2004 AMC 12B Problems/Problem 8

Problem

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?

$(\mathrm {A}) 5 \qquad (\mathrm {B}) 8 \qquad (\mathrm {C}) 9 \qquad (\mathrm {D}) 10 \qquad (\mathrm {E}) 11$

Solution

The sum of the first $n$ odd numbers is $n^2$ . As in our case $n^2=100$ , we have $n=\boxed{10}\Longrightarrow\mathrm{(D)}$ .

2004 AMC 12B (Problems • Answer Key • Resources)
Preceded by Problem 7	Followed by Problem 9
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2004 AMC 12B Problems/Problem 8

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