Difference between revisions of "2024 AMC 10B Problems/Problem 24"
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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? | ||
− | + | <math>\textbf{(A) } 0 \qquad\textbf{(B) } 1 \qquad\textbf{(C) } 2 \qquad\textbf{(D) } 3 \qquad\textbf{(E) } 4</math> | |
− | |||
− | <math>\textbf{(A) } | ||
− | ==Solution | + | ==Solution (The simplest way)== |
− | + | 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>. | |
− | + | 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. | ||
− | + | 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) }4}</math> possible values. | |
+ | ~elpianista227 | ||
− | ~[https://artofproblemsolving.com/wiki/index.php/User: | + | ==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 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== | ==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}} | ||
− | |||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 19:57, 14 November 2024
Contents
[hide]Problem
Let How many of the values , , , and are integers?
Solution (The simplest way)
First, we know that and must be integers since they are both divisible by .
Then Let’s consider the remaining two numbers. Since they are not divisible by , the result of the first term must be a certain number , and the result of the second term must be a certain number . Similarly, the remaining two terms must each be . Their sum is , so and are also integers.
Therefore, the answer is .
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 : becomes And in order for to be an integer, it's important to note that must be congruent to 0 modulo 8. Moreover, we know that . We can verify it by taking everything modulo 8 :
If , then -> TRUE If , then -> TRUE If , then it is obvious that the entire expression is divisible by 8. Therefore, it is true. If , then . Therefore, -> TRUE Therefore, there are possible values.
Addendum for certain China test papers : Note that . Therefore, taking everything modulo 8, whilst still maintaining the original expression, gives . This is true.
Therefore, there are possible values. ~elpianista227
Remark
On certain versions of the AMC in China, the problem was restated as follows:
LetHow many of the values , , , and are integers?
By identical reasoning, each term of is an integer, since is even.
Therefore, the answer is .
~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)
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.