find the last digit of $347^{61}$ I need help with the question, "find the last digit of $347^{61}$" . I don't know where to start, I know that it requires modulo arithmetic but I can't think where to start, this is all the question gave me. 
 A: $347^{61}\equiv 7^{61}$ (modulo 10).
$7^{61}=7\cdot 7^{60}=7\cdot 49^{30}\equiv 7\cdot (-1)^{30}$ (modulo 10).
Hence the answer is $7$.
A: hint: $347 = 7 \pmod {10}, 7^4 = 1 \pmod {10}$
A: HINT : Let $[n]$ be the last digit of $n$. Then,
$$[7^1]=7,[7^2]=9,[7^3]=3,[7^4]=1,[7^5]=7,\cdots$$
And $61\equiv 1\pmod 4$.
A: Will try and clarify a bit in depth without fancy words some other folks answers.
$347^{61} = (340+7)^{61}$ and we know from binomial theorem that all but the term $7^{61}$ will have at least one factor $34 \cdot 10$ in it. Now let us make a table of the powers of 7, saving only the last digit at each step, 1,7,4(9),6(3),2(1)
We see that 7,9,3,1, is followed by 7 and therefore the pattern must continue there onward. So we can reduce the exponent by clock arithmetics as well, the period being 4 and 61=4*15+1 ( i.e. 1 mod 4 ) We check our table above to see that 1 corresponds to 7, which therefore is our answer.
A: Euler's theorem implies that $a^{\phi(n)} \equiv 1 \text{ (mod } n)$ whenever $a$ is relatively prime to $n$. In this case, $\phi(10)=4$, so $a^{4}\equiv 1 \text{ (mod 10})$ whenever $\gcd(a, 10)=1$. So
$$
347^{61} = (\underbrace{347^{15}}_{a})^4 \cdot 347 \equiv 347 \equiv 7 \text{ (mod 10)}
$$
so the last digit is 7.
A: In general $a^{\varphi(n)} \equiv 1 \pmod{n}$ for every $a$ coprime to $n$, where $\varphi$ is Euler's totient function.
Clearly $347 \equiv 7 \pmod{10}$ is coprime with $10$. Furthermore, $\varphi(n) = 4$ and $61 \equiv 1 \pmod{4}$. Thus there is an integer $k$ such that
$$
347^{61} \equiv 7^{4k + 1} = (7^k)^4 \cdot 7 \equiv 7 \pmod{10}
$$
A: $ \color{#c00}{347^2}\equiv 7^2\equiv \color{#c00}{-1}\!\pmod{10},\ $ and  $\ \color{#c00}{i^2\equiv -1}\,\Rightarrow\, i^{\,\large 1+4n}\equiv i(\color{#c00}{i^{\large 2}})^{\large 2n}\equiv i(\color{#c00}{-1})^{\large 2n}\equiv i$
Remark $\ $ It can be viewed as modular analog of $\,\ i^{\,\large 1+4n} =\, i\,$ in $\,\Bbb C\,$ or, better, in the Gaussian integers $\,\Bbb Z[i] \cong \Bbb Z[x]/(x^2+1).\,$ In fact, we can map this calculation in the Gaussian integers into any ring with a sqrt of $\,-1\,$ via the universal properties of $\,\Bbb Z\,$ and said quotient ring. From this viewpoint $\,\Bbb Z[i]\,$ is the universal ring containing a sqrt of $\,-1.\,$ Any ring equality that holds in $\,\Bbb Z[i]\,$ maps to valid equality in any other ring containing a sqrt of $\,-1.\,$ For example, we have $\,2^2 = -1\, $in $\,\Bbb Z/5\,$ so we deduce $\,2^{1+4n} = 2,\,$ i.e. $\,2^{1+4n}\equiv 2\pmod{5}$
