(5) Arrange the seven digits $1,2,3,4,5,6,7$ appropriately to form a 7-digit number, such that it is a multiple of 11. The number of such numbers that can be formed is $\qquad$ .
(5) 576 Hint: Let $n$ be a number that satisfies 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 the only possibility is $a-b=0$, which means $a=b=14$. Among the numbers 1, 2, 3, 4, 5, 6, 7, the sum of any three numbers being 14 has only 4 cases: $\{1,6,7\},\{2,5,7\},\{3,4,7\},\{3,5,6\}$. In each case, 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 7-digit numbers are obtained.