The sum of three consecutive odd intergers is 39. What is the smallest of these intergers?
Let's represent the smallest odd integer as x.
Since there are consecutive odd integers, the second odd integer would be x + 2 and the third odd integer would be x + 4.
The sum of these three odd integers can be represented as:
x + (x + 2) + (x + 4) = 39
Simplifying the equation, we get:
3x + 6 = 39
Subtracting 6 from both sides, we have:
3x = 33
Dividing both sides by 3, we get:
x = 11
Therefore, the smallest odd integer is 11.
Let's assume the smallest odd integer as "x".
Since we have three consecutive odd integers, the next two odd integers will be "x+2" and "x+4".
According to the given information, the sum of these three consecutive odd integers is 39, so we can write the equation as:
x + (x+2) + (x+4) = 39
By simplifying the equation, we have:
3x + 6 = 39
Now, we can solve for x:
3x = 39 - 6
3x = 33
Dividing both sides by 3:
x = 33/3
x = 11
Therefore, the smallest odd integer is 11.