AMC10 2002 A

AMC10 2002 A · Q3

AMC10 2002 A · Q3. It mainly tests Exponents & radicals.

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

根据指数运算的标准约定，$2^{2^{2^2}} = 2^{(2^{(2^2)})} = 2^{16} = 65,536$。如果改变指数运算的顺序，可能得到多少其他值？

(A) 0 0

(B) 1 1

(D) 3 3

(E) 4 4

Answer

Correct choice: (B)

正确答案：（B）

Solution

(B) No matter how the exponentiations are performed, $2^{2^2}$ always gives 16. Depending on which exponentiation is done last, we have $(2^{2^2})^2 = 256,\quad 2^{(2^{2^2})} = 65,536,\quad \text{or}\quad (2^2)^{(2^2)} = 256,$ so there is one other possible value.

（B）无论如何进行幂运算，$2^{2^2}$ 总是等于 16。取决于最后进行的是哪一次幂运算，我们有 $(2^{2^2})^2 = 256,\quad 2^{(2^{2^2})} = 65,536,\quad \text{或}\quad (2^2)^{(2^2)} = 256,$ 因此还有另外一个可能的值。

Topics

AL.008 Exponents & radicals