AMC 10 · 2024 · #24
Grade 6 arithmeticProblem
Let
P(m)=2m+4m2+8m4+8m8
How many of the values P(2022), P(2023), P(2024), and P(2025) are integers?
Pick an answer.
AMC 10 2024 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
Try it yourself first — the explanation is most useful after you’ve attempted it.
Toolkit + CCSS Solution
Understand
Restated: Define $P(m) = \dfrac{m}{2} + \dfrac{m^2}{4} + \dfrac{m^4}{8} + \dfrac{m^8}{8}$. Of the four values $P(2022), P(2023), P(2024), P(2025)$, how many are integers?
Givens: $P(m) = \dfrac{m}{2} + \dfrac{m^2}{4} + \dfrac{m^4}{8} + \dfrac{m^8}{8}$; Four input values: $m = 2022, 2023, 2024, 2025$; $2022$ and $2024$ are even; $2023$ and $2025$ are odd; Answer choices: (A) $0$, (B) $1$, (C) $2$, (D) $3$, (E) $4$
Unknowns: How many of the four $P$-values are whole numbers
Understand
Restated: Define $P(m) = \dfrac{m}{2} + \dfrac{m^2}{4} + \dfrac{m^4}{8} + \dfrac{m^8}{8}$. Of the four values $P(2022), P(2023), P(2024), P(2025)$, how many are integers?
Givens: $P(m) = \dfrac{m}{2} + \dfrac{m^2}{4} + \dfrac{m^4}{8} + \dfrac{m^8}{8}$; Four input values: $m = 2022, 2023, 2024, 2025$; $2022$ and $2024$ are even; $2023$ and $2025$ are odd; Answer choices: (A) $0$, (B) $1$, (C) $2$, (D) $3$, (E) $4$
Plan
Primary tool: #7 Identify Subproblems
Secondary: #9 Solve an Easier Related Problem, #5 Look for a Pattern, #6 Guess and Check
The four inputs $2022, 2023, 2024, 2025$ look special but the denominators $2, 4, 8, 8$ only care about $m \bmod 8$. Tool #7 (Identify Subproblems) splits the question by parity: two even cases ($2022, 2024$) and two odd cases ($2023, 2025$). Tool #9 (Easier Related Problem) replaces the giant inputs with small representatives — $m = 2$ for the even case, $m = 1$ for the odd case — and computes $P$ exactly there. Tool #5 (Look for a Pattern) shows the integer/non-integer status is the same for every even (resp. odd) input, so the small representatives generalize. Tool #6 (Guess and Check) provides a direct check by computing $P(1)$ and $P(2)$ in seconds.
Execute — Answer: E
2.OA.C.3 Step 1 - Split the four inputs by parity.
- Even: $m = 2022, 2024$.
- Odd: $m = 2023, 2025$.
- We will show that $P(m)$ is an integer whenever $m$ is even, and also whenever $m$ is odd — so the integer-or-not behaviour depends only on parity.
💡 Grade 2 odd/even classification splits four cases into two pattern-classes.
6.EE.A.2 Step 2 - Even case via Tool #9: try the smallest non-zero even input, $m = 2$.
- $P(2) = \tfrac{2}{2} + \tfrac{4}{4} + \tfrac{16}{8} + \tfrac{256}{8} = 1 + 1 + 2 + 32 = 36$.
- Integer.
- Now check why this generalizes: write $m = 2k$.
- Then $\tfrac{m}{2} = k$, $\tfrac{m^2}{4} = \tfrac{4k^2}{4} = k^2$, $\tfrac{m^4}{8} = \tfrac{16k^4}{8} = 2k^4$, $\tfrac{m^8}{8} = \tfrac{256 k^8}{8} = 32 k^8$.
- Each piece is an integer, so $P(m)$ is an integer for every even $m$.
- In particular $P(2022)$ and $P(2024)$ are integers.
💡 Grade 6 substitute $m = 2k$ — each numerator inherits enough $2$s to cancel its denominator, leaving whole-number pieces.
5.NF.A.1 Step 3 - Odd case via Tool #9: try the smallest odd input, $m = 1$.
- $P(1) = \tfrac{1}{2} + \tfrac{1}{4} + \tfrac{1}{8} + \tfrac{1}{8} = \tfrac{4 + 2 + 1 + 1}{8} = \tfrac{8}{8} = 1$.
- Integer.
- The four fractional pieces $\tfrac{1}{2}, \tfrac{1}{4}, \tfrac{1}{8}, \tfrac{1}{8}$ already sum to exactly $1$, which is the key.
💡 Grade 5 add fractions with denominators $2, 4, 8$ via common denominator $8$ — the leftovers tile to one whole.
6.NS.B.4 Step 4 - Show the odd case generalizes.
- For any odd $m$, the four fractional parts of $\tfrac{m}{2}, \tfrac{m^2}{4}, \tfrac{m^4}{8}, \tfrac{m^8}{8}$ are exactly $\tfrac{1}{2}, \tfrac{1}{4}, \tfrac{1}{8}, \tfrac{1}{8}$ — same as in the $m = 1$ case.
- Reason: any odd $m$ satisfies $m \equiv 1 \pmod{2}$, $m^2 \equiv 1 \pmod{8}$ (every odd square ends in $1$ mod $8$ — check $1, 9, 25, 49, 81 \equiv 1 \pmod 8$), and therefore $m^4 = (m^2)^2 \equiv 1 \pmod 8$ and $m^8 = (m^4)^2 \equiv 1 \pmod 8$.
- So each numerator is $1$ more than a multiple of its denominator, and the fractional pieces match $m = 1$ exactly.
- Since they already sum to $1$, $P(m)$ is an integer for every odd $m$.
- In particular $P(2023)$ and $P(2025)$ are integers.
💡 Grade 6 multiples-and-remainders: every odd square sits one above a multiple of $8$, so all higher even powers stay at the same residue.
2.OA.C.3 Step 5 - Count.
- All four values $P(2022), P(2023), P(2024), P(2025)$ are integers, regardless of how big the inputs are — the structure depends only on parity, and both parities work.
💡 Grade 2 counting: two evens + two odds = four.
2.OA.C.3 Split the four inputs by parity. Even: $m = 2022, 2024$. Odd: $m = 2023, 2025$. 6.EE.A.2 Even case via Tool #9: try the smallest non-zero even input, $m = 2$. $P(2) = \t 5.NF.A.1 Odd case via Tool #9: try the smallest odd input, $m = 1$. $P(1) = \tfrac{1}{2} 6.NS.B.4 Show the odd case generalizes. For any odd $m$, the four fractional parts of $\t 2.OA.C.3 Count. All four values $P(2022), P(2023), P(2024), P(2025)$ are integers, regard Review
Reasonableness: Two free spot-checks. $P(2) = 36$ (Step 2) and $P(1) = 1$ (Step 3) are both integers, confirming the parity argument on the simplest representatives. For one more check, $P(3) = \tfrac{3}{2} + \tfrac{9}{4} + \tfrac{81}{8} + \tfrac{6561}{8} = \tfrac{12 + 18 + 81 + 6561}{8} = \tfrac{6672}{8} = 834$ — integer, agreeing with the odd-case claim. The answer $4$ is the upper end of the choice list, which feels right because the structure of $P$ was clearly built so that the fractional pieces $\tfrac{1}{2} + \tfrac{1}{4} + \tfrac{1}{8} + \tfrac{1}{8} = 1$ tile to one whole — the problem-setter chose denominators $2, 4, 8, 8$ deliberately for this.
Alternative: Tool #15 (Reorganize): write $P(m) = \tfrac{4m + 2m^2 + m^4 + m^8}{8}$ and ask only whether the numerator is a multiple of $8$. For $m$ even, every term $4m, 2m^2, m^4, m^8$ is a multiple of $8$ (since $m \ge 2$ even gives $m^k$ divisible by $2^k \ge 8$ for $k \ge 3$, and $4m$ is a multiple of $8$). For $m$ odd, $m^4 \equiv m^8 \equiv 1 \pmod 8$, $2 m^2 \equiv 2 \pmod 8$, $4m \equiv 4 \pmod 8$, summing to $1 + 1 + 2 + 4 = 8 \equiv 0 \pmod 8$. Either way the numerator is a multiple of $8$, so $P(m)$ is always an integer for integer $m$ — the answer is the full count $4$.
CCSS standards used (min grade 6)
2.OA.C.3Determine whether a group of objects (up to 20) has an odd or even number of members (Splitting the four inputs into two even ($2022, 2024$) and two odd ($2023, 2025$), and counting at the end.)5.NF.A.1Add and subtract fractions with unlike denominators by replacing given fractions with equivalent fractions in such a way as to produce an equivalent sum or difference of fractions with like denominators (Computing $P(1) = \tfrac{1}{2} + \tfrac{1}{4} + \tfrac{1}{8} + \tfrac{1}{8} = \tfrac{4 + 2 + 1 + 1}{8} = 1$ — common denominator $8$.)6.EE.A.2Write, read, and evaluate expressions in which letters stand for numbers (Substituting $m = 2k$ to rewrite $P(m) = k + k^2 + 2k^4 + 32 k^8$ and observe every piece is a whole number.)6.NS.B.4Find the greatest common factor of two whole numbers less than or equal to 100 and the least common multiple of two whole numbers less than or equal to 12 — extending to multiples and remainders modulo small primes (Using $m \text{ odd} \Rightarrow m^2 \equiv 1 \,(\bmod\,8)$ (every odd square sits one above a multiple of $8$, by inspection of $1^2, 3^2, 5^2, 7^2$), hence $m^4 \equiv m^8 \equiv 1 \,(\bmod\,8)$.)
⭐ Big inputs like $2024$ scare you for nothing: the denominators only care about parity. Test the smallest even $m = 2$ and the smallest odd $m = 1$ — both give whole-number $P(m)$, so every integer $m$ does, and the answer is all four.
⭐ Big inputs like $2024$ scare you for nothing: the denominators only care about parity. Test the smallest even $m = 2$ and the smallest odd $m = 1$ — both give whole-number $P(m)$, so every integer $m$ does, and the answer is all four.
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