Difference between revisions of "2024 AMC 10B Problems/Problem 24"
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<cmath>P(m)=\frac{4m + 2m^2 + m^4 + m^8}{8}</cmath> | <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. \\ | 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 (mod 8), 2023 \equiv ( | + | Moreover, we know that <math>2022 \equiv 6 (mod 8), 2023 \equiv 7 (mod 8), 2024 \equiv 0 (mod 8), 2025 \equiv 1 (mod 8)</math>. We can verify it by taking everything modulo 8 : |
\begin{itemize} | \begin{itemize} | ||
\item If <math>m = 2022</math>, then <math>4(6) + 2(4) + 0 + 0 = 24 + 8 = 32 \equiv 0 (mod 8)</math> -> TRUE | \item If <math>m = 2022</math>, then <math>4(6) + 2(4) + 0 + 0 = 24 + 8 = 32 \equiv 0 (mod 8)</math> -> TRUE | ||
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Therefore, there are <math>\boxed{\textbf{(E) }5}</math> possible values. | Therefore, there are <math>\boxed{\textbf{(E) }5}</math> possible values. | ||
~elpianista227 | ~elpianista227 | ||
+ | |||
==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 09:42, 14 November 2024
Problem
Let How many of the values , , , and are integers?
Certain China test papers: 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 .
Certain China test papers:
As explained above, numbers to are integers. The difference is that there is a new number, . Since it is divisible by 2, we can see that it is also an integer.
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 : \begin{itemize}
\item If , then -> TRUE \item If , then -> TRUE \item If , then it is obvious that the entire expression is divisible by 8. Therefore, it is true. \item If , then . Therefore, -> TRUE
\end{itemize} 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
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.