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AMC 10 · 2024 · #2

Easy mode Grade 5
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Problem

Remember that n!n! means n×(n−1)×(n−2)×⋯×2×1n \times (n-1) \times (n-2) \times \dots \times 2 \times 1. For example, 4!=4×3×2×14! = 4 \times 3 \times 2 \times 1.

What is the value of 10!−7!⋅6!10! - 7! \cdot 6!?

(A) −120(B) 0(C) 120(D) 600(E) 720\textbf{(A) } -120 \qquad\textbf{(B) } 0 \qquad\textbf{(C) } 120 \qquad\textbf{(D) } 600 \qquad\textbf{(E) } 720

Pick an answer.

(A)
-120
(B)
0
(C)
120
(D)
600
(E)
720
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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

#7 Identify Subproblems 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.
$$10! - 7! \cdot 6! = (10 \cdot 9 \cdot 8) \cdot 7! - 7! \cdot 6! = 7! \cdot (10 \cdot 9 \cdot 8 - 6!)$$

💡 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 $(a - b) \cdot c$.

#7 Identify Subproblems 4.NBT.B.5 Step 2
  • Evaluate $10 \cdot 9 \cdot 8$ inside the parenthesis.
  • Multiply in pairs: $10 \cdot 9 = 90$, then $90 \cdot 8 = 720$.
$$10 \cdot 9 \cdot 8 = 90 \cdot 8 = 720$$

💡 Multiplying $90$ by a one-digit $8$ is a Grade 4 multi-digit-by-one-digit calculation, easy to do without a pencil.

#5 Look for a Pattern 5.NBT.B.5 Step 3
  • Evaluate $6!$ inside the parenthesis.
  • Build it up the same way: $6 \cdot 5 = 30$, $30 \cdot 4 = 120$, $120 \cdot 3 = 360$, $360 \cdot 2 = 720$.
$$6! = 6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1 = 720$$

💡 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.

#7 Identify Subproblems 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.
$$7! \cdot (720 - 720) = 7! \cdot 0 = 0 \;\Rightarrow\; \textbf{(B)}$$

💡 Anything times zero is zero — a Grade 3 multiplication property. No need to compute $7! = 5040$.

[1] #7 3.OA.B.5 Pull $7!$ out of $10!$. Because $10! = 10 \cdot 9 \cdot 8 \cdot 7!$, the origina
[2] #7 4.NBT.B.5 Evaluate $10 \cdot 9 \cdot 8$ inside the parenthesis. Multiply in pairs: $10 \cd
[3] #5 5.NBT.B.5 Evaluate $6!$ inside the parenthesis. Build it up the same way: $6 \cdot 5 = 30$
[4] #7 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.5 Apply 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.5 Multiply 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.5 Fluently 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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