2019 AMC 10A Problem 23

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 $\boxed{\textbf{(C)}\ 5979}$.

Two layers of cumulative sums: one to find which of Tadd's turns contains his $2019$th number, and one to find where that round begins in the overall count. Each layer is an arithmetic series; estimating the square root ($3m^2/2 \approx 2019$ gives $m \approx 37$) tells you where to check. The $10{,}000$ cap is a red herring, since round $109$ ends well before it.

• Off-by-one errors: mixing up how many numbers Tadd said before round 109 (1926) with the position inside that round (93), or miscounting the round's first number.
• Assuming Tadd says roughly one third of the numbers and answering about $3 \cdot 2019 = 6057$, or taking round $108$ or $110$ as Tadd's.