AMC 10 · 2011 · #16
Grade 7 countingPick an answer.
The words "ends of each diagonal differ" hide a clean shape, so Tool #1 (Draw a Diagram) is the first move: draw only the diagonals and follow them. They form a five-pointed star, and walking the star visits all five vertices and returns to the start — it is one closed loop (a ring) of 5. That turns the problem into "color a ring of 5 so neighbors differ, using 6 colors." Tool #2 (Make a Systematic List) counts the easy open-chain version with the multiplication counting principle. Tool #7 (Identify Subproblems) and Tool #16 (Count the Complement) then repair the one link the chain ignored by subtracting the bad cases, which shrinks the ring by one and gives a short recursion.
Turn the diagonals into a ring
The diagonals redraw as a single ring.
Every diagonal is one link of the star, and the star closes up into a single loop, so "diagonal ends differ" just means "ring neighbors differ."
7.SP.C.8Draw A DiagramCount the open chain first
An open chain counts easily.
Down a chain each new vertex dodges only its single left neighbor, so it always has 5 safe colors.
Down an open chain each new vertex dodges only its single left neighbour, so it always has the same number of safe colours.
▸ Why?
Each choice is made without regard to the earlier ones beyond that neighbour, so the counts multiply.
▸ Why?
Only touching vertices are forbidden the same colour, so the constraint reaches exactly one step back.
Spot the one broken rule
Only the closing pair can break the rule.
The open chain over-counts by allowing the two ends to clash, so subtract just the clashing ones.
7.SP.C.8Identify SubproblemsGlue the ends: a smaller ring
Gluing the ends leaves a smaller ring.
Forcing the two loose ends to match fuses them into one vertex, shrinking the ring by exactly one.
7.SP.C.8Change Focus Count The ComplementWork down to the 4-ring
That smaller ring counts to 630.
The same subtract-the-smaller-ring rule bottoms out at the triangle, whose count is a plain product.
6.EE.A.1Identify SubproblemsFinish the 5-ring
Subtracting gives 3120, choice (C).
One subtraction closes the loop: start from the free chain, remove the colorings the last diagonal forbids.
7.SP.C.8Identify SubproblemsHidden inside the pentagon, the diagonals form one five-point ring, so the puzzle is just "color a loop of 5 with 6 colors so neighbors differ": count the easy open chain (6 · 5⁴ = 3750), then subtract the 630 that break the last link to land on 3120.
- Turn the diagonals into a ring
- Count the open chain first
- Spot the one broken rule
- Glue the ends: a smaller ring
- Work down to the 4-ring
- Finish the 5-ring