AMC 10 · 2012 · #24
Grade 11 patternalgebraPick an answer.
The whole problem is the sorted order of the 2011 terms; once that order is known the counting is a one-line equation. Computing the terms is hopeless, so the order has to come from structure. Tool #4 (Introduce a Variable) names the bases c_k and locates all of them relative to one fixed number L = 0.2010101…, which is what makes the bases comparable at all. Tool #1 (Draw a Diagram) supplies the two monotonicity facts read off the graph of y = c^x for 0 ＜ c ＜ 1: the curve falls as the exponent grows, and it rises as the base grows. Tool #9 (Solve an Easier Related Problem) settles a₁, a₂, a₃, a₄ by hand. Then Tool #5 (Look for a Pattern) does the load-bearing work — but a pattern spotted in four terms is not a proof, and here it would be a bad bet, since the same recursion with the bases ordered differently produces a different order. So the zigzag a₁ ＜ a₃ ＜ … ＜ a₂₀₁₁ ＜ a₂₀₁₀ ＜ … ＜ a₂ is proved by induction, each step using both monotonicity facts together with the exact side of L that each base sits on. Tool #15 (Organize Information in More Ways) rewrites the sorted list as an index formula, and Tool #13 (Convert to Algebra) turns "stays in the same position" into a linear equation. Finally Tool #3 (Eliminate Possibilities) cross-checks the result against the answer list, where each wrong choice matches a specific miscount.
Pin every base to one number
Every base sits near one limiting value.
Every base is the same repeating decimal chopped short, so they all line up in order on the two sides of it.
5.NBT.A.3Introduce A VariableTwo levers on a power
A power moves with the base and against the exponent.
A number below 1 shrinks when the exponent grows and grows when the base grows, and those are the only two moves available.
A number below one shrinks as its exponent grows and grows as its base grows, and those are the only two moves.
▸ Why?
An exponent counts how many times the base is used, and each extra use of a small base shrinks the value.
▸ Why?
Because neither trade ever turns around, changing one at a time settles any comparison.
Settle the first four by hand
The first four terms settle by hand.
Change the base first, then the exponent, and each single comparison collapses into two easy ones.
11.N-RN.A.1Solve An Easier Related ProblemProve the zigzag instead of guessing it
Induction proves the zigzag continues.
Each new term is trapped between the two before it, and the trap keeps working because the bases stay sorted the same way forever.
9.F-IF.A.3Look For A PatternWrite the sorted list as a formula
The sorted list then has a closed formula.
Sorting a zigzag means walking down one branch and then back up the other.
9.F-IF.A.3Organize Information In More WaysMatch the two index formulas
Matching the two index formulas leaves one survivor.
A term stays put exactly when its sorted position equals its own index, and that is a single linear equation.
9.A-REI.B.3Convert To AlgebraAdd the surviving indices
The sum is 1341, choice (C).
The question asks for a sum, but only one index stays in place, so the sum is that one index.
9.A-REI.B.3Eliminate PossibilitiesEvery new term lands between the two before it, so the list zigzags inward — the odd terms climb, the even terms fall — and once you know that order, finding which term keeps its own seat is just one linear equation.
- Pin every base to one number
- Two levers on a power
- Settle the first four by hand
- Prove the zigzag instead of guessing it
- Write the sorted list as a formula
- Match the two index formulas
- Add the surviving indices