The relation between three consecutive odd integers is that 3 times the first number is 3 more than twice the third number. Find the third number.
A. $9$
B. $11$
C. $13$
D. $15$

Hint: We first assume the three consecutive odd integers. Then, we try to express the given relation in an algebraic form with the first and third numbers. We get two equations based on the first and the third number. The relation states that these equations are the same. So, we get a linear equation of the unknown. We solve the equation to get the first number. Finally, we get to the third number.

Complete step-by-step solution:
We have three consecutive odd integers.
Let the numbers be $a,a+2,a+4$ where a is an odd integer..
As the numbers are consecutive and odd integers, that’s why we add 2 to get the next number.
Now, given that 3 times, the first number is 3 more than twice the third number.
We represent the given relation in an algebraic form with the first and third numbers.
For the first case 3 times of the first number can be expressed as $3\times a=3a$ …(i)
In the second case, 3 more than the twice of the third number can be expressed as $3+2(a+4)=3+2a+8=2a+11$ …(ii)
These two equations are of the same value.
So, equating these two we get $3a=2a+11$.
Now, we solve the equation to get the value of a.
$\begin{align}
  & 3a=2a+11 \\
 & \Rightarrow 3a-2a=11 \\
 & \Rightarrow a=11 \\
\end{align}$
So, the first number is 11.
The second and third numbers are respectively $11+2=13$ and $11+4=15$.
So, the third number is 15.
The correct option is (D).

Note: We need to pay attention when we are assuming the three consecutive odd integers. We can’t take only consecutive numbers by adding 1 as then we will get a mixed number of odd and even numbers. We also can take the three numbers in some cases in the form of $a-2,a,a+2$. It doesn’t help in this problem but in case of a problem of progression series this form is very beneficial.