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2004 AMC 10B Problem 10

2004 AMC 10B Problem 10AMC 10 Step by Step

Name: Date: September 13, 2026

Sequences & Series · difficulty 2 of 5 · about 2 min · position P1-10

A grocer makes a display of cans in which the top row has one can and each lower row has two more cans than the row above it. If the display contains $100$ cans, how many rows does it contain?

A)
$5$
B)
$8$
C)
$9$
D)
$10$
E)
$11$

Answer

Key insight

The rows hold 1, 3, 5, ..., and the sum of the first n odd numbers is n^2, so n^2 = 100 gives n = 10.

Solution

The rows contain $1, 3, 5, 7, \dots$ cans: the odd numbers in order. With $n$ rows, the last row has $2n - 1$ cans, and the total is an arithmetic series with $n$ terms, first term $1$ and last term $2n - 1$ :
$1 + 3 + \dots + (2n - 1) = n \cdot \frac{1 + (2n - 1)}{2} = n \cdot n = n^2.$

Setting $n^2 = 100$ gives $n = 10$ .

Quick check: ten rows end with $19$ cans, and $10 \cdot \frac{1 + 19}{2} = 100$ .

The answer is $\boxed{\textbf{(D)}\ 10}$ .

Why this works

Any arithmetic series equals (number of terms) $\times$ (average of first and last term). For the odd numbers this collapses to $n^2$ , a fact worth memorizing on its own; a picture of nested L-shapes filling an $n \times n$ square shows why. Recognizing " $1, 3, 5, \dots$ sums to $100$ " as " $n^2 = 100$ " turns the problem into one line.

Alternative approach

Test the choices: $9$ rows give $9^2 = 81$ cans and $11$ rows give $121$ , so only $10$ rows produce exactly $100$ .

The trap

Using a wrong series formula (such as n(n+1)/2) or assuming each row adds one can, which leads to a neighboring choice.

Common mistakes

Using a wrong series formula (such as n(n+1)/2) or assuming each row adds one can, which leads to a neighboring choice.
Solving $n^2 = 100$ and then adding or subtracting a row "for the top," when the top row is already the first odd number.

Techniques

Apply an identity: SFFT, sum of squares, difference of cubes, Vieta