Question

How do you find the two greatest consecutive even integers whose sum is less than 217?

Answer 1

Hint: Here in this question, we have to read the question and find the relation by reading the question. By knowing the relation, we can have to write in the form of numerals. While writing in the numeral form we use the variables, constants and arithmetic operations. After that we have to determine the values.

Complete step-by-step answer:
From the given data of a question, we have to find the two numbers which are real and they are consecutive even numbers.
So let us consider the two even consecutive numbers to be “2n” and “2n+2”.
In the question they have given the sum of these two numbers is less than 217. Therefore, we have \[ \Rightarrow 2n + 2n + 2 < 217\]----(1)
On simplifying the terms which are present in LHS we get
\[ \Rightarrow 4n + 2 < 217\]
Take 2 from LHS to RHS we get
\[ \Rightarrow 4n < 217 - 2\]
On simplifying we get
\[ \Rightarrow 4n < 215\]
When we divide the above equation by 2 we get
\[ \Rightarrow 2n < 107.5\]
Therefore the value of 2n should not exceed 107.5 and therefore the value of 2n will be 106 and 106 is an even number.
The other even number is 108. Because in the question we have two consecutive even integers. Therefore the two greatest even consecutive real numbers are 106 and 108.
So, the correct answer is “106 and 108”.

Note: Since in question they have given the two consecutive even integers we consider the number as 2n and 2n+2, because they represent the consecutive even numbers. While simplifying we use the simple arithmetic operations like addition, subtraction, multiplication and division.