# Difference between revisions of "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, and define a turn as the portion when one child says his numbers (similar to how a game is played).

We create a table to keep track of what numbers each child says for each round. $\begin{tabular}{||c c c c||} \hline Round & Tadd & Todd & Tucker \\ [0.5ex] \hline\hline 1 & 1 & 2-3 & 4-6 \\ \hline 2 & 7-10 & 11-15 & 16-21 \\ \hline 3 & 22-28 & 29-36 & 37-45 \\ \hline 4 & 46-55 & 56-66 & 67-78 \\ [1ex] \hline \end{tabular}$

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 $1 + 4 + 7 + \dots + (3n - 2) = \frac{n(3n-1)}{2}$, by the sum of arithmetic series formula.

We find that $\dfrac{37(110)}{2}=2035$, so Tadd says his 2035th number at the end of his turn in round 37. That also means that Tadd says his 2019th number in round 37. At the end of Tadd's turn in round 37, the children have, in total, completed $36+36+37=109$ turns. In general, at the end of turn $n$, the nth triangular number is said, or $\dfrac{n(n+1)}{2}$. So at the end of turn 109 (or the end of Tadd's turn in round 37), Tadd says the number $\dfrac{109(110)}{2}=5995$. Recalling that this was the 2035th number said by Tadd, so the 2019th number he said was $5995-16=5979$.

Thus, the answer is $\boxed{\textbf{(C) }5979}$.

## 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.)

## Video Solutions

### Video Solution 2 by TheBeautyofMath

~IceMatrix

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