Question

the product of three consecutive positive integers is divisible by 6 rest to the power 'n' is this statement true or false? justify your answer

Answer 1

Let us three consecutive  integers be, n, n + 1 and n + 2.
 Whenever a number is divided by 3 the remainder obtained is either 0 or 1 or 2.
 let n = 3p or 3p + 1 or 3p + 2,
where p is some integer. 
If n = 3p,
then n is divisible by 3. 
If n = 3p + 1,
then n + 2 = 3p + 1 + 2 = 3p + 3 = 3(p + 1) is divisible by 3.
 If n = 3p + 2,
then n + 1 = 3p + 2 + 1 = 3p + 3 = 3(p + 1) is divisible by 3.
 So that n, n + 1 and n + 2 is always divisible by 3. ⇒
n (n + 1) (n + 2) is divisible by 3.  
 Similarly, whenever a number is divided 2 we will get the remainder is 0 or 1. 
∴ n = 2q or 2q + 1, where q is some integer. 
If n = 2q, then n and n + 2 = 2q + 2 = 2(q + 1) are divisible by 2. 
If n = 2q + 1, then n + 1 = 2q + 1 + 1 = 2q + 2 = 2 (q + 1) is divisible by 2. 
So that n, n + 1 and n + 2 is always divisible by 2. ⇒ n (n + 1) (n + 2) is divisible by 2.   
But n (n + 1) (n + 2) is divisible by 2 and 3.
   ∴ n (n + 1) (n + 2) is divisible by 6.

Answer 2

Answer:

The statement is true that the product of any three consecutive positive numbers can be divisible by 6.

Solution:

Let take any 3 consecutive integers, say 3,4,5.

Product of 3,4,5 = 3 × 4 × 5 = 60, which is divisible by six

(\frac {60}{6}=10).

Taking another set of 3 consecutive integers, say 13, 14, 15

Product of 13, 14, 15 = 13 × 14 × 15 = 2730, which is divisible by six

(\frac {2730}{6}=433).

Thus, it can be observed that any 3 consecutive numbers chosen randomly has 1 or more even numbers (i.e. divisible by 2) and has 1 or more numbers that is divisible by three . This collectively leads to the product of the three consecutive integers to be divisible by six as (2 × 3 = 6).