7. Among the ten-digit natural numbers in octal, the natural numbers that can be divided by 7 and have all digits as 0 or 5 are $\qquad$ in number.
7.84.

It is easy to know that the necessary and sufficient condition for a number to be divisible by 7 in octal is that the sum of its digits is divisible by 7.

Therefore, this ten-digit number consists of seven 5s and three 0s, and the first digit must be 5, leaving the remaining nine digits with six 5s, which results in $\mathrm{C}_{9}^{6}=84$ cases.