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AMC12 2021 A

AMC12 2021 A · Q9

AMC12 2021 A · Q9. It mainly tests Exponents & radicals.

Which of the following is equivalent to \[(2+3)(2^2+3^2)(2^4+3^4)(2^8+3^8)(2^{16}+3^{16})(2^{32}+3^{32})(2^{64}+3^{64})?\]
下列哪一项等价于 \[(2+3)(2^2+3^2)(2^4+3^4)(2^8+3^8)(2^{16}+3^{16})(2^{32}+3^{32})(2^{64}+3^{64})?\]
(A) 3^{127} + 2^{127} 3^{127} + 2^{127}
(B) 3^{127} + 2^{127} + 2 \cdot 3^{63} + 3 \cdot 2^{63} 3^{127} + 2^{127} + 2 \cdot 3^{63} + 3 \cdot 2^{63}
(C) 3^{128}-2^{128} 3^{128}-2^{128}
(D) 3^{128} + 2^{128} 3^{128} + 2^{128}
(E) 5^{127} 5^{127}
Answer
Correct choice: (C)
正确答案:(C)
Solution
By multiplying the entire equation by $3-2=1$, all the terms will simplify by difference of squares, and the final answer is $\boxed{\textbf{(C)} ~3^{128}-2^{128}}$. Additionally, we could also multiply the entire equation (we can let it be equal to $x$) by $2-3=-1$. The terms again simplify by difference of squares. This time, we get $-x=2^{128}-3^{128} \Rightarrow x=3^{128}-2^{128}$. Both solutions yield the same answer. Note: Also notice when you multiply it by the first pair $(2+3)$, it immediately factors. Notice how it "domino" effects to the others, ultimately collapsing into $\boxed{\textbf{(C)} ~3^{128}-2^{128}}$.
将整个表达式乘以 $3-2=1$,所有项通过平方差公式简化,最终答案为 $\boxed{\textbf{(C)} ~3^{128}-2^{128}}$。 另外,也可以将整个表达式(设为 $x$)乘以 $2-3=-1$。项同样通过平方差公式简化,这次得到 $-x=2^{128}-3^{128} \Rightarrow x=3^{128}-2^{128}$。两种方法得到相同答案。 注意:乘以第一对 $(2+3)$ 时立即因式分解,并像“多米诺效应”一样传播到其他项,最终坍缩为 $\boxed{\textbf{(C)} ~3^{128}-2^{128}}$。
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