2019 AMC 10A Problem 23

Call each child's speaking time a turn. Turn $1$ has $1$ number, turn $2$ has $2$ numbers, turn $3$ has $3$, and so on. Tadd speaks in turns $1, 4, 7, 10, \ldots$, every third turn.

Step 1: how many numbers has Tadd said after each of his turns?

1. In his turns he says $1, 4, 7, 10, \ldots$ numbers, going up by $3$ each time. His $m$th turn is overall turn $3m - 2$, so in it he says $3m - 2$ numbers.
2. After $m$ of his turns, he has said
$$1 + 4 + 7 + \cdots + (3m - 2).$$
This adds $m$ numbers that go up evenly, so it is $m$ times the average of the first and last: $m \cdot \dfrac{1 + (3m - 2)}{2} = \dfrac{m(3m - 1)}{2}$.
3. Check with $m = 2$: $1 + 4 = 5$, and the formula gives $\frac{2 \cdot 5}{2} = 5$. Good.

Step 2: find the turn that holds his 2019th number.

1. We want $\frac{m(3m-1)}{2}$ to just reach $2019$. Roughly, $\frac{3m^2}{2} \approx 2019$ gives $m^2 \approx 1346$, so $m$ is around $37$.
2. $m = 36$: $\frac{36 \cdot 107}{2} = 1926$. That is less than $2019$.
3. $m = 37$: $\frac{37 \cdot 110}{2} = 2035$. That passes $2019$.
4. So his $2019$th number is in his $37$th turn. That is overall turn $3 \cdot 37 - 2 = 109$.
5. Before that turn he had said $1926$ numbers, so his $2019$th number is number $2019 - 1926 = 93$ in turn $109$.

Step 3: find where turn 109 starts.

1. Turns $1$ through $108$ use $1 + 2 + \cdots + 108 = \frac{108 \cdot 109}{2} = 5886$ numbers. So the last number before turn $109$ is $5886$.
2. Turn $109$ starts at $5887$, and its $93$rd number is $5886 + 93 = 5979$.

The answer is $\boxed{\textbf{(C)}\ 5979}$.

There are two running totals here, one for Tadd alone and one for everyone together, and each is a sum of evenly spaced numbers. A rough estimate ($\frac32 m^2 \approx 2019$) tells you which $m$ to check, so you only need two tries. The $10{,}000$ in the problem does not matter: turn $109$ ends long before that.

Call the $k$th block of numbers "round $k$"; it contains $k$ numbers, and rounds $1$ through $k$ together contain $1 + 2 + \dots + k = \frac{k(k+1)}{2}$ numbers. Tadd speaks in rounds $1, 4, 7, \dots$, i.e. round $3j - 2$ on his $j$th turn, saying $3j - 2$ numbers.

After $m$ turns Tadd has said
$$1 + 4 + 7 + \dots + (3m - 2) = \frac{m(3m - 1)}{2}$$
numbers. We need the first $m$ with $\frac{m(3m-1)}{2} \ge 2019$. Since $\frac{36 \cdot 107}{2} = 1926 < 2019$ and $\frac{37 \cdot 110}{2} = 2035 \ge 2019$, the $2019$th number is in Tadd's $37$th turn, which is round $3 \cdot 37 - 2 = 109$. Within that round it is the $2019 - 1926 = 93$rd number.

Rounds $1$ through $108$ contain $\frac{108 \cdot 109}{2} = 5886$ numbers, so round $109$ starts at $5887$, and its $93$rd number is $5886 + 93 = 5979$.

The answer is $\boxed{\textbf{(C)}\ 5979}$.