AMC 10 · 2016 · #1

Grade 6 arithmetic
factorialfraction-arithmetic identify-subproblems ↑ Prerequisites: factorial
📏 Short solution 💡 2 insights
Problem
A difference of two factorials is divided by a smaller factorial. Find the value.

Pick an answer.

(A)
99
(B)
100
(C)
110
(D)
121
(E)
132
How to solve
Strategy Identify Subproblems

Computing 11!, 10!, and 9! separately means multiplying huge numbers, which is slow and error-prone. Tool #7 (Identify Subproblems) breaks the work into safe pieces: rewrite each factorial so the shared 9! shows up, factor that 9! out of the top, then cancel it with the bottom. Tool #5 (Look for a Pattern) supplies the key observation that each factorial is just a few extra factors on top of 9!, so nothing ever has to be fully multiplied out.

1STEP 1

Expose the shared 9!

Both terms carry the same factorial.

11! = 11 · 10 · 9! = 110 · 9!, 10! = 10 · 9!
2STEP 2

Factor 9! out of the top

Factoring it out leaves a plain difference.

11!-10! = 110 · 9! - 10 · 9! = (110-10) · 9! = 100 · 9!
3STEP 3

Cancel and read the answer

Cancelling gives 100, choice (B).

(11!-10!)/9! = (100 · 9!)/9! = 100 → (B)
Answer
100
The answer never needed the actual size of 9!, which is a good sign — it cancelled completely, leaving a clean whole number 100 that sits right among the choices. A quick sanity check on the size: 11! is 11 times 10!, so 11!-10! is about 10 copies of 10!, and 10! is 10 copies of 9!; that is roughly 10 × 10 = 100 copies of 9!, matching 100 exactly.
💡Key takeaway

When factorials are subtracted or divided, find the biggest one they all share, pull it out, and watch it cancel — you almost never multiply the giant numbers.

  • Expose the shared 9!
  • Factor 9! out of the top
  • Cancel and read the answer