2004 AMC 12B Problems/Problem 8

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The following problem is from both the 2004 AMC 12B #8 and 2004 AMC 10B #10, so both problems redirect to this page.

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 • Resources)
Preceded by Problem 7	Followed by Problem 9
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25

2004 AMC 10B (Problems • Resources)
Preceded by Problem 9	Followed by Problem 11
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25

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2004 AMC 12B Problems/Problem 8

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