A virus is spreading in a country. The number of infected cases, $I$, after $t$ days, can be modelled by $I = \frac{a}{b + 2^{20-t}}$, where $a$ and $b$ are constants. The number of infected cases was 3600 after 16 days. Four days later, the number of infected cases increased to 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.