Question

1. what is the smallest of three consecutive odd integers that have a total sum of 51?

type a response

Explanation:

Step1: Define the variables

Let the three consecutive odd integers be \( x \), \( x + 2 \), and \( x + 4 \), where \( x \) is the smallest one.

Step2: Set up the equation

The sum of these three integers is 51. So, we have the equation:

$$ x + (x + 2) + (x + 4) = 51 $$

Step3: Simplify the equation

Combine like terms:

$$ 3x + 6 = 51 $$

Step4: Solve for \( x \)

Subtract 6 from both sides:

$$ 3x = 51 - 6 $$

$$ 3x = 45 $$

Divide both sides by 3:

$$ x = \frac{45}{3} $$

$$ x = 15 $$

Answer:

The smallest of the three consecutive odd integers is 15.

(Note: There seems to be a typo in the original problem statement where it says "517" instead of "51" based on the context of consecutive odd integers summing to a reasonable value. If it is indeed 517, the steps would be similar:

Step1: Define the variables

Let the three consecutive odd integers be \( x \), \( x + 2 \), and \( x + 4 \), where \( x \) is the smallest one.

Step2: Set up the equation

The sum of these three integers is 517. So, we have the equation:

$$ x + (x + 2) + (x + 4) = 517 $$

Step3: Simplify the equation

Combine like terms:

$$ 3x + 6 = 517 $$

Step4: Solve for \( x \)

Subtract 6 from both sides:

$$ 3x = 517 - 6 $$

$$ 3x = 511 $$

Divide both sides by 3:

$$ x = \frac{511}{3} \approx 170.33 $$

But this is not an integer, so likely a typo. Assuming it's 51, the answer is 15.)