Difference between revisions of "2006 AMC 12A Problems/Problem 12"
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== Solution == | == Solution == | ||
| − | The sum of the [[consecutive]]ly increasing [[integers]]s from 3 to 20 is <math>\frac{1}{2} | + | The sum of the [[consecutive]]ly increasing [[integers]]s from 3 to 20 is <math> \frac{1}{2}(18)(3+20) = 207 </math>. However, the 17 [[intersection]]s between the rings must also be subtracted, so we get <math> 207 - 2(17) = 173 \Rightarrow B </math>. |
== See also == | == See also == | ||
Revision as of 18:39, 31 January 2007
Problem
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A number of linked rings, each 1 cm thick, are hanging on a peg. The top ring has an outside diameter of 20 cm. The outside diameter of each of the outer rings is 1 cm less than that of the ring above it. The bottom ring has an outside diameter of 3 cm. What is the distance, in cm, from the top of the top ring to the bottom of the bottom ring?
Solution
The sum of the consecutively increasing integerss from 3 to 20 is 
. However, the 17 intersections between the rings must also be subtracted, so we get 
.
See also
| {{{header}}} | ||
| Preceded by Problem 11 |
AMC 12A 2006 |
Followed by Problem 13 |
