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 $n=\boxed{10}\Longrightarrow\mathrm{(D)}$.

See Also

2004 AMC 12B (ProblemsAnswer KeyResources)
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
All AMC 12 Problems and Solutions
2004 AMC 10B (ProblemsAnswer KeyResources)
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
All AMC 10 Problems and Solutions

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