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2016 AMC 10A Problem 1

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2016 AMC 10A Problem 1AMC 10 Step by Step
Name:                             Date: September 13, 2026
Algebraic Manipulation · difficulty 1 of 5 · about 1 min · position P1-10

What is the value of 11!−10!9!\dfrac{11!-10!}{9!} ?

  1. A)

    9999

  2. B)

    100100

  3. C)

    110110

  4. D)

    121121

  5. E)

    132132

Answer

B

Key insight

Factor 10! out of the numerator: 11! - 10! = 10!(11 - 1), and 10!/9! = 10.

Solution

Both terms of the numerator contain 10!10!, since 11!=11⋅10!11! = 11 \cdot 10!. Factor it out:
11!−10!=11⋅10!−10!=10⋅10!.11! - 10! = 11 \cdot 10! - 10! = 10 \cdot 10!.

Now divide by 9!9!. Because 10!=10⋅9!10! = 10 \cdot 9!, we get
10⋅10!9!=10⋅10=100.\frac{10 \cdot 10!}{9!} = 10 \cdot 10 = 100.

The answer is (B) 100\boxed{\textbf{(B)}\ 100}.

Why this works

Factorials of neighboring integers differ by a single multiplier, so n!n! is a common factor of (n+1)!(n+1)! and n!n!. Pull out the largest common factorial before doing any arithmetic; the leftover integers are small. Never compute 11!11! by hand on a contest.

Alternative approach

Split the fraction: 11!9!−10!9!=11⋅10−10=110−10=100\frac{11!}{9!} - \frac{10!}{9!} = 11 \cdot 10 - 10 = 110 - 10 = 100.

The trap

Expanding the factorials into huge numbers, or writing 11! - 10! as 1! and answering something tiny.

Common mistakes

  • Expanding the factorials into huge numbers, or writing 11! - 10! as 1! and answering something tiny.
  • Computing 11!9!=110\frac{11!}{9!} = 110 and forgetting to subtract the 10!9!=10\frac{10!}{9!} = 10 term, which gives choice (C).

Techniques

Set up the equation/formula and compute; no special trick needed