Find three consecutive odd numbers whose sum is 219.

Hint: An odd number is a number which is not divisible by 2. In other words, it gives a remainder of 1 when divided by 2.
An odd number is of the form $2n+1$ , where $n$ is an integer.
Consecutive odd numbers differ by 2.
Represent the numbers in terms of a variable $x$ and form an equation.

Complete step-by-step answer:
Let's say that the three consecutive odd numbers are $2x+1$, $2x+3$ and $2x+5$ .
According to the question:
 $(2x+1)+(2x+3)+(2x+5)=219$
Adding the terms together, we get:
⇒ $6x+9=219$
Subtracting 9 from both sides of the equation, we get:
⇒ $6x=210$
Dividing both sides by 6, we get:
⇒ $x=\dfrac{210}{6}=35$
Therefore, the numbers are $2(35)+1=70+1=71$ , $2(35)+3=70+3=73$ and $2(35)+5=70+5=75$ .
Check: $71+73+75=219$ .

Note: Even numbers are of the form $2n$ , where $n$ is an integer.
Consecutive even numbers differ by 2.
In general, any number which is a multiple of $k$ , can be assumed to be $kn$ , where $n$ is an integer. Consecutive values of $n$ will give consecutive such numbers. They will be of the form $...,k(n-1),kn,k(n+1),k(n+2),...$
Any number which gives a remainder of $r$ when divided by $k$ , can be assumed to be $kn+r$ , where $n$ is an integer. Consecutive values of $n$ will give consecutive such numbers. They will be of the form $...,k(n-1)+r,kn+r,k(n+1)+r,k(n+2)+r,...$