How to find a large modulo? I'm studying for an algebra test, and a question that comes up on every paper is one in a format like the following: 
Compute the element $(2·4·6·8·10·12·14·16)^3$ in $Z_{17}$.
Our lecturer hasn't really let us know how to do this. Any help would be really appreciated. 
 A: Define $[k]_{m}:=min\{n\in \mathbb{N}_{0} : k \equiv n \pmod m\}$ where $k\in \mathbb{Z}$ and $m>1$. The trick is that $[a*b]_{m}=[[a]_{m}*[b]_{m}]_{m}$ and $[a+b]_{m}=[[a]_{m}+[b]_{m}]_{m}$. From multiplicative formula it is easy to see by induction that also for all $t\in \mathbb{Z}_+$: $[a^t]_{m}=[a*a*..*a]_{m}=[[a]_m*[a]_m*..*[a]_m]_m=[([a]_{m})^t]_{m}$.
Easy interpretation of $[\ ]_m$ is that if any formula with variables over $\mathbb{Z}$ combined from adding, multiplication and raisiong to integer power, for example $a*b+c^d$ is given, then for any $a',b',c'\in \mathbb{Z}$ fulfilling $$a \equiv a' \pmod m\land b \equiv b' \pmod m\land c \equiv c' \pmod m$$ 
following congruence holds:
$$ a*b+c^d \equiv a'*b'+c'^d \pmod m$$
Congruence of powers is a bit more tricky, but if $p$ is prime then Fermat's little theorem gives formula:
$$ c^p\equiv c\pmod p$$
and as a consequence, for any $c,d,c',d'\in \mathbb{Z},\ p\nmid c$ such that:
$$ c \equiv c' \pmod p\land d \equiv d' \pmod {p-1}$$
following congruence holds:
$$ c^d \equiv c'^{d'}  \pmod p$$
For example if $k=(2*3*4*5*6)^4$ and $m=17$, then $((2*3*4)*(5*6))^4=(24*30)^4\equiv (7*13)^4= 91^4 \equiv 6^4=24 \equiv 7 \pmod {17}$. So finally $(2*3*4*5*6)^4\equiv 7 \pmod {17}$.
In your example $m=17$ and $k=(2*4*6*8*10*12*14*16)^3$. Try to solve it now.
A: A couple ways to do it:
Note $16 = -1$ and $35 = 1$
So dividing out powers of $2$ and multiples of $5$ and $7$
$2*4*6*8*10*12*14*16=2^8(1*2*3*4*(5*7)*8)= 16^2(2*3*4*8)= (-1*3*4)=-12 = 5$
And $5^3 = 25*5= 8*5= 40= 6$
A few things worth noting.  $17$ is prime so by Fermat's little theorem $a^{16} \equiv 1 \mod 17$ for all $a$.  If $\gcd(a,16)=1$ then the order is $a^{16} = 1$ and $a^8=-1$.  If $a=2$ then $2^8=1; 2^4=-1$.
