1. Proof by Induction (Divisibility)

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.

2. Originally Posted by page929
Prove by induction the following:

10^(n+1) + 3*10^n + 5 is divisible 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

At this point you should be seperating out.... $\displaystyle 10^{k+1}+(3)10^k+5$

to get... $\displaystyle 10\left(10^{k+1}+(3)10^k\right)+5=9\left(10^{k+1}+ (3)10^k\right)+10^{k+1}+(3)10^k+5$

(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.
Then if P(k) is really true, all of that will be divisible by 9.

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