[tex]\large{\sf{Question:-}} \\ \\[/tex]
The sum of 3 consecutive whole numbers is 66. What is the greatest of these numbers?
Solution:
The integers are [tex]\sf{20}[/tex], [tex]\sf{22}[/tex] and [tex]\sf{24}[/tex].
You can see that, [tex]\sf{20+22+24 = 66} \\ \\[/tex]
Find:
How did you get that? There's a bunch of ways. The simplest is just inspect.
We need 3 integers that are all close together to make [tex]\bf{66}[/tex].
Then,
[tex]\qquad\large{\dfrac{66}{3} \:= \:22} \\[/tex]
So the integers:
[tex]\qquad\qquad\sf{22, \:22, \:22} \\[/tex] add upto 66. But we want them to be consecutive and even.
Find:
Why don't we just take two from the first 22, and give it to last one?
Then the sum will be same,
[tex]\qquad\sf{22-2, \:22, \:22+2} \\[/tex]
And there we got,
[tex]\qquad\qquad\sf{20, \:22, \:24} \\[/tex]
We can also set up equation to do the job. Let x be the first of those three integer.
Then we know that
[tex]\qquad\qquad\sf{x+2 \:x+4} \\[/tex] are the next two consecutive numbers.
Then, we want the sum of those to be 66:
[tex]\qquad{\dfrac{x+(x+2)+(x+4)=66}{ \dfrac{3x+6=66}{3x=60}}} \\[/tex]
[tex]\qquad\qquad\qquad{\sf{x=20}} \\[/tex]
So our first integer is 20 and next are 22, 24.
[tex]\rule{230pt}{2pt}[/tex]