Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2024 AMC 10B Problems/Problem 24"
m (→‎Solution 4 (Tedious))
 
(One intermediate revision by the same user not shown)
Line 49: Line 49:
 
Taking a closer look at the terms, we notice that each term builds off of the previous one. There is <math>\frac{m}{2}</math>, <math>\frac{m^{2}}{4}</math>, which is equal to <math>\left(\frac{m}{2}\right)^{2}</math>, <math>\frac{m^{4}}{8}</math>, which is equal to <math>2\left(\frac{m^{2}}{4}\right)^{2}</math>, and <math>\frac{m^{8}}{8}</math>, which is equal to <math>8\left(\frac{m^{4}}{8}\right)^{2}</math>. Ultimately, this means that there are only two cases that we need to check for: the case in which <math>m</math> is even and the case in which <math>m</math> is odd.
 
Taking a closer look at the terms, we notice that each term builds off of the previous one. There is <math>\frac{m}{2}</math>, <math>\frac{m^{2}}{4}</math>, which is equal to <math>\left(\frac{m}{2}\right)^{2}</math>, <math>\frac{m^{4}}{8}</math>, which is equal to <math>2\left(\frac{m^{2}}{4}\right)^{2}</math>, and <math>\frac{m^{8}}{8}</math>, which is equal to <math>8\left(\frac{m^{4}}{8}\right)^{2}</math>. Ultimately, this means that there are only two cases that we need to check for: the case in which <math>m</math> is even and the case in which <math>m</math> is odd.
  
−
If <math>m</math> is even, <math>\frac{m}{2}</math> will be an integer, which means the rest of the terms will be an integer. This means in the problem, <math>P(2022)</math> and <math>P(2024)</math> will yield an integer result.
+
If <math>m</math> is even, <math>\frac{m}{2}</math> will be an integer, which means the rest of the terms will be an integer. This means in the problem, <math>P(2022)</math> and <math>P(2024)</math> will yield an integer result. (For certain Chinese versions, this also proves that <math>P(2026)</math> is an integer.)
  
 
If <math>m</math> is odd, <math>\frac{m}{2}</math> will result in <math>\frac{1}{2}</math>. Building off of each term will give you <math>\frac{1}{4}</math>, <math>\frac{1}{8}</math>, and <math>\frac{1}{8}</math>, and summing those up will grant <math>1</math>, an integer. This means in the problem, <math>P(2023)</math> and <math>P(2025)</math> will also result in integer answers.
 
If <math>m</math> is odd, <math>\frac{m}{2}</math> will result in <math>\frac{1}{2}</math>. Building off of each term will give you <math>\frac{1}{4}</math>, <math>\frac{1}{8}</math>, and <math>\frac{1}{8}</math>, and summing those up will grant <math>1</math>, an integer. This means in the problem, <math>P(2023)</math> and <math>P(2025)</math> will also result in integer answers.
  
−
This will get us to <math>\boxed{\textbf{(E) }4}</math> possible values.
+
In general, all integer <math>m</math> will make <math>P(m)</math> give an integer answer, but for this question, this will get us to <math>\boxed{\textbf{(E) }4}</math> integer values (<math>\boxed{\textbf{(E) }5}</math> for certain Chinese versions of the test).
  
 
~unpogged
 
~unpogged

Latest revision as of 01:36, 20 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 $2022 \equiv 6 \pmod 8, 2023 \equiv 7 \pmod 8, 2024 \equiv 0 \pmod 8, 2025 \equiv 1 \pmod 8$. We can verify it by taking everything modulo 8 :

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

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

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

Solution 3 (Factoring)

We can rewrite the expression as \[\frac{1}{8}\cdot m \cdot (m^7+m^3+2m+4).\] If $m$ is even, then $m$ gives a factor of $2$ and $m^7+m^3+2m+4$ will give a factor of $4,$ so the result will be an integer. If $m$ is odd, then notice that for any $m$ we have $m^2\equiv 1 \pmod{8}.$ Then $m^7+m^3+2m+4\equiv m+m+2m+4 \equiv 4(m+1) \equiv 0 \pmod{8}.$ So any integer $m$ will result in an integer, meaning the answer is $\boxed{\textbf{(E) }5}$.

-nevergonnagiveup

~flyingkinder123 (minor edits)

Solution 4 (Tedious)

Taking a closer look at the terms, we notice that each term builds off of the previous one. There is $\frac{m}{2}$, $\frac{m^{2}}{4}$, which is equal to $\left(\frac{m}{2}\right)^{2}$, $\frac{m^{4}}{8}$, which is equal to $2\left(\frac{m^{2}}{4}\right)^{2}$, and $\frac{m^{8}}{8}$, which is equal to $8\left(\frac{m^{4}}{8}\right)^{2}$. Ultimately, this means that there are only two cases that we need to check for: the case in which $m$ is even and the case in which $m$ is odd.

If $m$ is even, $\frac{m}{2}$ will be an integer, which means the rest of the terms will be an integer. This means in the problem, $P(2022)$ and $P(2024)$ will yield an integer result. (For certain Chinese versions, this also proves that $P(2026)$ is an integer.)

If $m$ is odd, $\frac{m}{2}$ will result in $\frac{1}{2}$. Building off of each term will give you $\frac{1}{4}$, $\frac{1}{8}$, and $\frac{1}{8}$, and summing those up will grant $1$, an integer. This means in the problem, $P(2023)$ and $P(2025)$ will also result in integer answers.

In general, all integer $m$ will make $P(m)$ give an integer answer, but for this question, this will get us to $\boxed{\textbf{(E) }4}$ integer values ($\boxed{\textbf{(E) }5}$ for certain Chinese versions of the test).

~unpogged

Remark

On certain versions of the AMC in China, the problem was restated 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? $\textbf{(A) } 1 \qquad\textbf{(B) } 2 \qquad\textbf{(C) } 3 \qquad\textbf{(D) } 4 \qquad\textbf{(E) } 5$

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

Video Solution 2 by SpreadTheMathLove

https://www.youtube.com/watch?v=24EZaeAThuE

Video Solution 3 by sevenblade(modular arithmetic)

https://youtu.be/5BXclh_DLEg

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

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

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2024_AMC_10B_Problems/Problem_24&oldid=235127"