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

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Now to find the number of rotations Tadd and his siblings go through before Tadd says his 2019th word, we know the inequality <math>\frac{3n^2-n}{2}<2019</math> must be satisfied, and testing numbers gives <math>n=36</math>.
 
Now to find the number of rotations Tadd and his siblings go through before Tadd says his 2019th 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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The main insight here to simplify the computation process is to notice that the 2019th number Tadd says is just the amount of numbers Todd and Tucker says plus 2019, which is our answers since Tadd goes first.
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The main insight here to simplify the computation process is to notice that the 2019th number Tadd says is just the amount of numbers Todd and Tucker says plus the 2019 Tadd says, which is our answer since Tadd goes first.
  
 
Thus, at this point, carrying out this calculation is quite simple:
 
Thus, at this point, carrying out this calculation is quite simple:

Revision as of 20:14, 10 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

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, $5886 + 93 = \boxed{\textbf{(C) }5979}$

Solution 2

Let's list how many words Tadd says, Todd says, and Tucker says each round.

Tadd: 1, 4, 7, 10, 13...

Todd: 2, 5, 8, 11, 14...

Tucker: 3, 6, 9, 12, 15...

We can find a general formula for the amount of numbers each of the kids say after the $n$th round (a round is defined the same as in the first solution). Let's just do it for Tadd at the moment

Tadd: $\sum_{i=1}^n 3n-2=-2n+3\sum_{i=1}^n n=-2n+\frac{3n(n+1)}{2}=\frac{3n^2-n}{2}$.

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

The main insight here to simplify the computation process is to notice that the 2019th number Tadd says is just the amount of numbers Todd and Tucker says plus the 2019 Tadd says, which is our answer since Tadd goes first.

Thus, at this point, carrying out this calculation is 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 $\boxed{\textbf{(C) }5979}$. Calculating out the answer confirms 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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