AMC 10 · 2019 · #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 $\frac{m(3m-1)}{2}$ . 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

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

T(n) =

\frac{n(3n - 1)}{2}

3STEP 3

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

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

2019 - 1926 = 93

5STEP 5

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

turn = 3 · 37 - 2 = 109

6STEP 6

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

\frac{108 · 109}{2}

= 5886 → turn 109 starts at 5887

7STEP 7

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 $\frac{m(3m-1)}{2}$ ; round 37 crosses 2019 at position 93; turn 109 opens at 5887; answer = 5887 + 92 = 5979.

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