2006 AMC 10A Problem 14

Below is the professionally curated solution for Problem 14 of the 2006 AMC 10A, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2006 AMC 10A solutions, or check the answer key.

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Concepts:arithmetic sequencesummation

Difficulty rating: 1330

14.

A number of linked rings, each 11 cm thick, are hanging on a peg. The top ring has an outside diameter of 2020 cm. The outside diameter of each of the other rings is 11 cm less than that of the ring above it. The bottom ring has an outside diameter of 33 cm. What is the distance, in cm, from the top of the top ring to the bottom of the bottom ring?

171171

173173

182182

188188

210210

Solution:

The top ring contributes its full outside diameter, 2020 cm. Because the rings are 11 cm thick, each ring hangs 22 cm below the top of the ring above it, so each lower ring adds its outside diameter minus 2.2.

The outside diameters run 20,19,,3,20, 19, \ldots, 3, so the added distances are 17,16,,1.17, 16, \ldots, 1. The total is 20+(17+16++1)=20+17182=20+153=173. 20 + (17 + 16 + \cdots + 1) = 20 + \frac{17 \cdot 18}{2} = 20 + 153 = 173.

Thus, the correct answer is B.

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