Difference between revisions of "2019 AMC 10A Problems/Problem 23"

(Created page with "==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...")
 
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==Solution==
 
==Solution==
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A round will be defined as one complete rotation through each of the three children.
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We should create a table to keep track of what numbers each child says for each round.
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<math>
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\begin{tabular}{||c c c c||}
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\hline
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Round & Tadd & Todd & Tucker \\ [0.5ex]
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\hline\hline
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1 & 1 & 2-3 & 4-6 \\
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\hline
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2 & 7-10 & 11-15 & 16-21 \\
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\hline
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3 & 22-28 & 29-36 & 37-45 \\
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\hline
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4 & 46-55 & 56-66 & 67-78 \\ [1ex]
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\hline
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\end{tabular}
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</math>
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Notice that at the end of each round, the last number said is the <math>3n^{\text{th}}</math> triangular number where <math>n</math> is the round number.
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Tadd says <math>1</math> number in round 1, <math>4</math> numbers in round 2, <math>7</math> numbers in round 3, and in general <math>3n - 2</math> numbers in round n. At the end of round n, the number of numbers Tadd has said so far is <math>1 + 4 + 7 + ... + (3n - 2)</math> or <math>\frac{n(3n-1)}{2}</math>. We want the smallest positive integer <math>k</math> such that <math>2019 \leq \frac{k(3k-1)}{2}</math>. The value of <math>k</math> will tell us which round Tadd says his 2019th number. Through guess and check, <math>k = 37</math>.
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Using our formula <math>\frac{n(3n-1)}{2}</math>, Tadd says <math>1926</math> numbers in the first 36 rounds, meaning we are looking for the <math>93^{\text{rd}}</math> <math>(2019 - 1926)</math> number Tadd says in the 37th round.
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We found that the last number said at the very end of the <math>n^{\text{th}}</math> round is <math>3n^{\text{th}}</math> triangular number. In particular, for <math>n = 36</math>, the <math>108^{\text{th}}</math> triangular number is <math>5886</math>. Finally, <math>5886 + 93 = \boxed{\textbf{(C) }5979}</math>
  
 
==See Also==
 
==See Also==

Revision as of 20:01, 9 February 2019

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?

$\textbf{(A)}\ 5743 \qquad\textbf{(B)}\ 5885 \qquad\textbf{(C)}\ 5979 \qquad\textbf{(D)}\ 6001 \qquad\textbf{(E)}\ 6011$

Solution

A round will be defined as one complete rotation through each of the three children.

We should create a table to keep track of what numbers each child says for each round.

$\begin{tabular}{||c c c c||}   \hline  Round & Tadd & Todd & Tucker \\ [0.5ex]   \hline\hline  1 & 1 & 2-3 & 4-6 \\   \hline  2 & 7-10 & 11-15 & 16-21 \\  \hline  3 & 22-28 & 29-36 & 37-45 \\  \hline  4 & 46-55 & 56-66 & 67-78 \\ [1ex]   \hline \end{tabular}$

Notice that at the end of each round, the last number said is the $3n^{\text{th}}$ triangular number where $n$ is the round number.

Tadd says $1$ number in round 1, $4$ numbers in round 2, $7$ numbers in round 3, and in general $3n - 2$ numbers in round n. At the end of round n, the number of numbers Tadd has said so far is $1 + 4 + 7 + ... + (3n - 2)$ or $\frac{n(3n-1)}{2}$. We want the smallest positive integer $k$ such that $2019 \leq \frac{k(3k-1)}{2}$. The value of $k$ will tell us which round Tadd says his 2019th number. Through guess and check, $k = 37$.

Using our formula $\frac{n(3n-1)}{2}$, Tadd says $1926$ numbers in the first 36 rounds, meaning we are looking for the $93^{\text{rd}}$ $(2019 - 1926)$ number Tadd says in the 37th round.

We found that the last number said at the very end of the $n^{\text{th}}$ round is $3n^{\text{th}}$ triangular number. In particular, for $n = 36$, the $108^{\text{th}}$ triangular number is $5886$. Finally, $5886 + 93 = \boxed{\textbf{(C) }5979}$

See Also

2019 AMC 10A (ProblemsAnswer KeyResources)
Preceded by
Problem 22
Followed by
Problem 24
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AMC 10 Problems and Solutions

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