45. From the 10 numbers $0,1,2,3,4,5,6,7,8,9$, select 5 different digits to form a five-digit number, and this five-digit number can be divisible by $3,5,7,13$. Then, the largest five-digit number is $\qquad$
Answer: 94185.
Solution: Since this number is divisible by $3,5,7,13$, of course, this number is divisible by the least common multiple of $3,5,7,13$, which is $3 \times 5 \times 7 \times 13=1365$,

Therefore,
this number is a multiple of 1365.
Let this five-digit number be $1365a$,

Then
$1365a \leq 98765$,

Solving for $a$,
$$
a \leq 72 \text{. }
$$

Calculating sequentially, $72 \times 1365=98280, 71 \times 1365=96915, 70 \times 1365=95550, 69 \times 1365=94185$.
Only 94185 meets the criteria.
Therefore, the largest five-digit number composed of 5 different digits and divisible by $3,5,7,13$ is 94185.