http://www.sosmath.com/algebra/factor/fac11/fac11.html

The second way is a much nicer one. We notice that

100^{3}

10^{6}

, so we know for

n = 100

that

n^{3} + 11 n

is bigger than

10^{6}

, so we check

n = 99

and we get: 970398 which is smaller than

10^{6}

. So we have the answer:

n = 100

is the first

n

for which

n^{3} + 11 n

is bigger than

10^{6}

The next thing we have to prove is that $n^{3} + 11 n$ is always divisible by 6. This we will prove by using modular arithmetic. We will use modulus 6. For each $n$ , we can have a residue of either: 0,1,2,3,4 or 5. For $n^{3}$ we get the following residues:

0,1,2,3,4,5 respectively (to $n$ ). For $11 n$ we get the following residues: $0, 5, 4, 3, 2, 1$ respectively (to $n$ ).

Combining $n^{3}$ and $11 n$ (respectively) we get a 0 residue, because: 0+0=0 (mod 6), 1+5=6=0 (mod 6), 2+4=6=0 (mod 6), 3+3=6=0 (mod 6), 4+2=6=0 (mod 6), 5+1=6=0 (mod 6). This means that we get a zero residue when dividing by 6, or in other words, $(n^{3} + 11 n)$ is a multiple of 6 or 6 divides $n^{3} + 11 n$ .