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2013 AMC 10B Problem 13

2013 AMC 10B Problem 13AMC 10 Step by Step

Name: Date: September 13, 2026

Sequences & Series · difficulty 2 of 5 · about 3 min · position P11-15

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?

A)
$2$
B)
$3$
C)
$5$
D)
$6$
E)
$8$

Answer

Key insight

The k-th turn says k numbers, so after k turns 1+2+...+k numbers are spoken; 45 < 53 <= 55 puts the 53rd number in the 10th turn, at position 8.

Solution

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

Why this works

The sequence is built from blocks of lengths $1, 2, 3, \dots$ , and the block boundaries are triangular numbers. To locate the $n$ th term, find the largest triangular number below $n$ ; the leftover tells the position inside the block, and within each block the value equals the position.

The trap

Confusing the turn number with the number said, or miscounting the position within the turn (53 - 45 = 8, not 7 or 9).

Common mistakes

Confusing the turn number with the number said, or miscounting the position within the turn (53 - 45 = 8, not 7 or 9).
Assuming the answer is the block number $10$ or using $55 - 53 = 2$ (choice (A)) by counting from the wrong end of the block.

Techniques

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