Prove by induction the following:

10^(n+1) + 3*10^n + 5 is divisable by 9

I got the base step correct. Here is what I had for the inductive step:

Assume Pk is true, 9|10^(k+1) + 3*10^k + 5

Consider Pk+1

10^(k+2) + 3*10^(k+1) + 5

10(10^(k+1)) + 3*10^k(10) + 5

(10^(k+1) + 3*10^k + 5) (10*10)

9|(10^(k+1) + 3*10^k + 5) (10*10) because 9|10^(k+1) + 3*10^k + 5

Any help would be greatful.