Difference between revisions of "2004 AMC 12B Problems/Problem 8"
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== Solution == | == Solution == | ||
| − | The sum of the first <math>n</math> odd numbers is <math>n^2</math>. As in our case <math>n^2=100</math>, we have <math>n=\boxed{ | + | The sum of the first <math>n</math> odd numbers is <math>n^2</math>. As in our case <math>n^2=100</math>, we have <math>n=\boxed{\mathrm{(D)\ }10}</math>. |
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| + | == Video Solution 1== | ||
| + | https://youtu.be/2Yqy5G8G1IU | ||
| + | |||
| + | ~Education, the Study of Everything | ||
== See Also == | == See Also == | ||
Latest revision as of 18:24, 22 October 2022
- The following problem is from both the 2004 AMC 12B #8 and 2004 AMC 10B #10, so both problems redirect to this page.
Contents
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 
cans, how many rows does it contain?
Solution
The sum of the first 
odd numbers is 
. As in our case 
, we have 
.
Video Solution 1
~Education, the Study of Everything
See Also
| 2004 AMC 12B (Problems • Answer Key • 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 | |
| All AMC 12 Problems and Solutions | |
| 2004 AMC 10B (Problems • Answer Key • 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 | ||
| All AMC 10 Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
