# Thread: Induction Proof again :(

1. ## Induction Proof again :(

Hi all,
Once again I'm pretty stumped at this one.
3^3 = 27
3^4 = 81
3^5 =243
3^6 = 729
3^7 = 2187

Now these are the same practically. An even second to last digit and a seven as the last digit. So the pattern will loop. Just proving it seems a right pain! Any help at all would be appreciated.

2. Originally Posted by Jimbobobo

Hi all,
Once again I'm pretty stumped at this one.
3^3 = 27
3^4 = 81
3^5 =243
3^6 = 729
3^7 = 2187

Now these are the same practically. An even second to last digit and a seven as the last digit. So the pattern will loop. Just proving it seems a right pain! Any help at all would be appreciated.
I can't see your linked image, and I doubt anyone else can either.

RonL

3. Prove by induction that in the decimal form 3^n, the second from end digit is even.

4. Originally Posted by Jimbobobo
Prove by induction that in the decimal form 3^n, the second from end digit is even.
Base case n=3 then 3^3=27 and so it is true that the second least significant digit is even.

Suppose it true for some $\displaystyle k$, that is the second least significant digit is even.
Now $\displaystyle 3^k$ is odd, and not divisible by $\displaystyle 5$, so the least significant digit is one of:

$\displaystyle 1,\ 3,\ 7,\ 9.$

Now the carry when these are multiplied by $\displaystyle 3$ is even (either $\displaystyle 0$ or $\displaystyle 2$), and
$\displaystyle 3$ times the second least significant digit of $\displaystyle 3^k$ is even. Even plus an even is
even so the second least significant digit of $\displaystyle 3^{3+1}$ is even, which
completes the induction step.

So we have proven that for all $\displaystyle n>2$ the second least significant digit is even.

RonL

5. Hello, Jimbobobo!

I have a slight variation of RonL's solution.

Prove by induction that in the decimal form of $\displaystyle 3^n$, the tens-digit is even.
We note that $\displaystyle 3^n$ ends in: 1, 3, 7, or 9.

And $\displaystyle 3^{n+1}$ ends in: 03, 09, 21, or 27.
. . Hence, there is a "carry" of either 0 or 2.

Verify $\displaystyle S(1)\!:\;\;3^1 \:=\:03$ . . . true!

Assume $\displaystyle S(k)\!:\;\;3^k$ has an even tens-digit.

. . This means that: .$\displaystyle 3^k$ is of the form: .$\displaystyle 10(even) + \{1,3,7,9\}$

. . Hence: .$\displaystyle 3^k \:=\:10(even) + \{1,3,7,9\}$

Multiply by 3: .$\displaystyle 3^{k+1} \;=\;3\left[10(even) + \{1,3,7,9\}\right]$

. . $\displaystyle \text{and we have: }\;3^{k+1} \;=\;\underbrace{10\!\cdot\!3(even)}_{\text{even}} + \underbrace{3\!\cdot\!\{1,3,7,9\}}_{\text{carry: 0 or 2}}$

$\displaystyle \text{Hence: }\;3^{k+1} \;=\;10\cdot\underbrace{[3(even) + \{0,2\}]}_{\text{even}} + \{1,3,7,9\}$

Therefore, $\displaystyle 3^{k+1}$ has an even tens-digit.
. . The inductive proof is complete.

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

I had this wild idea of using binary, but I got nowhere.

Consider: .$\displaystyle 3 \:=\:11_2$

I thought that .$\displaystyle \left(11_2\right)^n$ would lead to a neat solution,
. . but I didn't find one.

Anyone care to give it a try?