If the sum of three consecutive even integers is 66 what are the integers?

3 Answers
Mar 31, 2018

Answer:

#20, 22, 24#

Explanation:

Take the first integer as #x#. Since we are talking about the next even integer, it will come when you add two to #x#.

For example, let's say that you have #2# as the value of #x#. The next even integer is #4#, which is #x+2#, and the next is #6#, which is #x+4#.

So eventually, the equation is

#(x)+(x+2)+(x+4) = 66#

#3x + 6 = 66#

#3x = 66-6#

#3x = 60#

#x = 60 ÷ 3#

#x = 20#

So

  • #x = 20#
  • #x+2 = 22#
  • #x+4 = 24#
Mar 31, 2018

Answer:

So the three numbers are: #20; 22; 24#

Explanation:

You have to 'jump' an odd number to get to the next even number. So even numbers occur every second place (counting in 2's)

Set the first even number as #n#
Then the second is #n+2#
The third is #n+4#

#(n)+(n+2)+(n+4) = 3n+6 = 66#

Subtract 6 from both sides

#3n=60#

Divide both sides by 3

#n=20#

So the three numbers are: #20; 22; 24#

Mar 31, 2018

Let us assume the first even number to be #color(purple)(x#.

So the other to Consecutive even numbers would be:

#color(purple)(x+2 and x+4.#

So, according to the question,

#color(green)(->(x)+(x+2)+(x+4)=66#

#->x+x+x+2+4=66#

#->3x+6=66#

#->3x+cancel6-cancel6=66-6#

#->3x=60#

#(cancel(3)x)/cancel3=60/3#

#color(magenta)(x=20#

And so, the three consecutive even numbers which sum up to 66 :

#color(red)(1. x=20#

#color(red)(2. x+2=22#

#color(red)(3. x+4=24#

As a check:

The sum of all three should be #66#

#color(blue)(=20+22+24#

#color(blue)(=66#

Hence it is proved that the 3 no.s are #color(Darkorange)(20, 22 and 24#

~Hope this helps! :)