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2024 AMC 10B Problems/Problem 24

Revision as of 09:43, 14 November 2024 by Elpianista227 (talk | contribs) (→‎Solution 2 (Specific))

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?

$\textbf{(A) } 0 \qquad\textbf{(B) } 1 \qquad\textbf{(C) } 2 \qquad\textbf{(D) } 3 \qquad\textbf{(E) } 4$

Certain China test papers: 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?

$\textbf{(A) } 1 \qquad\textbf{(B) } 2 \qquad\textbf{(C) } 3 \qquad\textbf{(D) } 4 \qquad\textbf{(E) } 5$

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

Certain China test papers:

As explained above, numbers $P(2022)$ to $P(2025)$ are integers. The difference is that there is a new number, $P(2026)$. Since it is divisible by 2, we can see that it is also an integer.

Therefore, the answer is $\boxed{\textbf{(E) }5}$.

~iHateGeometry

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 $2022 \equiv 6 (mod 8), 2023 \equiv 7 (mod 8), 2024 \equiv 0 (mod 8), 2025 \equiv 1 (mod 8)$. We can verify it by taking everything modulo 8 :

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

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

Therefore, there are $\boxed{\textbf{(E) }5}$ 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

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