2022/068) Find the smallest positive number n such that $3n^3-2019$ is divisible by 2016

 $3n^2-2019$ is divisible by 2016

Both are mutiple of 3 so $n^3-667$ is divisible by 672

Or  $n^3-667+672$ that is $n^3-1$ is divisible by 672

Now $n^3-1= (n-1)(n^2+n+1)$ and 672 = 32 * 21 (product of even and odd)

As $n^2+n+1 = n(n+1) +1 $ is always odd so we have (n-1) multiple of 32.

Futher we must have $n^3 \equiv 1 \pmod 3$ and  $n^3 \equiv 1 \pmod 7$

For $n^3 \equiv 1 \pmod 3$ we take mod 3 and get 0 for 0 1 or 1 and 2 for 2 so n must be 1 mod 3

Or f n-1 must be mutiple of 3

So $n \equiv 1 \pmod {96}$

Futher we must have $n^3 \equiv 1 \pmod 7$ and  $n^3 \equiv 1 \pmod 7$

for $n^3 \equiv 1 \pmod 7$ we take mod 7 and get 0 for 0,1 for 1,2,4,  and 6 for 3,5,6

So we must have n = 1/2/4 mod 7

n = 96k + 1

k =1 gives n = 97 , mod 7 = 6 so not a sulution

k = 2 gives n = 193, mod 7 = 4 so is a solution

So ans smallest n = 193