AMC 8 · 2020 · #12
Grade 4 algebranumber-theoryProblem
For a positive integer n, the factorial notation n! represents the product of the integers from n to 1. What value of N satisfies the following equation? 5!⋅9!=12⋅N!
Pick an answer.
AMC 8 2020 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: The factorial $n!$ means the product of all positive integers from $n$ down to $1$ (so $4! = 4 \times 3 \times 2 \times 1 = 24$). The equation $5! \cdot 9! = 12 \cdot N!$ asks: which positive integer $N$ makes the two sides equal?
Givens: Definition: $n! = n \times (n-1) \times \cdots \times 2 \times 1$; Equation: $5! \cdot 9! = 12 \cdot N!$; Answer choices: (A) $10$, (B) $11$, (C) $12$, (D) $13$, (E) $14$
Unknowns: The positive integer $N$ that satisfies the equation
Understand
Restated: The factorial $n!$ means the product of all positive integers from $n$ down to $1$ (so $4! = 4 \times 3 \times 2 \times 1 = 24$). The equation $5! \cdot 9! = 12 \cdot N!$ asks: which positive integer $N$ makes the two sides equal?
Givens: Definition: $n! = n \times (n-1) \times \cdots \times 2 \times 1$; Equation: $5! \cdot 9! = 12 \cdot N!$; Answer choices: (A) $10$, (B) $11$, (C) $12$, (D) $13$, (E) $14$
Plan
Primary tool: #7 Identify Subproblems
Secondary: #3 Eliminate Possibilities
Tool #7 (Identify Subproblems) splits the equation $5! \cdot 9! = 12 \cdot N!$ into three small, friendly pieces: (a) compute $5!$ since it is the only "small" factorial in sight, (b) divide both sides by $12$ to clear the loose coefficient, and (c) read off the surviving expression as a single factorial. Each subproblem is a Grade 3-4 arithmetic move; the big-looking factorial expression never has to be expanded. Tool #3 (Eliminate Possibilities) is a natural backup for any AMC multiple-choice question — once we have $10 \cdot 9! = N!$, the five choices $10, 11, 12, 13, 14$ can each be tested against the recursive rule $n! = n \cdot (n-1)!$ until only one fits.
Execute — Answer: A
4.NBT.B.5 Step 1 - Subproblem 1: replace the only "small" factorial, $5!$, by its numerical value.
- The other factorial, $9!$, is large but we will not need to expand it.
💡 Multiplying five one-digit numbers together is exactly the Grade 4 multi-digit multiplication skill — no factorial magic needed.
3.OA.A.4 Step 2 Substitute $5! = 120$ into the original equation so both sides become an unknown-factorial equation with a number coefficient on each side.
💡 Treating $9!$ and $N!$ as "mystery numbers" turns the problem into a familiar Grade 3 \"unknown factor\" equation.
3.OA.C.7 Step 3 - Subproblem 2: divide both sides by the loose coefficient $12$ to keep only one number multiplying each factorial.
- Notice $120 \div 12 = 10$.
💡 Dividing $120$ by $12$ to get $10$ is a basic Grade 3 multiplication/division fact within $100$.
4.OA.C.5 Step 4 - Subproblem 3: read $10 \cdot 9!$ using the recursive pattern of the factorial.
- The rule $n! = n \cdot (n-1)!$ (which is just the definition with the largest factor pulled out front: $10! = 10 \cdot 9 \cdot 8 \cdots 1 = 10 \cdot 9!$) collapses the left side into a single factorial.
💡 Spotting the rule \"next factorial = next number $\times$ previous factorial\" is the Grade 4 \"generate a number pattern following a given rule\" standard.
3.OA.A.4 Step 5 - If two factorials of positive integers are equal, the integers themselves must be equal (the factorial function is strictly increasing for $n \ge 1$).
- So $N = 10$, which is answer choice (A).
- As a sanity check against the other choices, $11! = 11 \cdot 10!$ is already $11$ times too big, and the gap only grows for $12, 13, 14$ — so (A) is the unique match.
💡 \"Same product, same factor list, so same top number\" is the Grade 3 \"find the unknown in a multiplication equation\" idea, used here to eliminate the other four choices.
4.NBT.B.5 Subproblem 1: replace the only "small" factorial, $5!$, by its numerical value. 3.OA.A.4 Substitute $5! = 120$ into the original equation so both sides become an unknown 3.OA.C.7 Subproblem 2: divide both sides by the loose coefficient $12$ to keep only one n 4.OA.C.5 Subproblem 3: read $10 \cdot 9!$ using the recursive pattern of the factorial. T 3.OA.A.4 If two factorials of positive integers are equal, the integers themselves must b Review
Reasonableness: The left side $5! \cdot 9! = 120 \cdot 9!$ already contains $9!$ as a factor, and $120 = 12 \cdot 10$ contributes one extra factor of $10$ after we cancel the $12$. So the left side is exactly $10 \cdot 9! = 10!$, a single factorial just one step beyond $9!$. An answer of $N = 10$ — the next integer up from $9$ — is therefore the most natural outcome; jumping all the way to $N = 14$ would require an extra factor of $14 \cdot 13 \cdot 12 \cdot 11 = 24024$ on the left, which is not there.
Alternative: Tool #3 (Eliminate Possibilities) on its own also works as a primary strategy: rewrite the equation as $\dfrac{5! \cdot 9!}{12} = N!$. Compute the left side: $120 \cdot 9! / 12 = 10 \cdot 9! = 3{,}628{,}800$. Then test the choices — $10! = 3{,}628{,}800$ matches, while $11! = 39{,}916{,}800$ is already ten times too large. Only (A) survives.
CCSS standards used (min grade 4)
4.NBT.B.5Multiply a whole number of up to four digits by a one-digit whole number (Computing $5! = 5 \times 4 \times 3 \times 2 \times 1 = 120$ — a chain of one-digit multiplications producing a three-digit product.)3.OA.A.4Determine unknown whole number in multiplication or division equation (Treating $9!$ and $N!$ as unknown whole-number factors in $120 \cdot 9! = 12 \cdot N!$, and later concluding $N = 10$ from $10! = N!$.)3.OA.C.7Fluently multiply and divide within 100 (Dividing both sides by $12$ using the basic fact $120 \div 12 = 10$.)4.OA.C.5Generate a number or shape pattern following a given rule (Applying the recursive factorial pattern $n! = n \cdot (n-1)!$ to rewrite $10 \cdot 9!$ as $10!$.)
⭐ This AMC 8 problem only needs Grade 4 multiplication patterns — like $10 \times 9! = 10!$ — that you already know!
⭐ This AMC 8 problem only needs Grade 4 multiplication patterns — like $10 \times 9! = 10!$ — that you already know!
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