2002 AMC 12A 第 3 题

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所有题目均经美国数学协会(MAA)官方合法授权使用。

3.

按照指数运算的标准约定, 如果改变指数运算的执行顺序,还可能得到多少个其他值? 2222=2(2(22))=216=65,536.2^{2^{2^2}} = 2^{\left(2^{\left(2^2\right)}\right)} = 2^{16} = 65{,}536.

According to the standard convention for exponentiation, 2222=2(2(22))=216=65,536.2^{2^{2^2}} = 2^{\left(2^{\left(2^2\right)}\right)} = 2^{16} = 65{,}536. If the order in which the exponentiations are performed is changed, how many other values are possible?

00

11

22

33

44

答案:B
知识点:运算顺序指数分类讨论
难度评级:1270
解答:

22222^{2^{2^2}} 的五种加括号方式给出 ((22)2)2=28,(222)2=28,(22)22=28, \begin{aligned} &\left(\left(2^2\right)^2\right)^2 = 2^8, \\ &\left(2^{2^2}\right)^2 = 2^8, \\ &\left(2^2\right)^{2^2} = 2^8, \end{aligned} 2(22)2=216,2222=216.2^{\left(2^2\right)^2} = 2^{16},\quad 2^{2^{2^2}} = 2^{16}.

因此只可能得到 216=65,5362^{16} = 65{,}53628=2562^8 = 256。 除标准值外,恰好还有 11 个其他值。

因此,正确答案是 B

The five parenthesizations of 22222^{2^{2^2}} give ((22)2)2=28,(222)2=28,(22)22=28, \begin{aligned} &\left(\left(2^2\right)^2\right)^2 = 2^8, \\ &\left(2^{2^2}\right)^2 = 2^8, \\ &\left(2^2\right)^{2^2} = 2^8, \end{aligned} 2(22)2=216,2222=216.2^{\left(2^2\right)^2} = 2^{16},\quad 2^{2^{2^2}} = 2^{16}.

So the only values are 216=65,5362^{16} = 65{,}536 and 28=256.2^8 = 256. Besides the standard value there is exactly 11 other.

Thus, the correct answer is B.

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