2024 AMC 10B Problem 24

P21-25~5 minPrint

Modular Arithmetic Number Properties Apply an identity: SFFT, sum of squares, difference of cubes, Vieta Split into exhaustive cases and handle each

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?

Show solution without answering

Key insight

Put everything over 8. For even m each top term is a multiple of 8. For odd m the top becomes 4(m + 1), a multiple of 8.

Solution

1. Use one denominator. Rewrite every fraction over $8$ :
$P(m) = \frac{4m}{8} + \frac{2m^2}{8} + \frac{m^4}{8} + \frac{m^8}{8} = \frac{4m + 2m^2 + m^4 + m^8}{8}.$
So $P(m)$ is a whole number exactly when the top, $4m + 2m^2 + m^4 + m^8$ , is a multiple of $8$ .
2. Even $m$ ( $2022$ and $2024$ ). Each term on top has plenty of $2$ s: $4m$ has $4 \cdot 2 = 8$ as a factor, $2m^2$ has $2 \cdot 4 = 8$ , and $m^4$ and $m^8$ have at least $2^4 = 16$ . So the top is a multiple of $8$ , and $P(m)$ is a whole number.
3. Odd $m$ ( $2023$ and $2025$ ): a fact about odd squares. Look at a few: $1^2 = 1$ , $3^2 = 9$ , $5^2 = 25$ , $7^2 = 49$ . Each is $1$ more than a multiple of $8$ ( $0, 8, 24, 48$ ). This always holds: an odd number is $2j + 1$ , and $(2j+1)^2 = 4j(j+1) + 1$ . One of $j$ and $j + 1$ is even, so $4j(j+1)$ is a multiple of $8$ .
4. Then $m^4 = (m^2)^2$ and $m^8 = (m^4)^2$ are squares of odd numbers too, so each of $m^2$ , $m^4$ , $m^8$ is $1$ more than a multiple of $8$ .
5. Replace each of them by $1$ (this does not change whether the top is a multiple of $8$ ):
$4m + 2 \cdot 1 + 1 + 1 = 4m + 4 = 4(m + 1).$
Since $m$ is odd, $m + 1$ is even, so $4(m + 1)$ is a multiple of $8$ . So $P(m)$ is a whole number.
6. All four values are whole numbers.

The answer is $\boxed{\textbf{(E)}\ 4}$ .

Why this works

When fractions with denominators $2$ , $4$ and $8$ are added, test them together over $8$ , not one at a time. The fact that every odd square is $1$ more than a multiple of $8$ makes the huge powers easy. The specific years do not matter at all: $P(m)$ is a whole number for every whole number $m$ .

Alternative approach

Track the leftover parts for odd $m$ , such as $m = 2023$ .

$\frac{m}{2}$ : an odd number over $2$ leaves $\frac12$ .
$\frac{m^2}{4}$ : an odd square is $1$ more than a multiple of $4$ , so it leaves $\frac14$ .
$\frac{m^4}{8}$ and $\frac{m^8}{8}$ : each is $1$ more than a multiple of $8$ , so each leaves $\frac18$ .

The leftovers add to $\frac12 + \frac14 + \frac18 + \frac18 = 1$ , a whole number. So $P(m)$ is a whole number, and together with the even cases, all $4$ values work.

Concise solution

A shorter write-up, the way contest solutions are usually written. It may take a different route from the solution above.

Put everything over $8$ :
$P(m) = \frac{4m + 2m^2 + m^4 + m^8}{8},$
so $P(m)$ is an integer exactly when $8 \mid 4m + 2m^2 + m^4 + m^8$ . Only the parity of $m$ matters.

$m$ even. Write $m = 2k$ : $4m = 8k$ , $2m^2 = 8k^2$ , $m^4 = 16k^4$ , $m^8 = 256k^8$ . Every term is a multiple of $8$ , so $P(m)$ is an integer. This covers $2022$ and $2024$ .

$m$ odd. The square of any odd number is $1 \bmod 8$ (since $(2j+1)^2 = 4j(j+1) + 1$ and $j(j+1)$ is even). Hence $m^2 \equiv m^4 \equiv m^8 \equiv 1 \pmod 8$ , and
$4m + 2m^2 + m^4 + m^8 \equiv 4m + 2 + 1 + 1 = 4(m + 1) \pmod 8.$
Since $m + 1$ is even, $4(m+1)$ is a multiple of $8$ , so $P(m)$ is an integer. This covers $2023$ and $2025$ .

All four values are integers.

The answer is $\boxed{\textbf{(E)}\ 4}$ .

Review in the AoPS books

Introduction to Number Theory, Chapter 12: Introduction to Modular Arithmetic
Introduction to Number Theory, Chapter 1: Integers: The Basics
Introduction to Counting & Probability, Chapter 2: Basic Counting Techniques (§2.2)
Introduction to Algebra, Chapter 11: Special Factorizations