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

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==Solution 1==
 
==Solution 1==
  
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A round will be defined as one complete rotation through each of the three children.
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Define a ''round'' 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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We create a table to keep track of what numbers each child says for each round.
  
 
<math>
 
<math>
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</math>
 
</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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Notice that at the end of the <math>n</math>, the last number said is the <math>3n^{\text{th}}</math> triangular 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 <math>2019^{\text{th}}</math> number. Through guess and check, <math>k = 37</math>.
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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 <math>n</math>. At the end of round <math>n</math>, the number of numbers Tadd has said so far is <math>1 + 4 + 7 + \cdots + (3n - 2) = \frac{n(3n-1)}{2}</math>, by the arithmetic series sum formula.  
  
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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 <math>37^{\text{th}}</math> round.
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We therefore 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 in which round Tadd says his <math>2019^{\text{th}}</math> number. Through guess and check (or by actually solving the quadratic inequality), <math>k = 37</math>.
  
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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>.
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Now, using our formula <math>\frac{n(3n-1)}{2}</math>, Tadd says <math>1926</math> numbers in the first 36 rounds, so we are looking for the <math>(2019 - 1926) = 93^{\text{rd}}</math> number Tadd says in the <math>37^{\text{th}}</math> round.
 +
 
 +
We found that the last number said at the very end of the <math>n^{\text{th}}</math> round is the <math>3n^{\text{th}}</math> triangular number. For <math>n = 36</math>, the <math>108^{\text{th}}</math> triangular number is <math>5886</math>. Thus the answer is <math>5886 + 93 = \boxed{\textbf{(C) }5979}</math>.
  
 
==Solution 2==
 
==Solution 2==
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Let's list how many words Tadd says, Todd says, and Tucker says each round.
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Firstly, as in Solution 1, we list how many words Tadd says, Todd says, and Tucker says in each round.
  
 
Tadd: <math>1, 4, 7, 10, 13 \cdots</math>
 
Tadd: <math>1, 4, 7, 10, 13 \cdots</math>
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Tucker: <math>3, 6, 9, 12, 15 \cdots</math>
 
Tucker: <math>3, 6, 9, 12, 15 \cdots</math>
  
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We can find a general formula for the amount of numbers each of the kids say after the <math>n</math>th round (a round is defined in the same way as in the Solution 1). Let's just do it for Tadd at the moment
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We can find a general formula for the number of numbers each of the kids say after the <math>n</math>th round. For Tadd, we can either use the arithmetic series sum formula (like in Solution 1) or standard summation results to get <math>\sum_{i=1}^n 3n-2=-2n+3\sum_{i=1}^n n=-2n+\frac{3n(n+1)}{2}=\frac{3n^2-n}{2}</math>.
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Tadd: <math>\sum_{i=1}^n 3n-2=-2n+3\sum_{i=1}^n n=-2n+\frac{3n(n+1)}{2}=\frac{3n^2-n}{2}</math>.
 
  
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Now to find the number of rotations Tadd and his siblings go through before Tadd says his <math>2019</math>th word, we know the inequality <math>\frac{3n^2-n}{2}<2019</math> must be satisfied, and testing numbers gives <math>n=36</math>.
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Now, to find the number of rotations Tadd and his siblings go through before Tadd says his <math>2019</math>th word, we know the inequality <math>\frac{3n^2-n}{2}<2019</math> must be satisfied, and testing numbers gives the maximum integer value of <math>n</math> as <math>36</math>.
  
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The main insight here to simplify the computation process is to notice that the <math>2019</math>th number Tadd says is just the amount of numbers Todd and Tucker says plus the <math>2019</math> Tadd says, which is our answer since Tadd goes first.
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The next main insight, in order to simplify the computation process, is to notice that the <math>2019</math>th number Tadd says is simply the number of numbers Todd and Tucker say plus the <math>2019</math> Tadd says, which will be the answer since Tadd goes first.
  
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Thus, at this point, carrying out this calculation is quite simple:
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Carrying out the calculation thus becomes quite simple:
  
 
<cmath>\left(\sum_{i=1}^{36} 3n+\sum_{i=1}^{36} 3n-1\right)+2019=\left(\sum_{i=1}^{36} 6n-1\right)+2019=(5+11+17...+255)+2019=\frac{36(260)}{2}+2019</cmath>
 
<cmath>\left(\sum_{i=1}^{36} 3n+\sum_{i=1}^{36} 3n-1\right)+2019=\left(\sum_{i=1}^{36} 6n-1\right)+2019=(5+11+17...+255)+2019=\frac{36(260)}{2}+2019</cmath>
  
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At this point, we can note that the last digit of the answer is <math>9</math>, which gives <math>\boxed{\textbf{(C) }5979}</math>. Calculating out the answer confirms the answer if you have time.
+
At this point, we can note that the last digit of the answer is <math>9</math>, which gives <math>\boxed{\textbf{(C) }5979}</math>. (Completing the calculation will confirm the answer, if you have time.)
  
 
==See Also==
 
==See Also==

Revision as of 01:00, 27 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 1

Define a round as one complete rotation through each of the three children.

We 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 the $n$, the last number said is the $3n^{\text{th}}$ triangular 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 $1 + 4 + 7 + \cdots + (3n - 2) = \frac{n(3n-1)}{2}$, by the arithmetic series sum formula.

We therefore want the smallest positive integer $k$ such that $2019 \leq \frac{k(3k-1)}{2}$. The value of $k$ will tell us in which round Tadd says his $2019^{\text{th}}$ number. Through guess and check (or by actually solving the quadratic inequality), $k = 37$.

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

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

Solution 2

Firstly, as in Solution 1, we list how many words Tadd says, Todd says, and Tucker says in each round.

Tadd: $1, 4, 7, 10, 13 \cdots$

Todd: $2, 5, 8, 11, 14 \cdots$

Tucker: $3, 6, 9, 12, 15 \cdots$

We can find a general formula for the number of numbers each of the kids say after the $n$th round. For Tadd, we can either use the arithmetic series sum formula (like in Solution 1) or standard summation results to get $\sum_{i=1}^n 3n-2=-2n+3\sum_{i=1}^n n=-2n+\frac{3n(n+1)}{2}=\frac{3n^2-n}{2}$.

Now, to find the number of rotations Tadd and his siblings go through before Tadd says his $2019$th word, we know the inequality $\frac{3n^2-n}{2}<2019$ must be satisfied, and testing numbers gives the maximum integer value of $n$ as $36$.

The next main insight, in order to simplify the computation process, is to notice that the $2019$th number Tadd says is simply the number of numbers Todd and Tucker say plus the $2019$ Tadd says, which will be the answer since Tadd goes first.

Carrying out the calculation thus becomes quite simple:

\[\left(\sum_{i=1}^{36} 3n+\sum_{i=1}^{36} 3n-1\right)+2019=\left(\sum_{i=1}^{36} 6n-1\right)+2019=(5+11+17...+255)+2019=\frac{36(260)}{2}+2019\]

At this point, we can note that the last digit of the answer is $9$, which gives $\boxed{\textbf{(C) }5979}$. (Completing the calculation will confirm the answer, if you have time.)

See Also

2019 AMC 10A (Problems • Answer Key • Resources)
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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