Question

What is the HCF of two consecutive even numbers?

Answer 1

Hint: First of all, assume two consecutive even numbers \[2n\] and \[2\left( n+1 \right)\] . We can observe that 2 is the common factor for the numbers \[2n\] and \[2\left( n+1 \right)\] . We know the property that the HCF of two numbers is the number that is common in the factors of both numbers. Now, solve it further and get the HCF of two consecutive even numbers.

Complete step by step answer:
According to the question, we are given two consecutive even numbers and we are asked to find its HCF.
First of all, we have to consider two consecutive even numbers.
Let us assume an even number \[2n\] where \[n\] can be any integer ………………………………….(1)
We know the property that the two consecutive even numbers differ by 2 ………………………………. (2)
Now, from equation (1) and equation (2), we get
The number next to \[2n\] = \[2n+2\] .
The above number can also be written as \[2\left( n+1 \right)\] .
The two consecutive numbers are \[2n\] and \[2\left( n+1 \right)\] …………………………………….(3)
The number \[2\left( n+1 \right)\] is also divisible by 2.
Now, in equation (3), we can observe that the above two consecutive numbers have 2 as common in their factors ………………………………….(4)
Also, we know the property that the HCF of two numbers is the number that is common in the factors of both numbers …………………………………..(5)
 From equation (4) and equation (5), we have 2 as a factor which is common in both consecutive even numbers, \[2n\] and \[2\left( n+1 \right)\] .
Therefore, the HCF of two consecutive even numbers is 2.

Note:
The best way to approach this type of question where we are asked to calculate the HCF is to use the property that the HCF of two numbers is the number that is common in the factors of both numbers. Using this property saves time and we don’t need to spend time calculating the HCF using the traditional method.