I3.1 If $183 a 8$ is a mu1tiple of 287 , find $a$.
I3.2 The number of positive factors of $a^{2}$ is $b$, find $b$.
I3.3 In an urn, there are $c$ balls, $b$ of them are either black or red, $(b+2)$ of them are either red or white and 12 of them are either black or white. Find $c$.

I3.4 Given $f(3+x)=f(3-x)$ for all values of $x$, and the equation $f(x)=0$ has exactly $c$ distinct roots. Find $d$, the sum of these roots.
$287=7 \times 41$ and $2614 \times 7=18298$
$183 a 8-18298=10(a+1)$, a multiple of 7
$$
a=6
$$

Method 2 The quotient $\frac{183 a 8}{287}$ should be a two digit number.
The first digit of $\frac{183}{3}$ (approximate value) is 6 . The last digit must be $4(\because 7 \times 4=28)$
$$
\begin{array}{l}
\therefore 287 \times 64=18368 \\
\Rightarrow a=6
\end{array}
$$

Remark: The original question is: The number of factors of $a^{2} \cdots \cdots$, which may include negative factors.
$$
6^{2}=2^{2} \times 3^{2}
$$

Factors of 36 are in the form $2^{x} \times 3^{y}$, where $0 \leq x \leq 2,0 \leq y \leq 2$.
The number of factors $=(1+2)(1+2)=9$
Suppose there are $x$ black balls, $y$ red balls, $z$ white balls.
$$
\begin{array}{l}
x+y=9 \ldots \ldots .(1) \\
y+z=11 \ldots \ldots(2) \\
z+x=12 \ldots \ldots .(3) \\
(1)+(2)+(3): 2(x+y+z)=32 \\
c=x+y+z=16
\end{array}
$$

Let one root be $3+\alpha$.
$$
f(3+\alpha)=0=f(3-\alpha)
$$
$\Rightarrow 3-\alpha$ is also a root.
$$
3+\alpha+3-\alpha=6
$$
$\therefore$ Sum of a pair of roots $=6$
There are 16 roots, i.e. 8 pairs of roots
Sum of all roots $=8 \times 6=48$