AMC 10 · 2019 · #23
Grade 6 countingProblem
Travis has to babysit the terrible Thompson triplets. Knowing that they love big numbers, Travis devises a counting game for them. First Tadd will say the number 1, then Todd must say the next two numbers (2 and 3), then Tucker must say the next three numbers (4, 5, 6), then Tadd must say the next four numbers (7, 8, 9, 10), and the process continues to rotate through the three children in order, each saying one more number than the previous child did, until the number 10,000 is reached. What is the 2019th number said by Tadd?
Pick an answer.
AMC 10 2019 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
Try it yourself first — the explanation is most useful after you’ve attempted it.
Toolkit + CCSS Solution
Understand
Restated: 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 $2019$th number that Tadd actually says?
Givens: Turn $1$: Tadd; turn $2$: Todd; turn $3$: Tucker; turn $4$: Tadd; … the cycle repeats; Turn $k$ contains exactly $k$ consecutive integers; Tadd takes turns $1, 4, 7, 10, \ldots$ — i.e. turn number $3m - 2$ for round $m$; Process stops at $10{,}000$; we want the $2019$th integer Tadd says; Answer choices: (A) $5743$, (B) $5885$, (C) $5979$, (D) $6001$, (E) $6011$
Unknowns: The $2019$th integer spoken by Tadd
Understand
Restated: 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 $2019$th number that Tadd actually says?
Givens: Turn $1$: Tadd; turn $2$: Todd; turn $3$: Tucker; turn $4$: Tadd; … the cycle repeats; Turn $k$ contains exactly $k$ consecutive integers; Tadd takes turns $1, 4, 7, 10, \ldots$ — i.e. turn number $3m - 2$ for round $m$; Process stops at $10{,}000$; we want the $2019$th integer Tadd says; Answer choices: (A) $5743$, (B) $5885$, (C) $5979$, (D) $6001$, (E) $6011$
Plan
Primary tool: #7 Identify Subproblems
Secondary: #5 Look for a Pattern, #9 Solve an Easier Related Problem, #6 Guess and Check
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 $\tfrac{m(3m-1)}{2}$. 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 $\ge 2019$.
Execute — Answer: C
4.OA.C.5 Step 1 - Find Tadd's per-round count.
- Round $1$: Tadd says $1$ number ($\{1\}$).
- Round $2$: $4$ numbers ($\{7,8,9,10\}$).
- Round $3$: $7$ numbers.
- So round-$m$ count is $3m - 2$ — arithmetic with first term $1$, common difference $3$.
💡 Each round Tadd gets $3$ more numbers than the previous round.
6.EE.A.2 Step 2 Cumulative Tadd count through round $n$: arithmetic series sum $\sum_{m=1}^{n} (3m - 2) = 3 \cdot \tfrac{n(n+1)}{2} - 2n = \tfrac{n(3n - 1)}{2}$.
💡 Sum of an arithmetic sequence.
6.EE.A.2 Step 3 - Find smallest $n$ with $T(n) \ge 2019$.
- Test $n = 36$: $T(36) = \tfrac{36 \cdot 107}{2} = 1926$.
- Test $n = 37$: $T(37) = \tfrac{37 \cdot 110}{2} = 2035$.
- So $T(36) = 1926 < 2019 \le 2035 = T(37)$ — Tadd's $2019$th number is in round $37$.
💡 Test candidate round numbers until cumulative count crosses $2019$.
4.OA.A.3 Step 4 Position inside round $37$: Tadd's $2019$th number is the $(2019 - 1926) = 93$rd number of his round-$37$ turn.
💡 Subtract numbers said in earlier rounds to get the in-round index.
4.OA.C.5 Step 5 - Identify the global turn number.
- Tadd's round-$m$ turn is global turn $3m - 2$ (round $1$: turn $1$; round $2$: turn $4$; round $3$: turn $7$).
- For $m = 37$: global turn $3 \cdot 37 - 2 = 109$.
💡 Tadd appears every $3$ turns starting at turn $1$.
6.EE.A.2 Step 6 - Find the global integer that opens turn $109$.
- After turn $108$ finishes, the last integer said is $1 + 2 + \cdots + 108 = \tfrac{108 \cdot 109}{2} = 5886$.
- So turn $109$ opens at the next integer, $5887$.
💡 Triangular number formula gives the running total of all integers spoken.
4.NBT.B.4 Step 7 - The $93$rd integer of turn $109$ is $5887 + (93 - 1) = 5979$.
- So Tadd's $2019$th number is $5979$.
- The answer is (C).
💡 Start at $5887$, advance $92$ steps to the $93$rd integer of this turn.
4.OA.C.5 Find Tadd's per-round count. Round $1$: Tadd says $1$ number ($\{1\}$). Round $2 6.EE.A.2 Cumulative Tadd count through round $n$: arithmetic series sum $\sum_{m=1}^{n} ( 6.EE.A.2 Find smallest $n$ with $T(n) \ge 2019$. Test $n = 36$: $T(36) = \tfrac{36 \cdot 4.OA.A.3 Position inside round $37$: Tadd's $2019$th number is the $(2019 - 1926) = 93$rd 4.OA.C.5 Identify the global turn number. Tadd's round-$m$ turn is global turn $3m - 2$ ( 6.EE.A.2 Find the global integer that opens turn $109$. After turn $108$ finishes, the la 4.NBT.B.4 The $93$rd integer of turn $109$ is $5887 + (93 - 1) = 5979$. So Tadd's $2019$th Review
Reasonableness: Confirm cumulative counts. $T(37) - T(36) = 2035 - 1926 = 109$, which should equal Tadd's round-$37$ count $= 3 \cdot 37 - 2 = 109$. ✓ 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 $5887 + 92 - 10 \cdot \lfloor \cdot \rfloor$: $5887$ ends in $7$, plus $92$ ending in $2$, last digit $9$ — only choice (C) $5979$ ends in $9$.
Alternative: Tool #13 (Convert to Algebra): set up the inequality $\tfrac{n(3n-1)}{2} \ge 2019$ and solve $3n^2 - n - 4038 \ge 0$. Quadratic formula gives $n \ge \tfrac{1 + \sqrt{1 + 48456}}{6} \approx 36.85$, so $n = 37$. Same path afterwards. The Guess-and-Check route (testing $n=36, 37$) is faster for AMC pacing.
CCSS standards used (min grade 6)
4.OA.A.3Solve multi-step word problems using four operations with whole numbers (Subtracting $2019 - 1926 = 93$ to locate the position inside round $37$.)4.OA.C.5Generate a number or shape pattern following a given rule (Recognizing Tadd's round-$m$ turn length $3m - 2$ and his global turn number $3m - 2$ — patterns generated by simple rules.)4.NBT.B.4Fluently add and subtract multi-digit whole numbers (Computing $5887 + 92 = 5979$ to land on the target integer.)6.EE.A.2Write, read, and evaluate expressions in which letters stand for numbers (Building and evaluating $T(n) = \tfrac{n(3n-1)}{2}$ and the triangular-number formula $\tfrac{k(k+1)}{2}$ at specific $n$ and $k$.)
⭐ 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 $\tfrac{m(3m-1)}{2}$; round $37$ crosses $2019$ at position $93$; turn $109$ opens at $5887$; answer $= 5887 + 92 = 5979$.
⭐ 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 $\tfrac{m(3m-1)}{2}$; round $37$ crosses $2019$ at position $93$; turn $109$ opens at $5887$; answer $= 5887 + 92 = 5979$.
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