5. Use the digits $1,2, \cdots, 7$ to form a seven-digit number such that it is a multiple of 11. The number of seven-digit numbers that can be formed is $\qquad$

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5. 576 .

Let $n$ be a seven-digit number satisfying the condition, and let $a$ and $b$ be the sums of the digits in the odd and even positions, respectively. Then $a+b=28$, and $a-b$ is a multiple of 11.
Since $a+b$ and $a-b$ have the same parity, they must both be even. Clearly, $|a-b| \neq 22$, so only $a-b=0$.
Thus, $a=b=14$.
Considering the sums of two numbers from $1,2, \cdots, 7$ equal to 14, there are only the following four scenarios:
$$
\{1,6,7\},\{2,5,7\},\{3,4,7\},\{3,5,6\} \text {. }
$$

In each scenario, these numbers can only be placed in the even positions, and the remaining four numbers, whose sum is also 14, should be placed in the odd positions. Therefore, a total of $4 \times 6 \times 24=576$ such seven-digit numbers are obtained.