n³-n = n n²-1)
= n (n+1)(n-1)
=( n-1) n(n+1)
= product of threeconsecutive positive integers
Now, we ave to show that the product of three consecutive positive integers is divisible by 6.
We know that any positive integer a is of the form 3q, 3q + 1 or 3q + 2 for some integer q.
Let a, a + 1, a + 2 be any three consecutive integers.
Case I: if  a=3q
a(a + l)(a + 2) = 3q(3q + 1)(3q + 2)
 = 3q (2r)
= 6qr, which is divisible by 6.
(∵ Product of two consecutive integers (3q + 1) and (3q + 2) is an even integer, say 2r)
Case II: If   a=3q + 1
∴ a(a + l)(a + 2) = (3q + 1)(3q + 2)(3q + 3)
= (2r) (3)(q + 1)
= 6r(q + 1),
which is divisible by 6
Case III: If    a=3q + 2
ஃ a(a + l)(a + 2) = (3q + 2)(3q + 3)(3q + 4)
= multiple of 6 for ever
= 6r (say),
which is divisible by 6.
Hence, the product of three consecutive integers is divisible by 6.

Factors of 1 to 10 numbers
1 = 1
2 = 1 x 2
3 = 1 x 3
4 = 1 x 2 x 2
5 = 1 x 5
6 = 1 x 2 x 3
7 = 1 x 7
8 = 1 x 2 x 2 x 2
9 = 1 x 3 x 3
10 = 1 x 2 x 5
\(\therefore \)  LCM of numbers 1 to 10
= LCM (1, 2, 3, 4, 5, 6, 7, 8, 9, 10)
= 2 x 2 x 2 x 3 x 3 x 5 x 7 = 2520

Let a be any positive integer.
We know that, any odd positive integer is of the form 2q + 1, where q is a whole number.
\(\therefore \)  a = 2q + 1
\(\Rightarrow \)  a² = (2q + 1)²           [squaring both sides]
\(\Rightarrow \)   a² = 4q(q + 1) + 1            ...(i)
Note that q(q + 1) is either '0' or even, for any whole number q.
So, let q(q + 1) = 2m where m is a whole number.
From Eq.(i), we get  a² = 4(2m) + 1 = 8m + 1

No, because by Euclid's division lemma, 
we have, a = 4q + r,  \(0\le r<4\)             ...(i)
a can be in the form 4q, 4q + 1, 4q + 2 or 4q + 3

True, because the product of any two consecutive numbers, say n(n + 1) will always be even as one out of n or (n+1) must be even.