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

Video Solution 1

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See Also

 2004 AMC 12B (Problems • Answer Key • Resources) Preceded byProblem 7 Followed byProblem 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 (Problems • Answer Key • Resources) Preceded byProblem 9 Followed byProblem 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.