AMC 10 · 2010 · #2

Grade 4 arithmetic
Archetype Arithmetic Expression Decomposition └ Sub-archetype Linear Rate Accumulation
sequences-arithmeticconsecutive-integer-countmental-arithmetic identify-subproblemsestimation ↑ Prerequisites: sequences-arithmetic
📏 Medium solution 💡 2 insights
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Problem
Hourly trips run from a first departure to a last one, each carrying one fewer passenger than the last. Find the total carried.

Pick an answer.

(A)
585
(B)
594
(C)
672
(D)
679
(E)
694
How to solve
Strategy Identify Subproblems

The total splits cleanly into two questions that can be answered one at a time — how many trips were there, and how many tourists rode each trip — so Tool #7 (Identify Subproblems) sets the shape of the work. The split matters because the two halves are not equally hard. The adding is routine; the trip count is the entire difficulty, and a wrong count silently corrupts every load that follows it. Tool #2 (Make a Systematic List) settles that count by writing the departures out and checking them against a rule for counting a run with both ends included. Tool #5 (Look for a Pattern) turns the step-by-step 'one fewer' rule into a direct formula for the load on any given trip, which is what pins the last trip at 95. Tool #14 (Extreme Principle) brackets the total between the smallest and largest possible six-load days before any exact adding happens. Tool #3 (Eliminate Possibilities) finishes with a test that does not repeat the addition at all: a divisibility check that every six-trip total must pass.

1STEP 1

Count the trips, not the hours

Counting the departures gives six trips.

{10,11,12,13,14,15} → 15-10+1=6 trips
2STEP 2

Get each trip's load

Each load drops by one from the first.

a_k = 100-k, k=0,1,2,3,4,5 → 100, 99, 98, 97, 96, 95
3STEP 3

Bracket the total before adding

Bracketing the total first is a useful check.

6 × 95 = 570 ＜ T ＜ 600 = 6 × 100
4STEP 4

Add by leaning on 100

Leaning on the first load gives 585.

T=Σ_k=0⁵(100-k)=6 × 100-(0+1+2+3+4+5)=600-15=585
5STEP 5

Check the shape of the total

The total's shape confirms 585, choice (C).

n+(n+1)+…+(n+5)=6n+15=3(2n+5) odd; 585=3 × 195, 679=7 × 97
Answer
585
Rebuild the total from the other end as an independent check. Every one of the six trips carries at least 95, which accounts for 6 × 95 = 570, and the amounts carried above that floor are 5,4,3,2,1,0, which total 15. That gives 570+15=585 again, from the bottom rather than the top. The two ways to go wrong here are both about counting, not arithmetic. Counting seven trips gives 679, which is offered as choice (D); counting five trips gives 100+99+98+97+96=490, which is not offered, so only the overcount is baited. It is also worth being honest about how much each check can carry on its own. The bracket 570 ＜ T ＜ 600 leaves both 585 and 594 alive, so estimation alone does not finish the problem, and the divisibility test is what removes 594: an even number can never be a sum of six consecutive integers. Finally the reading holds up. Only outbound trips are counted, which is what 'took to the island' asks for, and the answer does not depend on whether the same tourist rode twice, since the problem reports how many were on each trip and asks for the day's total.
💡Key takeaway

Count the boats, not the hours between them — 10 AM through 3 PM is six departures — then pretend all six left full at a hundred and give back the fifteen missing seats.

  • Count the trips, not the hours
  • Get each trip's load
  • Bracket the total before adding
  • Add by leaning on 100
  • Check the shape of the total

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