Remainder When Divided By 70 : Remainder Theorem Problem If $x$ = $16^3$ + $17^3$ + $18^3$ + $19^3$ , then $x$ divided by $70$ leaves a remainder of?
I tried to solve this problem by using the remainder theorem which states that remainder of $$Rem[\frac{a+b+c+....}{x}] = Rem[\frac{a}{x}] + Rem[\frac{b}{x}] + Rem[\frac{c}{x}] + .... $$ where 'Rem' means 'Remainder Of'.
Using this same logic I can write $$Rem[\frac{16^3+17^3+18^3+19^3}{70}] = Rem[\frac{16^3}{70}] + Rem[\frac{17^3}{70}] + Rem[\frac{18^3}{70}] + Rem[\frac{19^3}{70}] = 18 +13+11+69 = 111 \Rightarrow Rem[\frac{111}{70}] = 41$$
So my answer comes up as 41. Can someone please explain to me what I might be doing wrong or what I have understood wrong?

To determine the remainder of some computation, $C, \pmod{70}$, simply determine the remainder of $C, \pmod{2}, \pmod{5}$, and $\pmod{7}.$  Then employ Number Theory methods to solve the resultant congruencies.
Let $C = 16^3 + 17^3 + 18^3 + 19^3.$
Since the second and fourth terms of $C$ (only) are odd, it is immediate that $C \equiv 0\pmod{2}.$
You can prove, by binomial expansion, that if $r,s,k,n$ are all positive integers, then $(r+ns)^k \equiv r^k \pmod{n}.$
Therefore, the problem immediately reduces to the twin problems of
computing
$$D = 1^3 + 2^3 + 3^3 + 4^3 \pmod{5}$$
and computing
$$E = 2^3 + 3^3 + 4^3 + 5^3 \pmod{7}.$$
The problem permits shortcuts, by recognizing that if 
$a \equiv (-b) \pmod{n},$ then $a^3 \equiv (-b)^3 = (-1)^3b^3 = -(b^3) \pmod{n}.$
Using this approach, note that when applying $\pmod{5}$ congruencies against $D$,
$4 = (-1),$ and $3 = (-2).$
Similarly, when applying $\pmod{7}$ congruencies against $E$
$5 = (-2),$ and $4 = (-3).$
Therefore, it is immediate that $D \equiv 0\pmod{5}$ and $E\equiv 0\pmod{7}$.
Therefore, it is immediate that $C$ is a multiple of $2$, $5$, and $7$.
