AMC10 2021 A

AMC10 2021 A · Q10

AMC10 2021 A · Q10. It mainly tests Exponents & radicals, Manipulating equations.

Which of the following is equivalent to \[(2+3)(2^2+3^2)(2^4+3^4)(2^8+3^8)(2^{16}+3^{16})(2^{32}+3^{32})(2^{64}+3^{64})?\]

以下哪项等价于 \[(2+3)(2^2+3^2)(2^4+3^4)(2^8+3^8)(2^{16}+3^{16})(2^{32}+3^{32})(2^{64}+3^{64})?\]

(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}

(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 $\boxed{\textbf{(C)} ~3^{128}-2^{128}}$.

将整个等式乘以 $3-2=1$，所有项将通过平方差简化，最终答案是 $\boxed{\textbf{(C)} ~3^{128}-2^{128}}$。另外，我们也可以将整个等式（设为 $x$）乘以 $2-3=-1$。项再次通过平方差简化。这次得到 $-x=2^{128}-3^{128} \Rightarrow x=3^{128}-2^{128}$。两种方法得出相同答案。注意：乘以第一对 $(2+3)$ 时，它立即因式分解。注意它如何“多米诺效应”到其他项，最终坍缩为 $\boxed{\textbf{(C)} ~3^{128}-2^{128}}$。

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