Problem 1099

Among two-digit numbers, the multiples of 17 are
$$17,34,51,68,85 \text {; }$$
the multiples of 23 are
$$23,46,69,92 \text {. }$$

In the above 9 numbers, the units digits include $1,2,3,4,5,6,7,8,9$. In the tens place, 6 appears twice, 7 does not appear, and the other digits $1,2,3,4,5,8,9$ each appear once.

Obviously, 7 cannot be in the tens place, meaning 7 cannot appear in the first 1999 digits of the two-thousand-digit number, so 7 must be the last digit of this two-thousand-digit number.

Thus, the digit before 7 should be 1 (since 17 is a multiple of 17), and before that, there should be a 5 (since 51 is a multiple of 17), and so on, forming the following diagram:
This means that when a certain digit appears, the next digit will be the one indicated by the arrow in the diagram.
Clearly, in the two-thousand-digit number $a_{1} a_{2} \cdots a_{1999} a_{2000}$,
$$a_{2000}=7, \quad a_{1999}=1, \quad a_{1998}=5, \quad a_{1997}=8, \quad a_{1996}=6,$$

Starting from the 1996th position, the first 1996 digits, from $a_{1} \rightarrow a_{1996}=6$, cycle through $6 \rightarrow 9 \rightarrow 2 \rightarrow 3 \rightarrow 4 \rightarrow 6 \rightarrow 9 \rightarrow \cdots$, with each cycle consisting of five numbers.
Therefore, the sum of all the digits in this two-thousand-digit number is
$$\begin{aligned} & (6+9+2+3+4) \cdot 399+(6+8+5+1+7) \\ = & 24 \cdot 399+27 \\ = & 3 \cdot(8 \cdot 399+9) . \end{aligned}$$

Thus, this two-thousand-digit number is a multiple of 3, and therefore a composite number.