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

Latest revision as of 01:57, 20 December 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 $\boxed{\textbf{(E) }4}$ .

-nevergonnagiveup

~flyingkinder123 (minor edits)

Solution 4

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 by Pi Academy (Fast and Easy)

https://youtu.be/Xn1JLzT7mW4?si=borSg8xrYLAz7mNY

Video Solution 2 by SpreadTheMathLove

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

@@ Line 39: / Line 39: @@
 ==Solution 3 (Factoring)==
-We can rewrite the expression as <cmath>\frac{1}{8}\cdot m \cdot (m^7+m^3+2m+4).</cmath> If <math>m</math> is even, then <math>m</math> gives a factor of <math>2</math> and <math>m^7+m^3+2m+4</math> will give a factor of <math>4,</math> so the result will be an integer. If <math>m</math> is odd, then notice that for any <math>m</math> we have <math>m^2\equiv 1 \pmod{8}.</math> Then <math>m^7+m^3+2m+4\equiv m+m+2m+4 \equiv 4(m+1) \equiv 0 \pmod{8}.</math> So any integer <math>m</math> will result in an integer, meaning the answer is <math>\boxed{\textbf{(E) }5}</math>.
+We can rewrite the expression as <cmath>\frac{1}{8}\cdot m \cdot (m^7+m^3+2m+4).</cmath> If <math>m</math> is even, then <math>m</math> gives a factor of <math>2</math> and <math>m^7+m^3+2m+4</math> will give a factor of <math>4,</math> so the result will be an integer. If <math>m</math> is odd, then notice that for any <math>m</math> we have <math>m^2\equiv 1 \pmod{8}.</math> Then <math>m^7+m^3+2m+4\equiv m+m+2m+4 \equiv 4(m+1) \equiv 0 \pmod{8}.</math> So any integer <math>m</math> will result in an integer, meaning the answer is <math>\boxed{\textbf{(E) }4}</math>.
 -nevergonnagiveup

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

Page

Toolbox

Search