Question

The product of three consecutive positive integers is divisible by 6'. Is this statement true or false ? Justify your answer.

Answer 1

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).

Answer 2

Answer:

Step-by-step explanation:

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.