2021 AMC 10A Spring 第 10 题

Question

下列表达式等价于哪一个？

\begin{aligned} &(2+3)(2^2+3^2)\\ &\quad\cdot(2^4+3^4)(2^8+3^8)\\ &\quad\cdot(2^{16}+3^{16})(2^{32}+3^{32})\\ &\quad\cdot(2^{64}+3^{64}) \end{aligned}\text{？}

Which of the following is equivalent to

\begin{aligned} &(2+3)(2^2+3^2)\\ &\quad\cdot(2^4+3^4)(2^8+3^8)\\ &\quad\cdot(2^{16}+3^{16})(2^{32}+3^{32})\\ &\quad\cdot(2^{64}+3^{64})? \end{aligned}

3^{127} + 2^{127}

3^{127} + 2^{127} + 2 \cdot 3^{63}

+ 3 \cdot 2^{63}

3^{127} + 2^{127} + 2 \cdot 3^{63}

+ 3 \cdot 2^{63}

3^{128} - 2^{128}

3^{128} + 2^{128}

5^{127}

乘以

3-2

，它的值等于

1

。

Multiply by

3-2,

which is equal to

1

每一步都会用一个因子产生下一次平方差。

Each factor then creates the next difference of squares

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把这个乘积乘以

3-2=1

。反复应用平方差公式，得到

\begin{aligned} (3-2)(3+2)&=3^2-2^2,\\ (3^2-2^2)(3^2+2^2)&=3^4-2^4 \end{aligned}\text{，}

同样的相消一直进行到最后一个因子。因此这个乘积等于

3^{128}-2^{128}

。

所以正确答案是 C。

Multiply the product by

3-2=1.

Repeatedly applying the difference-of-squares identity gives

\begin{aligned} (3-2)(3+2)&=3^2-2^2,\\ (3^2-2^2)(3^2+2^2)&=3^4-2^4, \end{aligned}

and the same cancellation continues through the final factor. Therefore the product is

3^{128}-2^{128}.

Thus, C is the correct answer.

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