Question

Three times the first of three consecutive odd integers is 3 more than twice the third. What is the third integer?

• A. 15
• B. 9
• C. 11
• D. 5

Hint:

When dealing with consecutive integers (even or odd), clearly define their sequence before forming equations, and always verify the result with the problem statement.

Answer 1

The Correct Option is B

Step 1: Represent the integers Let the first odd integer be $x$.
Since they are consecutive odd integers, the second is $x+2$ and the third is $x+4$. Step 2: Translate the statement into an equation “Three times the first” means $3x$.
“Twice the third” means $2(x+4)$.
“Three times the first is 3 more than twice the third” means: \[ 3x = 2(x+4) + 3 \] Step 3: Solve for $x$ \[ 3x = 2x + 8 + 3 \] \[ 3x = 2x + 11 \] \[ x = 11 \] Step 4: Identify the third integer The three integers are $11$, $13$, $15$. Thus the “third” integer = $15$ — but this contradicts the answer key given in options. Step 5: Check the problem wording carefully If instead “third integer” refers to the largest when counting backwards (meaning $x$ is smallest odd integer that satisfies) — re-evaluating, we may need to shift indexing: If first is $x$, second $x+2$, third $x+4$, the above is consistent and $x=5$ gives integers $5, 7, 9$. Checking: $3(5) = 15$ and $2(9) + 3 = 21$. This fails. So the correct solution from our equation gives $x=11$, hence third integer $15$. But given option marking says (b) 9, suggesting possibly that the problem originally indexed differently or is misprinted. \[ \boxed{15} \quad \text{(per algebraic derivation)} \]