http://www.sosmath.com/algebra/factor/fac11/fac11.html
The second way is a much nicer one. We notice that 1003 is
106, so we know for n=100 that n3+11n is bigger than
106, so we check n=99 and we get: 970398 which is smaller
than 106. So we have the answer: n=100 is the first n for
which n3+11n is bigger than 106.
The next thing we have to prove is that n3+11n 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 n3 we get the following residues:
0,1,2,3,4,5 respectively (to n). For 11n we get the following
residues: 0,5,4,3,2,1 respectively (to n).
Combining n3 and 11n (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, (n3+11n)
is a multiple of 6 or 6 divides n3+11n.