14. The greatest integer not greater than $\frac{3^{100}+2^{100}}{3^{96}+2^{96}}$ is ( ).
(A) 80
(B) 81
(C) 96
(D) 97
(E) 625
14. A.

Notice,
$$
\begin{array}{l}
\frac{3^{100}+2^{100}}{3^{96}+2^{96}}=3^{4}-\frac{\left(3^{4}-2^{4}\right) 2^{96}}{3^{96}+2^{96}} \\
=81-\frac{65 \times 2^{96}}{3^{96}+2^{96}} .
\end{array}
$$

Since $0<\frac{2^{96}}{3^{96}+2^{96}}<\frac{1}{65}$, we have,
$$
0<\frac{65 \times 2^{96}}{3^{96}+2^{96}}<1 \text {. }
$$

Therefore, the required value is $\left[\frac{3^{100}+2^{100}}{3^{96}+2^{96}}\right]=80$.