AMC 10 · 2024 · #2
Grade 5 arithmeticProblem
What is 10!−7!⋅6!
[ONLY FOR CERTAIN CHINESE TESTPAPERS]
What is 10!−7!⋅6!−5!
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: Compute the value of $10! - 7! \cdot 6!$, where $n!$ means the product of all positive integers from $1$ up to $n$.
Givens: $10! = 10 \cdot 9 \cdot 8 \cdot 7 \cdot 6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1$; $7! = 7 \cdot 6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1$; $6! = 6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1$; Answer choices: (A) $-120$, (B) $0$, (C) $120$, (D) $600$, (E) $720$
Unknowns: The numerical value of $10! - 7! \cdot 6!$
Understand
Restated: Compute the value of $10! - 7! \cdot 6!$, where $n!$ means the product of all positive integers from $1$ up to $n$.
Givens: $10! = 10 \cdot 9 \cdot 8 \cdot 7 \cdot 6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1$; $7! = 7 \cdot 6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1$; $6! = 6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1$; Answer choices: (A) $-120$, (B) $0$, (C) $120$, (D) $600$, (E) $720$
Plan
Primary tool: #7 Identify Subproblems
Secondary: #5 Look for a Pattern
Computing $10!$ and $7! \cdot 6!$ from scratch is many digits of arithmetic — exactly when Tool #7 (Identify Subproblems) helps: notice $10!$ already contains $7!$ inside it, so factor $7!$ out and let the small parenthesis decide the whole answer. Tool #5 (Look for a Pattern) is the supporting insight — spotting that $10 \cdot 9 \cdot 8 = 720 = 6!$ collapses the parenthesis to zero, which is the whole punchline.
Execute — Answer: B
3.OA.B.5 Step 1 - Pull $7!$ out of $10!$.
- Because $10! = 10 \cdot 9 \cdot 8 \cdot 7!$, the original expression has a common factor of $7!$ in both terms.
💡 Splitting $10!$ into $(10 \cdot 9 \cdot 8) \cdot 7!$ and pulling $7!$ out front uses the distributive property — the same Grade 3 "properties of multiplication" idea that lets us turn $a \cdot c - b \cdot c$ into $(a - b) \cdot c$.
4.NBT.B.5 Step 2 - Evaluate $10 \cdot 9 \cdot 8$ inside the parenthesis.
- Multiply in pairs: $10 \cdot 9 = 90$, then $90 \cdot 8 = 720$.
💡 Multiplying $90$ by a one-digit $8$ is a Grade 4 multi-digit-by-one-digit calculation, easy to do without a pencil.
5.NBT.B.5 Step 3 - Evaluate $6!$ inside the parenthesis.
- Build it up the same way: $6 \cdot 5 = 30$, $30 \cdot 4 = 120$, $120 \cdot 3 = 360$, $360 \cdot 2 = 720$.
💡 The familiar pattern that $6! = 720$ matches the $10 \cdot 9 \cdot 8$ result — this matching is the heart of the problem and is exactly Grade 5 multi-digit multiplication fluency.
3.OA.B.5 Step 4 - Substitute the two values back into the parenthesis.
- Since both equal $720$, the parenthesis is exactly zero, so the whole expression is zero.
💡 Anything times zero is zero — a Grade 3 multiplication property. No need to compute $7! = 5040$.
3.OA.B.5 Pull $7!$ out of $10!$. Because $10! = 10 \cdot 9 \cdot 8 \cdot 7!$, the origina 4.NBT.B.5 Evaluate $10 \cdot 9 \cdot 8$ inside the parenthesis. Multiply in pairs: $10 \cd 5.NBT.B.5 Evaluate $6!$ inside the parenthesis. Build it up the same way: $6 \cdot 5 = 30$ 3.OA.B.5 Substitute the two values back into the parenthesis. Since both equal $720$, the Review
Reasonableness: Double-check by brute force: $10! = 3{,}628{,}800$. And $7! \cdot 6! = 5040 \cdot 720 = 3{,}628{,}800$. Subtracting gives exactly $0$. So $10!$ really does equal $7! \cdot 6!$ — the identity is real, not a coincidence — and the answer matches (B).
Alternative: Tool #3 (Eliminate Possibilities): the expression $10! - 7! \cdot 6!$ is the difference of two huge positive integers. If it weren't zero, it would be at least a multiple of $7! = 5040$ in magnitude, which would be vastly larger than any choice. The only "small" choice in $\{-120, 0, 120, 600, 720\}$ that a multiple of $7!$ can be is $0$ itself, pointing straight to (B).
CCSS standards used (min grade 5)
3.OA.B.5Apply properties of operations as strategies to multiply and divide (Using the distributive property to pull the common factor $7!$ out of $10! - 7! \cdot 6!$, and using the "anything times zero is zero" multiplication property to finish.)4.NBT.B.5Multiply a whole number of up to four digits by a one-digit whole number (Computing $10 \cdot 9 \cdot 8 = 90 \cdot 8 = 720$ as a chain of one-digit multiplications.)5.NBT.B.5Fluently multiply multi-digit whole numbers (Building up $6! = 720$ through repeated multi-digit multiplications and recognizing it equals the $10 \cdot 9 \cdot 8$ result.)
⭐ This AMC 10 problem only needs Grade 5 multi-digit multiplication plus the Grade 3 distributive property to spot that $10 \cdot 9 \cdot 8 = 6!$, making the whole thing zero!
⭐ This AMC 10 problem only needs Grade 5 multi-digit multiplication plus the Grade 3 distributive property to spot that $10 \cdot 9 \cdot 8 = 6!$, making the whole thing zero!
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