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

(Solution (The simplest way))
(Solution 2 (Specific))
Line 36: Line 36:
 
Therefore, there are <math>\boxed{\textbf{(E) }5}</math> possible values.   
 
Therefore, there are <math>\boxed{\textbf{(E) }5}</math> possible values.   
 
~elpianista227
 
~elpianista227
 +
 +
==Remark==
 +
On certain versions of the AMC in China, the problem was restates as follows:
 +
 +
Let<cmath>P(m)=\frac{m}{2}+\frac{m^2}{4}+\frac{m^4}{8}+\frac{m^8}{8}</cmath>How many of the values <math>P(2022)</math>, <math>P(2023)</math>, <math>P(2024)</math>, <math>P(2025),</math> and <math>P(2026)</math> are integers?\[10pt]
 +
<math>\textbf{(A) } 1 \qquad\textbf{(B) } 2 \qquad\textbf{(C) } 3 \qquad\textbf{(D) } 4 \qquad\textbf{(E) } 5</math>
 +
 +
By identical reasoning, each term of <math>P</math> is an integer, since <math>2026</math> is even.
 +
 +
Therefore, the answer is <math>\boxed{\textbf{(E) }5}</math>.
 +
 +
~[https://artofproblemsolving.com/wiki/index.php/User:iHateGeometry iHateGeometry], countmath1
  
 
==Video Solution 1 by Pi Academy (In Less Than 2 Mins ⚡🚀)==
 
==Video Solution 1 by Pi Academy (In Less Than 2 Mins ⚡🚀)==

Revision as of 12:35, 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)

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

Solution 2 (Specific)

Take everything modulo 8 and re-write the entire fraction with denominator 8. This means that we're going to transform the fraction as follows : \[P(m)=\frac{m}{2}+\frac{m^2}{4}+\frac{m^4}{8}+\frac{m^8}{8}\] becomes \[P(m)=\frac{4m + 2m^2 + m^4 + m^8}{8}\] And in order for $P(m)$ to be an integer, it's important to note that $4m + 2m^2 + m^4 + m^8$ must be congruent to 0 modulo 8. Moreover, we know that $2022 \equiv 6 (mod 8), 2023 \equiv 7 (mod 8), 2024 \equiv 0 (mod 8), 2025 \equiv 1 (mod 8)$. We can verify it by taking everything modulo 8 :

If $m = 2022$, then $4(6) + 2(4) + 0 + 0 = 24 + 8 = 32 \equiv 0 (mod 8)$ -> TRUE If $m = 2023$, then $4(7) + 2(1) + 1 + 1 = 28 + 2 + 1 + 1 = 32 \equiv 0 (mod 8)$ -> TRUE If $m = 2024$, then it is obvious that the entire expression is divisible by 8. Therefore, it is true. If $m = 2025$, then $2025 \equiv 1 (mod 8)$. Therefore, $4(1) + 2(1) + 1 + 1 = 8 \equiv 0 (mod 8)$ -> TRUE Therefore, there are $\boxed{\textbf{(E) }4}$ possible values.

Addendum for certain China test papers : Note that $2026 \equiv 2 (mod 8)$. Therefore, taking everything modulo 8, whilst still maintaining the original expression, gives $4(2) + 2(4) + 0 + 0 = 16 \equiv 0 (mod 8)$. This is true.

Therefore, there are $\boxed{\textbf{(E) }5}$ possible values. ~elpianista227

Remark

On certain versions of the AMC in China, the problem was restates as follows:

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?\[10pt] $\textbf{(A) } 1 \qquad\textbf{(B) } 2 \qquad\textbf{(C) } 3 \qquad\textbf{(D) } 4 \qquad\textbf{(E) } 5$

By identical reasoning, each term of $P$ is an integer, since $2026$ is even.

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

~iHateGeometry, countmath1

Video Solution 1 by Pi Academy (In Less Than 2 Mins ⚡🚀)

https://youtu.be/Xn1JLzT7mW4?feature=shared

~ Pi Academy

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png