2004 AMC 12B Problem 8

Below is the professionally curated solution for Problem 8 of the 2004 AMC 12B, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2004 AMC 12B solutions, or check the answer key.

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Concepts:sum of first n odd numbers

Difficulty rating: 1220

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?

$5$

$8$

$9$

$10$

$11$

Solution:

The rows contain $1, 3, 5, \ldots, (2n - 1)$ cans, and the sum of the first $n$ odd numbers is $n^2.$ Setting $n^2 = 100$ gives $n = 10.$

Thus, the correct answer is D.

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