Difference between revisions of "2004 AMC 12B Problems/Problem 8"
(New 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 ...) |
|||
| Line 1: | Line 1: | ||
| + | {{duplicate|[[2004 AMC 12B Problems|2004 AMC 12B #8]] and [[2004 AMC 10B Problems|2004 AMC 10B #10]]}} | ||
| + | |||
== Problem == | == 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? | 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? | ||
| Line 11: | Line 13: | ||
{{AMC12 box|year=2004|ab=B|num-b=7|num-a=9}} | {{AMC12 box|year=2004|ab=B|num-b=7|num-a=9}} | ||
| + | {{AMC10 box|year=2004|ab=B|num-b=9|num-a=11}} | ||
Revision as of 12:40, 7 February 2009
- 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?
Solution
The sum of the first 
odd numbers is 
. As in our case 
, we have 
.
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 | ||
