Preceded by
Problem 23 		Followed by
Problem 25
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

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