AMC 10 · 2016 · #1
Grade 6 arithmeticPick an answer.
AMC 10 2016 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
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.
Expose the shared 9!
Peel each factorial down to the shared 9!: 11! = 11· 10· 9! = 110· 9! and 10! = 10· 9!.
A factorial is just the next factorial with a couple of extra factors stacked in front, so 9! is hiding inside both.
5.NBT.B.5Look For A PatternFactor 9! out of the top
Both numerator terms carry 9!, so factor it out: 110· 9! - 10· 9! = (110-10)· 9! = 100· 9!.
If two amounts are both built from 9!, their difference is also built from 9! — just count how many 9!s are left.
6.EE.A.3Identify SubproblemsCancel and read the answer
Now it's (100· 9!)/9!; the 9! cancels top and bottom, leaving 100 — answer (B).
Dividing by 9! undoes multiplying by 9!, so the giant factor disappears and only 100 remains.
4.NF.A.1Identify SubproblemsWhen 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