2013 AMC 10B Problem 13

1. Write out the start of the list:
$$\underbrace{1}_{\text{turn 1}},\ \underbrace{1, 2}_{\text{turn 2}},\ \underbrace{1, 2, 3}_{\text{turn 3}},\ \underbrace{1, 2, 3, 4}_{\text{turn 4}},\ \ldots$$
Turn $k$ says the numbers $1$ up to $k$, so it has $k$ numbers in it.
2. Keep a running total of how many numbers have been said after each turn:

1. After turn $9$, $45$ numbers have been said. After turn $10$, $55$ have. Since $53$ is between them, the $53$rd number is said during turn $10$.
2. It is number $53 - 45 = 8$ within turn $10$.
3. Turn $10$ says $1, 2, 3, \ldots, 10$, and its $8$th number is $8$.

The answer is $\boxed{\textbf{(E)}\ 8}$.

The list is made of blocks of length $1, 2, 3, \ldots$, so the block endings are the running totals $1, 3, 6, 10, \ldots$ (the triangular numbers). To find a far-away term, find which block it lands in, then its place inside that block. Here each block counts up from $1$, so the value equals the place.

The full list of numbers said is
$$1,\; 1, 2,\; 1, 2, 3,\; 1, 2, 3, 4,\; \dots$$
Turn $k$ consists of the numbers $1$ through $k$, so it has $k$ entries, and the first $k$ turns together contain $1 + 2 + \cdots + k = \frac{k(k+1)}{2}$ numbers.

Find which turn holds the $53$rd number: $\frac{9 \cdot 10}{2} = 45 < 53$ and $\frac{10 \cdot 11}{2} = 55 \ge 53$. So the $53$rd number is in turn $10$, and it is the $53 - 45 = 8$th number of that turn. Turn $10$ says $1, 2, \dots, 10$, whose $8$th entry is $8$.

The answer is $\boxed{\textbf{(E)}\ 8}$.