Possibly there is a more direct way, but it is certainly possible to just compute the numbers Danny likes that end in a specified digit. For $1$, this is just $1$. For $2,3,5,7$, this is clearly just $7$. For $4,9$, this is clearly $\binom{8}{2}=28$ which is divisible by $7$. For $8$, this is $\binom{9}{3}=12 \cdot 7$ which is divisible by $7$. Finally, for $6$, this is just $1+3+5+7+9+11+13=49$ (counting how many digits $6$ we have) which is divisible by $7$. Hence the total count is $1 \pmod{7}$.

Above: there is a more direct way. Hint: Replace $a_1,\ldots,a_7$ by $a_1,a_2/a_1,a_3/a_2,\ldots$. This function is reversible. Then use cyclic permutations.