Question

How do you find consecutive odd integers whose sum is 116?

Answer 1

Hint: the odd integers are written in the form $2x + 1,\,x \in N$ where N is the natural numbers. Take two integers as the form $2x + 1,\,2x + 3$ add them and make them equal to the given number. Find x and other odd integers.

Complete step-by-step answer:
The objective of the problem is to find the consecutive odd integers whose sum is 116 .
 About Odd numbers : Odd numbers are whole numbers which when divided by 2 gives remainder as one. Odd numbers are written in the form $2x + 1,\,x \in N$ where N is the natural numbers. The odd integers are 1,3,5,7,9,11, and so on. One is the first positive odd integer.
 Given that the sum of two consecutive odd integers is 116.
The odd integers are written as $2x + 1$. We need two integers so one integer is $2x + 1$ next odd integer can be written as $2x + 1 + 2$ that is $2x + 3$.
It is given that the sum of two consecutive integers is 116.
Now we can write
 $\left( {2x + 1} \right) + \left( {2x + 3} \right) = 116$
Add all the terms that are on left hand side , we get
$\Rightarrow$ $\left( {4x + 4} \right) = 116$
Taking 4 as common we get
$\Rightarrow$ $4\left( {x + 1} \right) = 116$
Diving with four on both sides we get
$
  \dfrac{{4\left( {x + 1} \right)}}{4} = \dfrac{{116}}{4} \\
\Rightarrow x + 1 = 29 \\
$
Subtract one on both sides we get
$
\Rightarrow x + 1 - 1 = 29 - 1 \\
\Rightarrow x = 28 \\
$
Now substitute the x= 28 in $2x + 1$ and $2x + 3$
$\Rightarrow$ \[2\left( {28} \right) + 1\,,\,\,2\left( {28} \right) + 3\]
On solving above we get ,
$57,59$
The consecutive odd integers are $57,59$.
Now we can check the values by adding them
On adding 57 and 59 we get 116.
Therefore, the required two consecutive odd integers are $57,59$.

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