Difference between revisions of "2006 AMC 12A Problems/Problem 12"
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== Solution 2 == | == Solution 2 == | ||
− | Alternatively, the sum of the consecutive | + | Alternatively, the sum of the consecutive integers from 3 to 20 is <math> \frac{1}{2}(18)(3+20) = 207 </math>. However, the 17 [[intersection]]s between the rings must be subtracted, and we also get <math> 207 - 2(17) = \boxed{\textbf{(B) }173}</math>. |
== See Also == | == See Also == |
Latest revision as of 11:56, 17 December 2021
- The following problem is from both the 2006 AMC 12A #12 and 2006 AMC 10A #14, so both problems redirect to this page.
Contents
Problem
A number of linked rings, each cm thick, are hanging on a peg. The top ring has an outside diameter of cm. The outside diameter of each of the outer rings is cm less than that of the ring above it. The bottom ring has an outside diameter of cm. What is the distance, in cm, from the top of the top ring to the bottom of the bottom ring?
Solution 1
The inside diameters of the rings are the positive integers from to . The total distance needed is the sum of these values plus for the top of the first ring and the bottom of the last ring. Using the formula for the sum of an arithmetic series, the answer is .
Solution 2
Alternatively, the sum of the consecutive integers from 3 to 20 is . However, the 17 intersections between the rings must be subtracted, and we also get .
See Also
2006 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 11 |
Followed by Problem 13 |
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 |
2006 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 13 |
Followed by Problem 15 | |
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.