Theorem: If n is a positive integer, then n(n+1)(n+2) is divisible by 3

Proof by induction:

1*2*3 = 6, which is divisible by 6.

Assume that for n = k, k(k+1)(k+2) is divisible by 6

We need to show that, based on that assumption, (k+1)(k+2)(k+3) is also
divisible by 6.

(k+1)(k+2)(k+3) = (k+1)(k+2)k + (k+1)(k+2)3 = k(k+1)(k+2) + 3(k+1)(k+2).

By induction hypothesis, the first term is divisible by 6, 
and the second term 3(k+1)(k+2) is divisible by 6 because it contains
a factor 3 and one of the two consecutive integers k+1 or k+2 is
even and thus is divisible by 2.  Thus it is divisible by both 3 and
2, which means it is divisible by 6.  The theorem is proved since
the sum of two multiples of 6 is also a multiple of 6. 

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Theorem: If n is a positive integer, then n(n+1)(n+2)(n+3) is divisible 
by 24.

Proof by induction:

1*2*3*4 = 24, which is divisible by 24.

Assume that for n = k, k(k+1)(k+2)(k+3) is divisible by 24

We need to show that (k+1)(k+2)(k+3)(k+4), based on that ssumption,
is also divisible by 24.

(k+1)(k+2)(k+3)(k+4) = (k+1)(k+2)(k+3)k + (k+1)(k+2)(k+3)4 = k(k+1)(k+2)(k+3) + 4(k+1)(k+2)(k+3).

By induction hypothesis, the first term is divisible by 24, 
and the second term 4(k+1)(k+2)(k+3) is divisible by 24 because it contains
a factor 4 and by the preceding theorem (k+1)(k+2)(k+3) is divisible by
6, so 4(k+2)(k+3)(k+4) is divisible by 24.  Therefore (k+1)(k+2)(k+3)(k+4)
is divisible by 24 since it is the sum of two multiples of 24.

Edwin