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

(certain china test papers)
(Solution (The simplest way))
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==Solution (The simplest way)==
 
==Solution (The simplest way)==
Here is the English translation with selective LaTeX formatting:
 
 
 
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>.  
 
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>.  
  

Revision as of 08:10, 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?

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

Certain China test papers: Let \[P(m)=\frac{m}{2}+\frac{m^2}{4}+\frac{m^4}{8}+\frac{m^8}{8}\] How many of the values $P(2022)$, $P(2023)$, $P(2024)$, $P(2025)$ and $P(2026)$ are integers?

$\textbf{(A) } 1 \qquad\textbf{(B) } 2 \qquad\textbf{(C) } 3 \qquad\textbf{(D) } 4 \qquad\textbf{(E) } 5$

Solution (The simplest way)

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

Certain China test papers:

As explained above, numbers $P(2022)$ to $P(2025)$ are integers. The difference is that there is a new number, $P(2026)$. Since it is divisible by 2, we can see that it is also an integer.

Therefore, the answer is $\boxed{\textbf{(E) }5}$.

~iHateGeometry

See also

2024 AMC 10B (ProblemsAnswer KeyResources)
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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