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

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  </math>
 
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==Solution 1==
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==Solution==
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===Solution 1===
  
 
Define a ''round'' as one complete rotation through each of the three children.
 
Define a ''round'' as one complete rotation through each of the three children.
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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>.
 
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>.
  
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==Solution 2==
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===Solution 2===
 
Firstly, as in Solution 1, we list how many numbers Tadd says, Todd says, and Tucker says in each round.
 
Firstly, as in Solution 1, we list how many numbers Tadd says, Todd says, and Tucker says in each round.
  

Revision as of 18:08, 1 February 2020

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

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 each $n^{\text{th}}$ round, 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 numbers 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 number, 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:

\[\left(\sum_{i=1}^{36} 3n+\sum_{i=1}^{36} 3n-1\right)+2019=\left(\sum_{i=1}^{36} 6n-1\right)+2019=(5+11+17...+215)+2019=\frac{36(220)}{2}+2019\]

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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