Question

求不超过下式的最大整数： $\dfrac{3^{100}+2^{100}}{3^{96}+2^{96}}?$

What is the greatest integer less than or equal to $\dfrac{3^{100}+2^{100}}{3^{96}+2^{96}}?$

$80$

$81$

$96$

$97$

$625$

答案：A

知识点：指数取整函数极限情形界定

难度评级：1540

解答：

令 $a=3^{96}$ ， $b=2^{96}$ 。原式为 $\dfrac{81a+16b}{a+b}=16+\dfrac{65a}{a+b}$ ，所以小于 $16+65=81$ 。

要证明其最大整数为 $80$ ，还要证明原式大于 $80$ 。这等价于 $81a+16b>80a+80b$ ，也就是 $a>64b$ 。

因为 $\left(\dfrac32\right)^2=\dfrac94>2$ ，所以原式大于 $80$ 且小于 $81$ 。因此正确答案是 A。 $\dfrac{a}{b}=\left(\dfrac32\right)^{96}>2^{48}>64$ 。

Let $a=3^{96}$ and $b=2^{96}$ . The expression is $\dfrac{81a+16b}{a+b}=16+\dfrac{65a}{a+b}$ , so it is less than $16+65=81$ .

To show the floor is $80$ , we also need the expression to be greater than $80$ . This is equivalent to $81a+16b>80a+80b$ , or $a>64b$ .

Because $\left(\dfrac32\right)^2=\dfrac94>2,$ we have $\dfrac{a}{b}=\left(\dfrac32\right)^{96}>2^{48}>64.$ Hence the expression is greater than $80$ and less than $81.$ Thus, A is the correct answer.

2018 AMC 10A 第 14 题

答案：A

难度评级：1540

解答：

其他年份的第 14 题