AMC 10 · 2016 · #10

Grade 7 counting
logical-deductionpermutations-basic invariant-monovariantcasework ↑ Prerequisites: logical-deduction
📏 Medium solution 💡 3 insights
Problem
Everyone shifts seats while one person is away and only an end seat is left. Find her original seat.

Pick an answer.

(A)
1
(B)
2
(C)
3
(D)
4
(E)
5
How to solve
Strategy Introduce a Variable

Instead of hunting for the whole seating chart, tool #4 (Introduce a Variable) measures each person by a single signed number: how far they moved, with right as + and left as -. The key fact is that nobody enters or leaves the row, so every seat someone leaves is taken by someone else, which forces all five signed moves to add up to 0. Tool #1 (Draw a Diagram) keeps the left-right number line straight in mind, and once Ada's move is known, tool #3 (Eliminate Possibilities) uses the end-seat clue to rule out one of the two ends and pin down where she started.

1STEP 1

Turn each move into a signed number

Each move becomes a signed number.

Bea=+2, Ceci=-1, Dee+Edie=0
2STEP 2

All five moves add to zero

All five must add to zero.

2 + (-1) + 0 + a = 0
3STEP 3

Solve for Ada's move

So her own move is one left.

1 + a = 0 → a = -1
4STEP 4

Use the end-seat clue to find the start

The end-seat clue gives 2, choice (B).

end ∈ {1,5}, moved -1 → end=1, start=1-(-1)=2 → (B)
Answer
2
Build a chart that fits and check it. Start with Ada 2, Bea 1, Ceci 3, Dee 4, Edie 5. Then Bea 1→3, Ceci 3→2, Dee and Edie swap to 5 and 4. The taken seats become {2,3,4,5}, leaving seat 1 open for Ada, an end seat. Every rule holds and Ada started in seat 2, confirming (B).
💡Key takeaway

Tag each move + for right and - for left; since no one leaves the row, all the moves cancel to zero, so Ada's must be -1, putting her start at seat 2.

  • Turn each move into a signed number
  • All five moves add to zero
  • Solve for Ada's move
  • Use the end-seat clue to find the start