1971 AMC 12 第 27 题

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27.

一个盒子里装有筹码,每枚筹码都是红色、白色或蓝色。蓝色筹码的数量至少是白色筹码数量的一半,且至多是红色筹码数量的三分之一。白色或蓝色筹码总数至少为 5555。红色筹码的最少数量为:

A box contains chips, each of which is red, white, or blue. The number of blue chips is at least half the number of white chips and at most one-third the number of red chips. The number which are white or blue is at least 55.55. The minimum number of red chips is:

2424

3333

4545

5454

5757

答案:E
知识点:不等式最优化极限情形界定
难度评级:1880
小提示:

设三种筹码的数量分别为 r,w,br,w,b,把每个条件都写成不等式

Let the counts be r,w,br,w,b and translate every condition into an inequality

大提示:

w2bw\le2bw+b55w+b\ge55 求整数 bb 的最小可能值

From w2bw\le2b and w+b55w+b\ge55, find the least possible integer bb

解答:

设三种筹码的数量分别为 r,w,br,w,b。条件给出 w2b,r3b,w+b55 \begin{gathered} w\le2b,\\ r\ge3b,\\ w+b\ge55\text{。} \end{gathered} 因此 3bw+b553b\ge w+b\ge55,所以 b19b\ge19,且 r57r\ge57。取 (w,b,r)=(36,19,57)(w,b,r)=(36,19,57) 时可以达到等号,因此最小值为 5757

因此,正确答案为 E

Let the counts be r,w,b.r,w,b. The conditions give w2b,r3b,w+b55. \begin{gathered} w\le2b,\\ r\ge3b,\\ w+b\ge55. \end{gathered} Hence 3bw+b55,3b\ge w+b\ge55, so b19b\ge19 and r57.r\ge57. Equality is possible with (w,b,r)=(36,19,57),(w,b,r)=(36,19,57), so the minimum is 57.57.

Therefore, the correct answer is E.

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