<s>[INST] 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?
 [/INST] 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 $70$.
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