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?
 A: Another way, where there is no need to compute the numeric values of the four cubes.
Note that $16+19=17+18=35=70/2$, hence
\begin{align*}
16^3+17^3+18^3+19^3
&=16^3+(35-18)^3 +18^3+(35-16)^3\\
&\equiv 16^3+35^3-18^3+18^3+35^3-16^3\\
&=2\cdot 35^3\equiv 0\pmod{70}.
\end{align*}
In particular, when we expand the cube $(35-2n)^3$ we find
$$35^3-3\cdot 35^2\cdot 2n+3\cdot 35\cdot (2n)^2-(2n)^3\equiv 35^3-(2n)^3\pmod{70}$$
because $3\cdot 35^2\cdot 2n$ and $3\cdot 35\cdot (2n)^2$ are divisible by $70$.
A: 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 $7$.
A: I made a mistake in my solution with the remainders. Below is the correct approach and answer :-
Using this same logic I can write $$Rem[\frac{16^3+17^3+18^3+19^3}{70}] = Rem[\frac{16^3}{70}] + Rem[\frac{17^3}{70}] + Rem[\frac{18^3}{70}] + Rem[\frac{19^3}{70}] = 36 +13+22+69 = 140 \Rightarrow Rem[\frac{140}{70}] = 0$$
A: Hint: $\,a+b\mid a^3+b^3\Rightarrow 35\mid 16^3\!+\!19^3,\,17^3\!+\!18^3\Rightarrow 35\mid x.\,$ $x$ is even so  $\,2\cdot 35\mid x$
