【Question 13】  
A positive integer with $n+3$ digits $144 \cdots 430(n$ 4's$)$, is a multiple of 2015, the smallest positive integer $n$ is $\qquad$
【Analysis and Solution】
2015 decomposed into prime factors: $2015=5 \times 13 \times 31$;
and $1 \underbrace{44 \cdots 430}_{n \uparrow 4}$ can be broken down into: $1 \underbrace{44 \cdots 4}_{n \uparrow 4} 30=2 \times 5 \times 13 \times \underbrace{11 \cdots 1}_{n+1 \uparrow 1}$;
if $1 \underbrace{44 \cdots 430}_{n \uparrow 4}$ is a multiple of 2015;
then it only needs to satisfy that $\underbrace{11 \cdots 1}_{n+1 \uparrow 1}$ is a multiple of 31;
$\underbrace{11 \cdots 1}_{15 \uparrow 1} \div 31=3584229390681$;
that is, when $n+1$ is at least 15, $\underbrace{11 \cdots 1}_{n+1 \uparrow 1}$ is a multiple of 31;
so $n$ is at least $15-1=14$.