AMC 10 · 2024 · #5
Grade 6 number-theoryProblem
What is the least value of n such that n! is a multiple of 2024?
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: Find the smallest positive integer $n$ such that $n!$ is divisible by $2024$.
Givens: $n!$ means $1 \times 2 \times 3 \times \cdots \times n$; We need $2024 \mid n!$ (that is, $n!$ is a multiple of $2024$); Answer choices: (A) $11$, (B) $21$, (C) $22$, (D) $23$, (E) $253$
Unknowns: The least value of $n$ for which $n!$ is a multiple of $2024$
Understand
Restated: Find the smallest positive integer $n$ such that $n!$ is divisible by $2024$.
Givens: $n!$ means $1 \times 2 \times 3 \times \cdots \times n$; We need $2024 \mid n!$ (that is, $n!$ is a multiple of $2024$); Answer choices: (A) $11$, (B) $21$, (C) $22$, (D) $23$, (E) $253$
Plan
Primary tool: #7 Identify Subproblems
Secondary: #3 Eliminate Possibilities
The condition $2024 \mid n!$ looks intimidating until we split it. Tool #7 (Identify Subproblems) says: factor $2024$ into primes, then check each prime separately — $n!$ is a multiple of $2024$ exactly when $n!$ contains every prime power in that factorization. The largest prime in the factorization sets a hard floor on $n$. Tool #3 (Eliminate Possibilities) then walks the answer choices: any candidate smaller than that floor cannot work, so they drop out immediately and the smallest candidate that clears the floor is the answer.
Execute — Answer: D
4.OA.B.4 Step 1 - Subproblem 1: break $2024$ into primes.
- Pull out factors of $2$ until what is left is odd, then factor the odd part.
💡 Recognizing $253 = 11 \times 23$ is the Grade 4 "find factor pairs" move; both $11$ and $23$ are prime.
6.NS.B.4 Step 2 - Subproblem 2: turn $2024 \mid n!$ into a condition on $n$.
- Because $n!$ is the product of every integer from $1$ to $n$, its prime factorization contains every prime $p \le n$.
- So $n!$ has $11$ and $23$ as factors iff $n \ge 11$ and $n \ge 23$ — the larger requirement wins, forcing $n \ge 23$.
💡 Asking when a factorial contains a given prime is the same idea as asking for the LCM-style "smallest container" — a Grade 6 GCF/LCM mindset.
6.EE.A.1 Step 3 - Subproblem 3: check the $2^3$ requirement is automatic.
- Among $1,2,\ldots,23$ the even numbers $2$, $4$, $6$, $8$ alone contribute $2^1 \cdot 2^2 \cdot 2^1 \cdot 2^3 = 2^7$, far more than the $2^3$ that $2024$ needs.
- So once $n \ge 23$, the power-of-$2$ part is never the bottleneck.
💡 Adding the exponents of like bases is the Grade 6 exponent rule $2^a \cdot 2^b = 2^{a+b}$ — and $7 > 3$, so the $2$-part is comfortable.
6.NS.B.4 Step 4 - Use the floor $n \ge 23$ to eliminate answer choices.
- Choices (A) $11$, (B) $21$, (C) $22$ are all less than $23$, so their factorials skip the prime $23$ and cannot be multiples of $2024$.
- Choice (E) $253$ works but is not the least.
- Choice (D) $23$ clears all three requirements ($2^3$, $11$, $23$), so $23!$ is divisible by $2024$.
💡 Once the largest prime sets the floor, eliminating every smaller choice is a one-line check — the classic "largest prime is the bottleneck" move for least-multiple problems.
4.OA.B.4 Subproblem 1: break $2024$ into primes. Pull out factors of $2$ until what is le 6.NS.B.4 Subproblem 2: turn $2024 \mid n!$ into a condition on $n$. Because $n!$ is the p 6.EE.A.1 Subproblem 3: check the $2^3$ requirement is automatic. Among $1,2,\ldots,23$ th 6.NS.B.4 Use the floor $n \ge 23$ to eliminate answer choices. Choices (A) $11$, (B) $21$ Review
Reasonableness: Verify $n = 23$ works and $n = 22$ fails. $23! = 1 \cdot 2 \cdots 22 \cdot 23$ contains the factor $23$, contains $11$ (since $11 \le 23$), and contains $2^3$ (since $8 \le 23$); product of those gives $2024$, so $2024 \mid 23!$. For $n = 22$: $22!$ has no factor of $23$ because $23$ is prime and does not appear in the product $1 \cdot 2 \cdots 22$, so $23 \nmid 22!$, hence $2024 \nmid 22!$. Both directions match, confirming $23$ is the smallest.
Alternative: Tool #6 (Guess and Check) on the choices in order: test $n = 11$ (no factor of $23$ — fail), $n = 21$ (still no $23$ — fail), $n = 22$ (still no $23$ — fail), $n = 23$ (has $23$, $11$, and plenty of $2$s — succeeds). Stop at the first success, which is (D). This trades the cleaner subproblem split for a direct walk through the menu.
CCSS standards used (min grade 6)
4.OA.B.4Find all factor pairs and recognize multiples; determine prime or composite (Breaking $2024$ into prime factors and recognizing that $253 = 11 \times 23$ with both factors prime.)6.NS.B.4Find greatest common factor and least common multiple of two numbers (Turning $2024 \mid n!$ into the requirement that $n!$ contain every prime of $2024$, the same "smallest container" thinking behind LCM.)6.EE.A.1Write and evaluate numerical expressions involving whole-number exponents (Tracking the power of $2$ inside $n!$ by adding exponents: $2 \cdot 4 \cdot 6 \cdot 8 = 2^{1+2+1+3} = 2^7 \ge 2^3$.)
⭐ Splitting $2024 = 2^3 \times 11 \times 23$ turns this AMC 10 problem into a Grade 6 question — once you spot the biggest prime $23$, the answer has nowhere left to hide.
⭐ Splitting $2024 = 2^3 \times 11 \times 23$ turns this AMC 10 problem into a Grade 6 question — once you spot the biggest prime $23$, the answer has nowhere left to hide.
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