2004 AMC 12B Problems/Problem 8

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

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Invalid username
Login to AoPS