I think i know how to solve it but is that the best way? Is there a better way (using number theory). What i do is: knowing that
1st power last digit: 3
2nd power last digit: 9
3rd power last digit: 7
4rh power last digit: 1
5th power last digit: 3
$3^{347} = 3^{5\cdot69+2} = (3^5)^{69} \cdot3^2 = 3\cdot3^2=3^3=27 $ so the result is $7$.
How about $$ 3^2 \equiv -1\pmod {10} $$ so $$ 3^{347} \equiv 3^{2\cdot 173+1} \equiv 3 \cdot(-1)^{173} \equiv -3 \equiv 7 \pmod {10} $$
I don't know why none of the current answers talk about why your methods are wrong and you just happened to get the right answer, so I guess I'll touch on that.
First off, it doesn't make any sense to "factor out" $3^5$ when the last digit cycles with a length of $4$. Regardless, it's not mathematically wrong, so here's how you would do it using that:
$3^{347} \equiv 3^{5*69 + 2} \equiv (3^5)^{69} * 3^2 \pmod {10}$
The next step is where you make your first mathematically incorrect error. You reduce $(3^5)^{69}$ to $3$, but actually it reduces to $3^{69}$. Then, you would say $3^{69} \equiv 3^{4 * 17 + 1} \equiv (3^4)^{17} * 3 \equiv (1)^{17} * 3 \equiv 3 \pmod {10}$. Luckily for you, this happens to reduce to $3$, so you got the right answer. Now we can safely conclude $(3^5)^{69} * 3^2 \equiv 3 * 3^2 \equiv 7 \pmod {10}$
As I said before, it makes much more sense to factor out $3^4$. Here's what it would look like using that method:
$3^{347} \equiv (3^4)^{86} *3^3 \equiv 1^{86}* 3^3 \equiv 7 \pmod {10}$
Much easier, yes?
The last digit rotates in a cycle of 4, not 5. You would be better off applying your method starting from 0:
0th power last digit: 1
1st power last digit: 3
2nd power last digit: 9
3rd power last digit: 7
4th power last digit: 1
And continue from there.
Your reasoning is wrong; you claimed $(3^5)^{69} \equiv 3$, but you were lucky that this was the case; it does not work in general. Since $3^5\equiv 3\pmod{10}$, you could have simplified $(3^5)^{69}$ to $3^{69}$ and proceeded from there.
It may be simpler to reduce by $3^4$, since $3^4\equiv 1\pmod{10}$: $$3^{347} = \big(3^4\big)^{86}\cdot3^3 \equiv 1^{86}\cdot3^3 = 3^3 = 27 \equiv 7\pmod{10}.$$
$$3^{4n+3}=(3^4)^n\cdot3^3=(1+80)^n\cdot(20+7)$$
Using Binomial Expansion, $$(1+80)^n=1+\binom n180+\cdots+\binom n{n-1}80^{n-1}++80^n\equiv1\pmod{10}$$
bc(1) agrees with me that the last digit is 7.
– vonbrand
Apr 17 '14 at 15:40
Yes there are other ways. Clearly you need to work modulo 10 (you need to say so in your answer otherwise the equality doesn't make any sense.
Then notice that $3^2 = 9 \equiv -1 \mod 10$.
$347 = 173\times2 + 1$ ie you obtain :
$$3^{347} \equiv (-1)^{173}\times 3 \equiv -3 \mod 10$$
Therefore last digit is $7$.
You need to be a bit more careful. You have $$3^4=81\equiv 1 \mod 10$$ You should use equivalences rather than equals when doing such a reduction, because it makes your workings clearer.
You have also noted $$3^5\equiv 3$$ And used this to reduce $(3^5)^{69}\times 3^2$ - what you should get is $$(3^5)^{69}\times 3^2 \equiv 3^{69}\times 3^2$$ and you need to reduce again.
If you were using a multiple of $4$ for reduction in the exponent rather than a multiple of $5$ you'd be taking a power of $1$ and that would simplify things..
The exponent $4$ works because there are $4$ integers less than $10$ and co-prime to $10$ - we say the Euler function $\varphi (10)=4$. You might what to investigate the Euler-Fermat theorem (an extension of Fermat's Little Theorem) which says that if $(n,r)=1$ (they are relatively prime) then $r^{\varphi(n)}\equiv 1 \mod n$.
You want: $$ 3^{347} \bmod 10 $$ Luckily for you, $\gcd(3, 10) = 1$, so Euler's theorem helps out. $\phi(10) = (2 - 1) (5 - 1) = 4$, and: $$ 3^{347} \equiv 3^3 \equiv 7 \pmod{10} $$