2019 AMC 10A Problems/Problem 23

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 $2019^{\text{th}}$ 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 $37^{\text{th}}$ 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: $1, 4, 7, 10, 13 \cdots$

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

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

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 in the same way as in the Solution 1). 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 $2019$ th 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 $2019$ th 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.

2019 AMC 10A (Problems • Answer Key • Resources)
Preceded by Problem 22	Followed by Problem 24
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2019 AMC 10A Problems/Problem 23

Contents

Problem

Solution 1

Solution 2

See Also