AMC12 2006 A

AMC12 2006 A · Q12

AMC12 2006 A · Q12. It mainly tests Arithmetic sequences basics, Arithmetic misc.

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?

有若干个相连的环，每个环厚 $1$ cm，挂在一个挂钩上。最上面的环外径为 $20$ cm。其余每个外侧环的外径都比它上面的环小 $1$ cm。最下面的环外径为 $3$ cm。从最上面环的顶部到最下面环的底部的距离是多少（单位：cm）？

(A) 171 171

(B) 173 173

(D) 188 188

(E) 210 210

Answer

Correct choice: (B)

正确答案：（B）

Solution

The inside diameters of the rings are the positive integers from $1$ to $18$. The total distance needed is the sum of these values plus $2$ 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 $\frac{18 \cdot 19}{2} + 2 = \boxed{\textbf{(B) }173}$.

这些环的内径是从 $1$ 到 $18$ 的正整数。所需的总距离是这些数的和，再加上最上面第一个环的顶部和最下面最后一个环的底部所对应的 $2$。利用等差数列求和公式，答案是 $\frac{18 \cdot 19}{2} + 2 = \boxed{\textbf{(B) }173}$。

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