5. (10 points) $A=3+\frac{1^{2}}{6+\frac{3^{2}}{5^{2}}}$, the integer closest to $A \times 100$ is
【Analysis】First, let the original expression be $A=$
$$
6+\frac{5^{2}}{6+\frac{7^{2}}{6+\frac{9^{2}}{x}}}
$$
$\qquad$ , then $x=6+\frac{11^{2}}{6+\cdots}>6$, so $x>6$,
$$
x=6+\frac{11^{2}}{6+\cdots}<6+\frac{11^{2}}{6}<27 \text {, }
$$

When $x=6$,
$$
\text { the original expression }=3+\frac{1^{2}}{6+\frac{3^{2}}{6+\frac{5^{2}}{6+\frac{7^{2}}{6+\frac{9^{2}}{6}}}}}=3+\frac{1^{2}}{6+\frac{3^{2}}{6+\frac{5^{2}}{6+\frac{7^{2}}{39}}}}=3+\frac{1^{2}}{6+\frac{3^{2}}{6+\frac{975}{332}}}=3+\frac{1^{2}}{6+\frac{996}{989}}=3.142,
$$

When $x=27$,
$$
\text { the original expression }=3+\frac{1^{2}}{6+\frac{3^{2}}{6+\frac{5^{2}}{6+\frac{81}{27}}}}=3+\frac{1^{2}}{6+\frac{927}{843}}=3.1408,
$$

Then, $3.142 \times 100=314.2$,
$$
3.1408 \times 100=314.08 \text {, }
$$

Then, $314.08<A \times 100<314.2$, the only integer between them is 314,
Therefore, the integer closest to $A \times 100$ is 314.
The answer is: 314.