# Thread: Prove that 9 is a factor of...

1. ## Prove that 9 is a factor of...

Prove that 9 is a factor of:
P(n): $\displaystyle 10^(^n^+^1^) + 3 * 10^n + 5$ for all positive integer n

Using the induction method,
I have proven P(1) to be true

P(k): $\displaystyle 10^k (10+3) + 5 = 9a$
P(k+1): $\displaystyle 10^(^k^+^1^) (10+3) + 5 = ?$

How do I continue? I am new at doing proofs

2. Do you have to use induction? Sometimes that's tedious for problems like this. I would try to rewrite 10 as 9+1 and then use the binomial theorem.

3. Yes, unfortunately it is required to use the induction method.

4. Originally Posted by mushroom
Prove that 9 is a factor of:
P(n): $\displaystyle 10^(^n^+^1^) + 3 * 10^n + 5$ for all positive integer n

Using the induction method,
I have proven P(1) to be true

P(k): $\displaystyle 10^k (10+3) + 5 = 9a$
P(k+1): $\displaystyle 10^(^k^+^1^) (10+3) + 5 = ?$

How do I continue? I am new at doing proofs
$\displaystyle 10^{k + 1} (10 + 3) + 5$

$\displaystyle = 10 \cdot 10^k (10 + 3) + 5$

Now use the P(k) step in the form of $\displaystyle 10^k(10 + 3) = 9a - 5$
$\displaystyle = 10 (9a - 5) + 5$

And I'm sure you can take it from here.

-Dan

5. Hello, mushroom!

$\displaystyle \text{Prove that 9 is a factor of: }\:10^{n+1} + 3\!\cdot\!10^n + 5\,\text{ for all positive integers }n$

$\displaystyle \text{Verify }S(1)\!:\; 10^2 + 3\!\cdot10 + 5 \:=\:135,\,\text{ a multiple of 9 . . . True!}$

$\displaystyle \text{Assume }S(k)\!:\;10^{k+1} + 3\!\cdot\!10^k + 5 \:=\:9a\,\text{ for some integer }a$

$\displaystyle \text{Add }(9\!\cdot\!10^{k+1} + 27\!\cdot\!10^k)\,\text{ to both sides:}$

. . $\displaystyle 10^{k+1} + {\bf9\!\cdot\!10^{k+1}} + 3\!\cdot\!10^k + {\bf27\!\cdot\!10^k} + 5 \;=\;9a + {\bf9\!\cdot\!10^{k+1}} + {\bf27\!\cdot\!10^k}$

. . . . . . . .$\displaystyle 10^{k+1}(1 + 9) + 3\!\cdot\!10^k(1+9) + 5 \;=\;9a + 9\!\cdot\!10^{k+1} + 27\!\cdot\!10^k$

. . . . . . . . . . . . $\displaystyle 10^{k+1}\cdot 10 + 3\!\cdot\!10^k\!\cdot\!10 + 5 \;=\;9\left(a + 10^{k+1} + 3\!\cdot\!10^k\right)$

. . . . . . . . . . . . . . . .$\displaystyle 10^{k+2} + 3\!\cdot\!10^{k+1} + 5 \;=\;\underbrace{9\left(a + 10^{k+1} + 3\!\cdot\!10^k\right)}_{\text{a multiple of 9}}$

We have proved statement $\displaystyle S(k+1).$

The inductive proof is complete.