AMC 10 · 2006 · #25
Grade 11 countingPick an answer.
The obstacle is that condition (2) refers to k=|S|, so the two rules cannot be applied in either order to the whole family at once. Tool #7 (Identify Subproblems) removes that obstacle in one move: slice the family by size. A set has exactly one size, so the slices are disjoint and complete, and inside a slice k is a constant, which turns condition (2) into the ordinary sentence "every element is at least k". Then Tool #4 (Introduce a Variable) does the real work. Both surviving conditions are about spare room — room below the first element, room between neighbours — so instead of tracking where the elements sit, name the slack in each gap. That single change of variable converts both conditions into "all variables are non-negative" and produces one clean identity: the slack always totals 17-3k. Tool #11 (Work Backwards) is not optional decoration here. The change of variable is only allowed to be used for counting if it is a bijection, so the construction is run in reverse and every rule is re-verified on the rebuilt subset; a map that merely sends valid subsets somewhere would give an inequality, not a count. Tool #14 (Extreme Principle) then reads the range of k straight off the slack identity — slack cannot be negative — which is more honest than eyeballing a "tightest packing", since that picture would itself need proof. Finally Tool #15 (Organize Information in More Ways) counts the slack lists by re-encoding them as arrangements of units and dividers, and the same re-encoding, pushed one step further in the review, collapses all five cases into a single recursion that confirms the total by a completely different route.
Fix the size before counting
Fixing the size first makes the second rule usable.
Condition (2) only means something once you know how big S is, so decide the size first and the condition turns into an ordinary floor.
10.S-CP.A.1Identify SubproblemsMeasure the gaps, not the elements
Measuring the gaps instead of the elements absorbs both rules.
Both rules are really about leftover room, so stop tracking where the numbers sit and start tracking how much slack each gap holds.
6.EE.B.6Introduce A VariableRebuild the subset from the slack
The subset can be rebuilt from those gaps, so nothing is lost.
A count is only as trustworthy as its inverse — if you can rebuild the subset from the gap list and always land back inside the rules, the two collections are the same size.
6.EE.B.6Work BackwardsThe budget decides which sizes exist
The leftover slack is a budget, and it must not go negative.
Every element you add eats three units of room — one to lift the floor, two to keep its distance — and there are only 17 units in the budget.
7.EE.B.4Extreme PrincipleCount the ways to spend the slack
That caps the size at five.
Sharing a fixed pile of identical units among labelled boxes is the same as choosing where to drop the dividers.
Sharing a fixed pile of identical units among labelled boxes is the same as choosing where to drop the dividers.
▸ Why?
Each way of dropping the dividers gives exactly one sharing and each sharing gives exactly one drop.
▸ Why?
The divider positions are chosen together out of one row of places, so counting them counts every sharing once.
Add the five cases
Splitting the budget gives one binomial per size, summing to 405, choice (E).
The five sizes never overlap, so the separate counts simply add.
11.S-CP.B.9Identify SubproblemsEvery element you add costs three units of room — one to lift the floor, two to keep its distance from the last one — so with a fixed budget of 17, the only question left is how to spread the leftover slack among the gaps.
- Fix the size before counting
- Measure the gaps, not the elements
- Rebuild the subset from the slack
- The budget decides which sizes exist
- Count the ways to spend the slack
- Add the five cases
More like this
Same archetype — closest grade level first.
