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

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{{duplicate|[[2024 AMC 10B Problems/Problem 24|2024 AMC 10B #24]] and [[2024 AMC 12B Problems/Problem 18|2024 AMC 12B #18]]}}
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==Problem==
 +
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>, and <math>P(2025)</math> are integers?
  
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==Problem 18==
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<math>\textbf{(A) } 0 \qquad\textbf{(B) } 1 \qquad\textbf{(C) } 2 \qquad\textbf{(D) } 3 \qquad\textbf{(E) } 4</math>
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The Fibonacci numbers are defined by <math>F_1=1,</math> <math>F_2=1,</math> and <math>F_n=F_{n-1}+F_{n-2}</math> for <math>n\geq 3.</math> What is<cmath>\dfrac{F_2}{F_1}+\dfrac{F_4}{F_2}+\dfrac{F_6}{F_3}+\cdots+\dfrac{F_{20}}{F_{10}}?</cmath>
 
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<math>\textbf{(A) }318 \qquad\textbf{(B) }319\qquad\textbf{(C) }320\qquad\textbf{(D) }321\qquad\textbf{(E) }322</math>
 
  
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==Solution #1==
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==Solution (The simplest way)==
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The first 20 terms F_n = 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, 610, 987, 1597, 2584, 4181, 6765
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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>.
  
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so ans = 1 + 3 + 4 + 7 + 11 + 18 + 29 + 47 + 76 + 123 = <math>\boxed{(B) 319} </math>.  
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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.
  
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~[https://artofproblemsolving.com/wiki/index.php/User:Cyantist luckuso]
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Therefore, the answer is <math>\boxed{\textbf{(E) }4}</math>.
 +
 
 +
~[https://artofproblemsolving.com/wiki/index.php/User:Athmyx 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 :
 +
<cmath>P(m)=\frac{m}{2}+\frac{m^2}{4}+\frac{m^4}{8}+\frac{m^8}{8}</cmath>
 +
becomes
 +
<cmath>P(m)=\frac{4m + 2m^2 + m^4 + m^8}{8}</cmath>
 +
And in order for <math>P(m)</math> to be an integer, it's important to note that <math>4m + 2m^2 + m^4 + m^8</math> must be congruent to 0 modulo 8.
 +
Moreover, we know that <math>2022 \equiv 6 \pmod 8, 2023 \equiv 7 \pmod 8, 2024 \equiv 0 \pmod 8, 2025 \equiv 1 \pmod 8</math>. We can verify it by taking everything modulo 8 :
 +
 
 +
If <math>m = 2022</math>, then <math>4(6) + 2(4) + 0 + 0 = 24 + 8 = 32 \equiv 0 \pmod 8</math> -> TRUE
 +
If <math>m = 2023</math>, then <math>4(7) + 2(1) + 1 + 1 = 28 + 2 + 1 + 1 = 32 \equiv 0 \pmod 8</math> -> TRUE
 +
If <math>m = 2024</math>, then it is obvious that the entire expression is divisible by 8. Therefore, it is true.
 +
If <math>m = 2025</math>, then <math>2025 \equiv 1 \pmod 8</math>. Therefore, <math>4(1) + 2(1) + 1 + 1 = 8 \equiv 0 \pmod 8</math> -> TRUE
 +
Therefore, there are <math>\boxed{\textbf{(E) }4}</math> possible values.
 +
 
 +
Addendum for certain China test papers :
 +
Note that <math>2026 \equiv 2 \pmod 8</math>. Therefore, taking everything modulo 8, whilst still maintaining the original expression, gives <math>4(2) + 2(4) + 0 + 0 = 16 \equiv 0 \pmod 8</math>. This is true.
 +
 
 +
Therefore, there are <math>\boxed{\textbf{(E) }5}</math> possible values. 
 +
~elpianista227
 +
 
 +
==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>.
 +
 
 +
-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 <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. (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.
 +
 
 +
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
 +
 
 +
==Remark==
 +
On certain versions of the AMC in China, the problem was restated 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?
 +
<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 by chicken_rice (super clear and quick + subscribe plz)
 +
 
 +
https://www.youtube.com/watch?v=kQGbyhze_mY
 +
 
 +
~chicken rice friend
 +
 
 +
==Video Solution 2 by SpreadTheMathLove==
 +
https://www.youtube.com/watch?v=24EZaeAThuE
 +
 
 +
Video Solution 3
 +
by chicken_rice (SUPER CLEAR AND QUICK!!! SUBSCRIBE!!!)
 +
https://www.youtube.com/watch?v=kQGbyhze_mY
  
 
==See also==
 
==See also==
 
{{AMC10 box|year=2024|ab=B|num-b=23|num-a=25}}
 
{{AMC10 box|year=2024|ab=B|num-b=23|num-a=25}}
−
{{AMC12 box|year=2024|ab=B|num-b=17|num-a=19}}
 
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 19:04, 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

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 chicken_rice (super clear and quick + subscribe plz)

https://www.youtube.com/watch?v=kQGbyhze_mY

~chicken rice friend

Video Solution 2 by SpreadTheMathLove

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

Video Solution 3 by chicken_rice (SUPER CLEAR AND QUICK!!! SUBSCRIBE!!!) https://www.youtube.com/watch?v=kQGbyhze_mY

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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