Question

The product of two consecutive natural number is always
(A) An even number
(B) An odd number
(C) A prime number
(D) Divisible by 3

Answer 1

Hint: Two consecutive natural numbers are $n$ and $\left( {n + 1} \right)$. The product of two consecutive natural numbers is $n\left( {n + 1} \right)$. This product is an even number if it has $2$ as one of the factors otherwise it is an odd number. If the product is divisible by one $1$ and number itself then it is a prime number. If either of two consecutive numbers $n$ or $\left( {n + 1} \right)$ is divisible by $3$ then the product must be divisible by $3$.

Complete step-by-step solution:
Given, the two consecutive natural numbers are $n$ and $\left( {n + 1} \right)$.
Now, the product of two consecutive natural numbers is $n \times \left( {n + 1} \right) = n\left( {n + 1} \right)$.
There are two possible cases:
 Case 1: if a natural number $n$ is an even number.
Since $n$ is an even number, it can be written as $2m$ because every even number is divisible by $2$.
So, the product of two consecutive natural numbers is $2m\left( {2m + 1} \right)$.
Now, it is clearly visible that the product has $2$ as one of the factors. So, the product is an even number.
Case 2: if a natural number $n$ is an odd number.
Since $n$ is an odd number, it can be written as $2m + 1$ because every odd number is $1$ more than its previous even number.
So, the product of two consecutive natural number is $\left( {2m + 1} \right)\left( {2m + 1 + 1} \right) = \left( {2m + 1} \right)\left( {2m + 2} \right)$.
Now, taking $2$ as common. we get $\left( {2m + 1} \right)2\left( {m + 1} \right)$
Now, it is clearly visible that the product has $2$ as one of the factors. So, the product is an even number.
Thus, the product of two consecutive natural numbers is always an even number.

Hence, option (A) is correct.

Note: The product and sum of two even numbers is always an even number whereas the product of two odd numbers is always an odd number but the sum of two odd numbers is always an even number.