I4.2 When the integers $1108+a$, $1453$, $1844+2a$ and 2281 are divided by some positive integer $n(>1)$, they all get the same remainder $b$. Determine the value of $b$.
$$
\begin{array}{l}
1108=p n+b \\
1453=q n+b \\
1844=r n+b \\
2281=s n+b
\end{array}
$$
$\qquad$
$\qquad$
$\qquad$
$\qquad$
$p, q, r, s$ are non-negative integers and $0 \leq b<n$.
(2) - (1): $345=(q-p) n$
(3) $-(2): 391=(r-q) n$
(4) - (3): $437=(s-r) n$ $\qquad$
$\therefore n$ is the common factor of 345, 391, and 437.
$$
\begin{array}{l}
345=3 \times 5 \times 23, 391=17 \times 23, 437=19 \times 23 \\
\therefore n=1 \text { or } 23
\end{array}
$$

When $n=1, b=0$. (rejected)
When $n=23$, sub. $n=23$ into (1): $1108=23 \times 48+4$
$$
b=4
$$

Remark: original question: $\qquad$ divided by some positive integer $n$, they all get the same remainder $b$.
There are two possible answers for $b: 0$ or 4.