Question

How do you find four consecutive even integers whose sum is $42$?

Answer 1

Hint: In order to solve this question, let the four even consecutive integers be $x,x + 2,x + 4$ and $x + 6$. According to the question, on adding these four numbers we get our sum as $42$. We solve the question further as given. We get our answer to be a fraction, but according to the question, only an integer is acceptable. Hence, we conclude that no consecutive even integers add up to give this sum.

Complete step-by-step solution:
In this question, we are asked to find four consecutive integers whose sum is equal to $42$. Integers refer to any number whole number which can be written without any fractional part. As we are asked to find even integers, that would mean that each number shall have a difference of $2$, while if they were just consecutive integers then their difference would be $1$.
Thus, let the numbers be $x,x + 2,x + 4$ and $x + 6$
According to the given:
$ \Rightarrow x + x + 2 + x + 4 + x + 6 = 42$
Adding up the variables and constants, we get:
$ \Rightarrow 4x + 12 = 42$
Adding $ - 12$ to both sides of the equation, we get:
$ \Rightarrow 4x = 42 - 12 = 30$
On dividing both sides of the equation with $4$, we get:
$ \Rightarrow x = \dfrac{{\not{{30}}}}{{\not{4}}} = \dfrac{{15}}{2}$
Since, we already know that integers cannot be fractional hence it is verified that no consecutive even integers can add up to $42$
Let us take a few random even numbers to further verify our answer,
$ \Rightarrow 8 + 10 + 12 + 14 = 44$
$ \Rightarrow 6 + 8 + 10 + 12 = 36$

Note: The difference between any two consecutive either odd or even is 2.
The term consecutive numbers is used to frame a word problem.
The sum of any two consecutive numbers is always odd.
For example $4 + 5 = 9$ and $ - 8 + \left( { - 7} \right) = 15$