5. For $n$ consecutive natural numbers, if each number is written in its standard prime factorization form, and each prime factor appears an odd number of times, such $n$ consecutive natural numbers are called a “consecutive $n$ strange group” (for example, when $n=3$, $22=2^{1} \times 11^{1}$, $23=23^{1}$, $24=2^{3} \times 3^{1}$, then $22, 23, 24$ form a “consecutive 3 strange group”). The maximum possible value of $n$ in a consecutive $n$ strange group is . $\qquad$
5.7.

For example: $29, 30, 31, 32, 33, 34, 35$ is a consecutive 7 peculiar group.

Below is the proof: When $n \geqslant 8$, there does not exist a consecutive $n$ peculiar group.

Otherwise, one of them (denoted as $m$) is divisible by 8, so among these $n$ numbers, there must be $m-4$ or $m+4$, which is divisible by 4 but not by 8, contradicting the condition that "each prime factor appears an odd number of times."