16. Given $A=\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8}+\frac{1}{9}$, then the integer part of $A$ is
Analysis: Fractions with denominators in multiple relationships can be added together first.
That is: $A=\left(\frac{1}{3}+\frac{1}{6}+\frac{1}{9}\right)+\left(\frac{1}{4}+\frac{1}{8}+\frac{1}{5}+\frac{1}{7}\right)=\frac{11}{18}+\left(\frac{1}{4}+\frac{1}{8}+\frac{1}{5}+\frac{1}{7}\right)$,
By the Squeeze Theorem, we get: $\frac{11}{18}+\frac{1}{8} \times 4<A<\frac{11}{18}+\frac{1}{4} \times 4$
That is: $1<\frac{11}{18}+\frac{1}{2}<A<1 \frac{11}{18}$, so the integer part of $\mathrm{A}$ is 1.