某种病毒正在一个国家传播。$t$ 天后，感染病例数 $I$ 可以用模型 $I = \frac{a}{b + 2^{20-t}}$ 来表示，其中 $a$ 和 $b$ 是常数。16 天后，感染病例数为 3600。4 天后，感染病例数增加到 9000。

1. Form the first equation using the data for $t = 16$.

When $t = 16$, $I = 3600$.
$3600 = \frac{a}{b + 2^{20-16}}$
$3600 = \frac{a}{b + 2^4}$
$3600 = \frac{a}{b + 16}$
$a = 3600(b + 16)$

2. Form the second equation using the data for $t = 20$.

Four days later, $t = 16 + 4 = 20$, and $I = 9000$.
$9000 = \frac{a}{b + 2^{20-20}}$
$9000 = \frac{a}{b + 2^0}$
$9000 = \frac{a}{b + 1}$
$a = 9000(b + 1)$

3. Solve the simultaneous equations to find $a$ and $b$.

Equating the expressions for $a$:
$3600(b + 16) = 9000(b + 1)$
$3600b + 57600 = 9000b + 9000$
$5400b = 48600$
$b = 9$
Substitute $b = 9$ into $a = 9000(b + 1)$:
$a = 9000(9 + 1) = 90000$
Thus, $a = 90000$ and $b = 9$.

1. Set up the inequality using the given model and constants from part (a)

$\frac{90000}{9 + 2^{20-t}} > 100$

2. Rearrange the inequality to isolate the exponential term

$90000 > 100(9 + 2^{20-t})$
$900 > 9 + 2^{20-t}$
$891 > 2^{20-t}$

3. Use logarithms to solve for t

$\ln(891) > \ln(2^{20-t})$
$\ln(891) > (20-t)\ln 2$
$20 - t < \frac{\ln 891}{\ln 2}$
$20 - t < 9.799$
$t > 10.201$

4. Determine the smallest integer value for n

Since $t > 10.201$ and $n$ is an integer, the number of infected cases first exceeds 100 after 11 days.
$n = 11$

1. Determine the behaviour of the exponential term as $t$ becomes very large

In the long run, $t \to \infty$.
As $t \to \infty$, $20 - t \to -\infty$, so $2^{20-t} \to 0$.

2. Calculate the limiting value of $I$

$I \to \frac{90000}{9 + 0} = 10000$
Therefore, the number of infected cases approaches 10000.