AMC 10 · 2019 · #23

Grade 6 counting

Archetype Arithmetic Expression Decomposition

triangular-numberssequences-arithmeticpattern-recognitionsystematic-enumeration pattern-recognitionidentify-subproblems ↑ Prerequisites: triangular-numberssequences-arithmetic

📏 Long solution 💡 3 insights

Problem

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 $2019$ th number said by Tadd?

Pick an answer.

(A)

5743

(B)

5885

(C)

5979

(D)

6001

(E)

6011

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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

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

#5 Look for a Pattern 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 $3$.

$$\text{round-}m \text{ count} = 3m - 2$$

💡 Each round Tadd gets $3$ more numbers than the previous round.

#7 Identify Subproblems 6.EE.A.2 Step 2

Cumulative Tadd count through round $n$: arithmetic series sum $\sum_{m=1}^{n} (3m - 2) = 3 \cdot \tfrac{n(n+1)}{2} - 2n = \tfrac{n(3n - 1)}{2}$.

$$T(n) = \dfrac{n(3n - 1)}{2}$$

💡 Sum of an arithmetic sequence.

#6 Guess and Check 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 $37$.

$$T(36) = 1926, \quad T(37) = 2035$$

💡 Test candidate round numbers until cumulative count crosses $2019$.

#7 Identify Subproblems 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.

$$2019 - 1926 = 93$$

💡 Subtract numbers said in earlier rounds to get the in-round index.

#5 Look for a Pattern 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 $3 \cdot 37 - 2 = 109$.

$$\text{turn} = 3 \cdot 37 - 2 = 109$$

💡 Tadd appears every $3$ turns starting at turn $1$.

#9 Solve an Easier Related Problem 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, $5887$.

$$\dfrac{108 \cdot 109}{2} = 5886 \;\Rightarrow\; \text{turn 109 starts at } 5887$$

💡 Triangular number formula gives the running total of all integers spoken.

#7 Identify Subproblems 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).

$$5887 + 92 = 5979$$

💡 Start at $5887$, advance $92$ steps to the $93$rd integer of this turn.

[1] #5 4.OA.C.5 Find Tadd's per-round count. Round $1$: Tadd says $1$ number ($\{1\}$). Round $2

[2] #7 6.EE.A.2 Cumulative Tadd count through round $n$: arithmetic series sum $\sum_{m=1}^{n} (

[3] #6 6.EE.A.2 Find smallest $n$ with $T(n) \ge 2019$. Test $n = 36$: $T(36) = \tfrac{36 \cdot

[4] #7 4.OA.A.3 Position inside round $37$: Tadd's $2019$th number is the $(2019 - 1926) = 93$rd

[5] #5 4.OA.C.5 Identify the global turn number. Tadd's round-$m$ turn is global turn $3m - 2$ (

[6] #9 6.EE.A.2 Find the global integer that opens turn $109$. After turn $108$ finishes, the la

[7] #7 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.3 Solve multi-step word problems using four operations with whole numbers (Subtracting $2019 - 1926 = 93$ to locate the position inside round $37$.)
4.OA.C.5 Generate 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.4 Fluently add and subtract multi-digit whole numbers (Computing $5887 + 92 = 5979$ to land on the target integer.)
6.EE.A.2 Write, 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 $= 5887 + 92 = 5979$.

Problem

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: C

Review

CCSS standards used (min grade 6)

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