Question

Show that the product of any two consecutive positive integers is disable by 2.

Answer 1

Hint: We know that if a number is divisible two then the reminder will be zero. We will take one number as x and another number will definitely be x+1. The number which we have taken can be an even number or odd so we will have two cases. We will check when x is even and when x is taken odd.

Complete step by step answer:
Let the two consecutive numbers be x and x+1.
Then the product of both the numbers will be x(x+1).
Case1:
If x is taken as even number i.e. x = 2k
Then x+1 = 2k+1
The product of two number will be 2k(2k+1)
We will now solve the above expression
$\Rightarrow 4{{k}^{2}}+2k$
We will take 2 common from the above expression
$\Rightarrow 2(2{{k}^{2}}+k)$
We can clearly see that above expression is divisible by 2
Case2:
If x is taken as odd number i.e. x=2k+1
Then x+1 = (2k+1) +1
The product of two number will be 2k+1 {(2k+1) +1}
We will now solve the above expression
$\begin{align}
  & \Rightarrow {{\left( 2k+1 \right)}^{2}}+2k+1 \\
 & \Rightarrow 4{{k}^{2}}+1+4k+2k+1 \\
 & \Rightarrow 4{{k}^{2}}+6k+2 \\
 & \Rightarrow 2(2{{k}^{2}}+3k+1) \\
\end{align}$
We can clearly see that above expression is divisible by 2

Note:
The numbers which are divisible by 2 are known as the even numbers; example:2, 4, 6, 8…. The numbers which reminder as 1 on dividing with 2 are known as odd numbers; example: 3, 5, 7, 9. The numbers which are only divisible by 1 and itself are known as prime numbers example: 1, 3, 5, 7, 11.