We write $$\frac{4^{100}+3^{100}}{4^{96}+3^{96}}=\frac{4^{96}}{4^{96}+3^{96}}\cdot \frac{4^{100}}{4^{96}}+\frac{3^{96}}{4^{96}+3^{96}}\cdot \frac{3^{100}}{3^{96}}=\frac{4^{96}}{4^{96}+3^{96}}\cdot 256+\frac{3^{96}}{4^{96}+3^{96}}\cdot 81$$. Hence we see that our number is less than $256$ but more than $81$. So the answer is $$\boxed{\rm (A)\textasciitilde255}$$. 
不超过 $$\frac{4^{100}+3^{100}}{4^{96}+3^{96}}$$ 的最大整数是? (改编自 2018 AMC10 A 第$$14$$题)
Options:
A) $$255$$ 
B) $$256$$ 
C) $$270$$ 
D) $$264$$ 
E) $$265$$ 