Prove that the product of 3 consecutive positive integers is divisible by 6

Asked by Sthitaprajna Mishra | 13th Sep, 2014, 07:25: AM

Expert Answer:

Let three consecutive positive integers be, n, n + 1 and n + 2. 
When a number is divided by 3, the remainder obtained is either 0 or 1 or 2.  
∴ 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,  n + 2 = 3p + 1 + 2 = 3p + 3 = 3(p + 1) is divisible by 3. 
If n = 3p + 2,  n + 1 = 3p + 2 + 1 = 3p + 3 = 3(p + 1) is divisible by 3. 
 
So, we can say that one of the numbers among n, n + 1 and n + 2 is always divisible by 3.  
⇒ n (n + 1) (n + 2) is divisible by 3. 
 
Similarly, when a number is divided 2, the remainder obtained is 0 or 1. 

∴ n = 2q or 2q + 1, where q is some integer. 
If n = 2q  n and n + 2 = 2q + 2 = 2(q + 1) are divisible by 2. 
If n = 2q + 1 ⇒ n + 1 = 2q + 1 + 1 = 2q + 2 = 2 (q + 1) is divisible by 2. 
 
So, we can say that one of the numbers among n, n + 1 and n + 2 is always divisible by 2. 
⇒ n (n + 1) (n + 2) is divisible by 2. 
 
Hence n (n + 1) (n + 2) is divisible by 2 and 3.
∴ n (n + 1) (n + 2) is divisible by 6. 

Answered by Mili Hariyani | 14th Sep, 2014, 03:56: AM