Question

The sum of three consecutive odd numbers is more than 207, how do you find the minimum values of these integers?

Answer 1

Hint: We start the problem by considering three consecutive odd numbers. Let the first odd number be x, the second consecutive odd number after x is x+2, the third consecutive odd number after x is x+4. Then we find the value of x according to the condition given to get the values of the consecutive odd numbers.

Complete step by step solution:
Odd numbers are not exactly divisible by 2 and cannot be divided exactly divided into pairs.
The phrase “consecutive odd numbers” given in the question means that we are supposed to find the successive odd integers.
Let us suppose the first odd number is x.
The consecutive odd integer after odd number x is x+2
Reason: We need to skip an even number x+1 in between.
The consecutive odd integer after odd number x+2 is x+4
Reason: We need to skip an even number x+3 in between.
As mentioned above,
First odd number: x
Second odd number: x+2
Third odd number: x+4
Sum of three consecutive odd numbers = First odd number + Second odd number + Third odd number.
The Sum of three consecutive odd numbers is given by,
$\Rightarrow x+\left( x+2 \right)+\left( x+4 \right)$
$\Rightarrow \left( x+x+x \right)+2+4$
$\Rightarrow 3x+6$
As per the question,
The sum of three consecutive odd numbers is more than 207 but to find the value of x easily let us assume that sum is equal to 207.
Now,
$\Rightarrow 3x+6=207$
Now evaluate further.
$\Rightarrow 3x+6-6=207-6$
$\Rightarrow 3x=201$
Now divide the entire equation with 3
$\Rightarrow x=\dfrac{201}{3}$
$\Rightarrow x=67$
Now substitute them back into the expressions we have assumed for consecutive odd integers.
x=67
x+2 = 67 +2 = 69
x+4 = 67 +4 = 71
Verify by adding them all together and getting the sum as 207.
67 + 69 + 71 = 207.
But we need to find the values whose sum is greater than 207.
Since $3x+6>207$ we need to choose $x>67$
The three consecutive odd numbers can be found out by changing the value of x to the next odd integer which is 69.
Likewise,
x = 69
x + 2 = 69 + 2 = 71
x + 4 = 69 + 4 = 73
$\Rightarrow x+\left( x+2 \right)+\left( x+4 \right)>207$
The sum, 69 + 71 + 73 is 213 which is greater than 207.
The three consecutive integers whose sum is greater than 207 are 69, 71, 73

Note: We have assumed the three consecutive odd integers as x, x+2, and x+4. There is also another way to assume the three consecutive integers which are 2x+1, 2x+3, and 2x+5. Either way, if we solve, we get the same values of x.