Competition · AMC preparation · step 4 of 4

AMC 10 · 2019A · #23

Grade 6 counting
Archetype Arithmetic Expression Decomposition └ Sub-archetype Linear Rate Accumulation
triangular-numberssequences-arithmeticpattern-recognitionsystematic-enumeration pattern-recognitionidentify-subproblems ↑ Prerequisites: triangular-numberssequences-arithmetic
📏 Long solution 💡 3 insights
Problem
Three kids Tadd, Todd, Tucker take turns counting in order. Turn k contributes k consecutive integers: Tadd says 1, Todd says 2, 3, Tucker says 4, 5, 6, Tadd says 7, 8, 9, 10, and so on. What is the 2019th number that Tadd actually says?

Pick an answer.

(A)
5743
(B)
5885
(C)
5979
(D)
6001
(E)
6011

AMC 10 2019 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve
Strategy Identify Subproblems

Tool #7 (Subproblems): split into (a) find which round contains Tadd's 2019th number, (b) find what global turn number it is, (c) translate that turn's offset to a global integer. Tool #5 (Pattern): Tadd's round-m turn length is 3m - 2 (arithmetic progression), cumulative is (m(3m-1))/2. Tool #9 (Easier Problem): try small rounds first (m=1, 2, 3) to confirm the pattern and the offset rule. Tool #6 (Guess and Check): test m = 36, 37, 38 to find the smallest m with cumulative count ≥ 2019.

1STEP 1

Count Tadd's numbers per round

Tadd's per-round count grows arithmetically (rounds give 1, 4, 7, … numbers), so round m has 3m - 2 numbers.

round-m count = 3m - 2
2STEP 2

Sum the rounds so far

Sum the round counts: the cumulative through round n is Tadd's arithmetic-series total n(3n−1)2\frac{n(3n - 1)}{2}.

T(n) = (n(3n - 1))/2
3STEP 3

Find the round that passes 2019

Testing rounds, T(36) = 1926 ＜ 2019 ≤ 2035 = T(37), so Tadd's 2019th number is in round 37.

T(36) = 1926, T(37) = 2035
4STEP 4

Locate the spot inside round 37

Subtract earlier rounds — 2019 - 1926 = 93 — so it's the 93rd number of Tadd's round-37 turn.

2019 - 1926 = 93
5STEP 5

Number the turn overall

Tadd's round-m turn is global turn 3m - 2, so round 37 is global turn 109.

turn = 3 · 37 - 2 = 109
6STEP 6

Find the turn's first number

After turn 108 the total said is 108⋅1092\frac{108 · 109}{2} = 5886, so turn 109 opens at 5887.

(108 · 109)/2 = 5886 → turn 109 starts at 5887
7STEP 7

Step to the 93rd number

The 93rd number of turn 109 is 5887 + (93 - 1) = 5979 — Tadd's 2019th number. The answer is (C).

5887 + 92 = 5979
Answer
5979
Confirm cumulative counts. T(37) - T(36) = 2035 - 1926 = 109, which should equal Tadd's round-37 count = 3 · 37 - 2 = 109. ✓ Turn-109 length is 109 integers (turn k has k numbers), so positions 1 through 109 inside turn 109 correspond to global integers 5887 to 5995. Position 93 gives 5979, which is between 5887 and 5995 — consistent. Also check we are well under the 10,000 cutoff. Finally, the last digit of the answer must match 5887 + 92 - 10 · ⌊ · ⌋: 5887 ends in 7, plus 92 ending in 2, last digit 9 — only choice (C) 5979 ends in 9.
💡Key takeaway

This AMC 10 problem only needs Grade 6 expression-building (and a touch of multi-digit arithmetic) you already know — round-m Tadd count 3m-2 sums to m(3m−1)2\frac{m(3m-1)}{2}; round 37 crosses 2019 at position 93; turn 109 opens at 5887; answer = 5887 + 92 = 5979.

  • Count Tadd's numbers per round
  • Sum the rounds so far
  • Find the round that passes 2019
  • Locate the spot inside round 37
  • Number the turn overall
  • Find the turn's first number
  • Step to the 93rd number

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