AMC10 2018 A

AMC10 2018 A · Q14

AMC10 2018 A · Q14. It mainly tests Exponents & radicals, Vieta / quadratic relationships (basic).

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

$\frac{3^{100} + 2^{100}}{3^{96} + 2^{96}}$ 的最大整数部分是多少？

(A) 80 80

(B) 81 81

(D) 97 97

(E) 625 625

Answer

Correct choice: (A)

正确答案：（A）

Solution

Answer (A): Because the powers-of-3 terms greatly dominate the powers-of-2 terms, the given fraction should be close to $\frac{3^{100}}{3^{96}}=3^4=81$. Note that $(3^{100}+2^{100})-81(3^{96}+2^{96})=2^{100}-81\cdot 2^{96}=(16-81)\cdot 2^{96}<0$, so the given fraction is less than $81$. On the other hand $(3^{100}+2^{100})-80(3^{96}+2^{96})=3^{96}(81-80)-2^{96}(80-16)=3^{96}-2^{102}$. Because $3^2>2^3$, $3^{96}=(3^2)^{48}>(2^3)^{48}=2^{144}>2^{102}$, it follows that $(3^{100}+2^{100})-80(3^{96}+2^{96})>0$, and the given fraction is greater than $80$. Therefore the greatest integer less than or equal to the given fraction is $80$.

答案（A）：由于 $3$ 的幂项远远主导 $2$ 的幂项，所给分式应接近 $\frac{3^{100}}{3^{96}}=3^4=81$。注意 $(3^{100}+2^{100})-81(3^{96}+2^{96})=2^{100}-81\cdot 2^{96}=(16-81)\cdot 2^{96}<0$，因此该分式小于 $81$。另一方面， $(3^{100}+2^{100})-80(3^{96}+2^{96})=3^{96}(81-80)-2^{96}(80-16)=3^{96}-2^{102}$。因为 $3^2>2^3$， $3^{96}=(3^2)^{48}>(2^3)^{48}=2^{144}>2^{102}$，所以 $(3^{100}+2^{100})-80(3^{96}+2^{96})>0$，从而该分式大于 $80$。因此，不超过该分式的最大整数为 $80$。

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