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.