2013 AMC 10B Problem 13

P11-15~3 minPrint

Sequences & Series Set up the equation/formula and compute; no special trick needed Compute small cases, spot the pattern, generalize

Jo and Blair take turns counting from $1$ to one more than the last number said by the other person. Jo starts by saying " $1$ ", so Blair follows by saying " $1, 2$ " . Jo then says " $1, 2, 3$ " , and so on. What is the 53rd number said?

Show solution without answering

Key insight

Turn k says k numbers, so after 9 turns 45 numbers have been said. The 53rd number is the 8th number of turn 10, which is 8.

Solution

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:

Turn	1	2	3	4	5	6	7	8	9	10
Numbers said so far	1	3	6	10	15	21	28	36	45	55

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 $10$ .
It is number $53 - 45 = 8$ within turn $10$ .
Turn $10$ says $1, 2, 3, \ldots, 10$ , and its $8$ th number is $8$ .

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

Why this works

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.

Concise solution

A shorter write-up, the way contest solutions are usually written. It may take a different route from the solution above.

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 $\boxed{\textbf{(E)}\ 8}$ .

Review in the AoPS books

Introduction to Algebra, Chapter 21: Sequences & Series (§21.1, §21.2, §21.3, §21.4)
Prealgebra, Chapter 15: Problem-Solving Strategies (§15.1)
Intermediate Algebra, Chapter 17: More Sequences and Series (§17.1, §17.2)
Introduction to Number Theory, Chapter 12: Introduction to Modular Arithmetic (§12.6)