(5) Given the positive integer $n=a b c<10000$, where $a$, $b$, and $c$ are all prime numbers, and $2a+$ 
$3b=c$, $4a+c+1=4b$, then the value of $n$ is $\qquad$
(5) 1118 Hint: Since $b=6a+1, c=20a+3$, then $a(6a+1)(20a+3)<10000$, which means $120a^3<10000$, so $a<5$. Also, since $c=20a+3$ is a prime number, $a \neq 3$. Therefore, $a=2, b=13, c=43, n=1118$.