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

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 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.)

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

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