How can I prove that 6^n + 4 is divisible by 10? I want to write a direct proof.

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- March 20th 2014, 07:13 AMStuck Man6^n + 4 is divisible by 10
How can I prove that 6^n + 4 is divisible by 10? I want to write a direct proof.

- March 20th 2014, 07:39 AMPlatoRe: 6^n + 4 is divisible by 10
- March 20th 2014, 07:53 AMStuck ManRe: 6^n + 4 is divisible by 10
I should not use mathematical induction because the question is in an exercise on direct proof and mathematical induction is introduced later.

- March 20th 2014, 07:59 AMSlipEternalRe: 6^n + 4 is divisible by 10
For a direct proof:

Assume $n$ is a positive integer (as it is false otherwise).

$\displaystyle \begin{align*}6^n + 4 & = 5\left(1+\sum_{k=0}^{n-1} 6^k\right) \\ & = 10\left(1 + \sum_{k=1}^{n-1} 3^k2^{k-1} \right)\end{align*}$ - March 20th 2014, 08:57 AMStuck ManRe: 6^n + 4 is divisible by 10
What kind of expansion is done for 6^n? I don't think I've seen it before.

- March 20th 2014, 09:04 AMSlipEternalRe: 6^n + 4 is divisible by 10
- March 20th 2014, 11:02 AMStuck ManRe: 6^n + 4 is divisible by 10
Thanks.

- March 20th 2014, 11:17 AMStuck ManRe: 6^n + 4 is divisible by 10
The question said to consider the units digit. There is a hint: what does 6x6 have as a units digit?

- March 20th 2014, 11:38 AMPlatoRe: 6^n + 4 is divisible by 10
If $6^n$ has a unit digit of $6$ what is the unit digit of $6^n+4~?$

Look at this table. - March 20th 2014, 12:05 PMStuck ManRe: 6^n + 4 is divisible by 10
It would be necessary to prove that 6^n has a unit digit of 6.

- March 20th 2014, 12:13 PMSlipEternalRe: 6^n + 4 is divisible by 10
- March 20th 2014, 04:00 PMPlatoRe: 6^n + 4 is divisible by 10
@Stuck Man, I hope that you will tell us all what your instructor expected of you in an answer.

**Please.**

Here are some comments I would ask you to consider.

You posted this question in a number theory forum. It is unthinkable that someone doing number theory would not have induction as a tool. As posed then this is basic algebra.

If I am correct about that, then I suspect the hint you were given, $6\cdot 6=~?$, is right on. At a very basic level, any integer ending in a six when multiplied by six results in an integer also ending in six. Now I admit that without induction there is no rigorous proof for that. But without induction available a rigorous proof is not called for. - March 21st 2014, 05:01 AMStuck ManRe: 6^n + 4 is divisible by 10
The question is "By considering the units digit prove that 6^n + 4 is always divisible by 10 where n is a member of N." The topic of the exercise is direct proof. Proof by induction is later in the chapter.

- March 21st 2014, 11:44 AMJeffMRe: 6^n + 4 is divisible by 10
There is a direct but very long proof.

$Step\ 1:\ Prove\ that\ 5m\ ends\ in\ 5\ if\ m\ is\ odd.$

$Step\ 2:\ Prove\ that\ 4^k\ is\ even.$

$Step\ 3:\ Prove\ that\ 5^p = (4 + 1)^p\ is\ odd.$

$Step\ 4:\ Prove\ that\ 5^q = 5 * 5^{(q - 1)}\ ends\ in\ 5.$

$Step\ 5:\ Prove\ that\ 6^r\ is\ even.$

$Step\ 6:\ Prove\ that\ 6^s = (5 + 1)^s\ ends\ in\ 1\ or\ 6.$

$Step\ 7:\ Prove\ that\ 6^n\ does\ not\ end\ in\ 1.$ This step is the only one that is even a bit tricky.

$Step\ 8:\ Prove\ that\ 6^n + 4\ ends\ in\ 0\ and\ so\ is\ divisible\ by\ 10.$ Trivial

I must admit that I have not worked out each step in detail, but I believe that you never need to use induction or rely on any theorem that itself is proved by induction. Step 6 does rely on contradiction. Can you use that?

AFTERTHOUGHT: I am not sure, but I think I see a way to avoid step 7. I'll think about it more as I walk the dog. - March 21st 2014, 03:09 PMJeffMRe: 6^n + 4 is divisible by 10
OK. I am back from letting the dog do canine things and think there is a way to prove this exercise without utlilizing

*reductio ad absurdum*. I am not 100% sure. I was a history major, and my proofs are undoubtedly not up to modern standards of rigor.

Step 5 $\implies last\ digit\ of\ 6^n \in\ Set_A = \{0,\ 2,\ 4,\ 6,\ 8\}.$

Step 6 $\implies last\ digit\ of\ 6^n \in\ Set_B = \{1,\ 6\}.$

So Step 7 reduces to $last\ digit\ of\ 6^n \in Set_A \bigcap Set_B = \{6\}.$

Because I have never understood those mathematicians who deny the law of the excluded middle and consequently reject proofs by*reductio*, I don't know whether this is considered a direct argument or not, but it is not, at least not superficially, a proof by contradiction.

Hope this all helps.