Find the smallest value of $n$ such that $3n^3-2019$ is a multiple of 2016

Question

I've been struggling with this question for a while.

So far I've gotten up to n^3 ≡ 1 mod 672 and I am not sure what I can do from here.

Thank you!

Answer 1

Using basic modular arithmetic: $$\begin{align}3n^3-2019&\equiv 0\pmod{2016} \Rightarrow \\ 3n^3-3-2016&\equiv 0\pmod{2016} \Rightarrow \\ 3n^3-3&\equiv 0\pmod{2016} \Rightarrow \\ n^3-1&\equiv 0\pmod{672} \Rightarrow \\ (2a+1)^3-1&\equiv 0\pmod{672} \Rightarrow \\ 8a^3+12a^2+6a&\equiv 0\pmod{672} \Rightarrow \\ 4a^3+6a^2+3a&\equiv 0\pmod{336} \Rightarrow \\ 4(3b)^3+6(3b)^2+3(3b)&\equiv 0\pmod{336} \Rightarrow \\ 36b^3+18b^2+3b&\equiv 0\pmod{112} \Rightarrow \\ 12b^3+6b^2+b&\equiv 0\pmod{112} \Rightarrow \\ 12(2c)^3+6(2c)^2+(2c)&\equiv 0\pmod{112} \Rightarrow \\ 48c^3+12c^2+c&\equiv 0\pmod{56} \Rightarrow \\ 48(4d)^3+12(4d)^2+(4d)&\equiv 0\pmod{56} \Rightarrow \\ (54\cdot 14+12)d^3+(3\cdot 14+6)d^2+d&\equiv 0\pmod{14} \Rightarrow \\ 12(2e)^3+6(2e)^2+(2e)&\equiv 0\pmod{14} \Rightarrow \\ (6\cdot 7+6)e^3+(7+5)e^2+e&\equiv 0\pmod{7} \Rightarrow \\ 6e^3+5e^2+e&\equiv 0\pmod{7} \Rightarrow \\ \text{min } e&=2 \Rightarrow \\ \text{min } n&=2a+1=\\ &=2(3b)+1=\\ &=6(2c)+1=\\ &=12(4d)+1=\\ &=48(2e)+1=\\ &=193.\end{align}$$

Answer 2

Hint: $3n^3-2019$ is a multiple of $2016$ iff $3n^3 \equiv 2019 \equiv 3 \bmod 2016$ iff $n^3 \equiv 1 \bmod 672$

Write $672=2^5 \cdot 3 \cdot 7$. Then

• $n^3 \equiv 1 \bmod 32$, which happens iff $n \equiv 1 \bmod 32$

• $n^3 \equiv 1 \bmod 3$, which happens iff $n \equiv 1 \bmod 3$

• $n^3 \equiv 1 \bmod 7$, which happens iff $n \equiv 1,2,4 \bmod 7$

Now use the chinese reminder theorem for each of the three possibilities mod $7$.