AMC 10 · 2012 · #18
Grade 7 countingPick an answer.
Counting the 10-long lists head-on is hopeless, so we change what we count (tool #16). First shrink the problem (tool #9): for n = 2, 3, 4 the counts are 2, 4, 8, hinting at 2^ n-1. A number-line picture (tool #1) reveals the key structure — the values written so far always form one unbroken block that can only grow at its left or right end. That lets us re-describe each list as a string of 9 left/right moves, with the starting number forced by the requirement that the block end as [1, 10]. Counting those strings is a plain multiplication-principle list (tool #2), giving 2⁹, and matching the value to the options (tool #3) picks the choice.
Test tiny versions first
Tiny cases suggest doubling each time.
Shrinking the list makes the hidden doubling visible before we try to prove it.
4.OA.C.5Solve An Easier Related ProblemSee the numbers as a segment
The written numbers always form one stretch.
A new value differs by 1 from an old one, so the covered numbers stay glued into a single stretch.
Each new value differs from an old one by exactly one, so the covered numbers stay glued into a single stretch.
▸ Why?
Every step is the same fixed size, so a new number can only sit immediately beside the block already covered.
▸ Why?
A stretch extended at either end is still one stretch, so the property carries forward at every step.
Turn a list into moves
So each new number is a left or right choice.
Stop counting orderings; count the left/right decisions instead — the starting number takes care of itself.
7.SP.C.8Change Focus Count The ComplementCount the move strings
There are nine such choices in a row.
Nine independent two-way forks multiply into 2⁹ outcomes.
7.SP.C.8Make A Systematic ListEvaluate and pick the choice
That gives 512, choice (B).
The clean power of two, 2⁹, lands exactly on one listed choice.
6.EE.A.1Eliminate PossibilitiesWhen a rule only ever lets you add to the left end or the right end of a block, forget the numbers and just count the left/right choices — here that turns a scary ordering problem into 2⁹ = 512.
- Test tiny versions first
- See the numbers as a segment
- Turn a list into moves
- Count the move strings
- Evaluate and pick the choice