Question

Find three consecutive even numbers whose sum is 66. Help pls and do not spam . Thank You .

Answer 1

Answer:

So the three numbers are: 20,22,24.

Step-by-step explanation:

If the second number is

n

, then the first is

n−2

and the third n+2

, so we have:

66=(n−2)+n+(n+2)=3n

Dividing both ends by

3 , we find n=22.

So the three numbers are:

20,22,24 .

The smallest of these is

20

.

Answer 2

Question:

Find three consecutive even numbers whose sum is 66.

Answer:

$The \: numbers \: are \:\underline\bold{22, \: 24 \: and \: 26}.$

Step-by-step explanation:

Given data :

Let the smallest number be x

The other 2 numbers are (x + 2) and (x + 4)

Therefore the numbers are x , (x + 2) , (x + 4)

Their sum is 66

$\longrightarrow x + (x + 2) + (x + 4) = 66$

$\longrightarrow 3x + 6 = 66$

$\longrightarrow 3x = 66$

$\longrightarrow \boxed{x = 22}$

Now, finding the numbers

${1}^{st} \: number = x = 22$

${2}^{nd} \: number = (x + 2) = 22 + 2 = 24$

${3}^{rd} \: number = (x + 4) = 22 + 4 = 26$

Therefore, the three numbers are 22, 24 and 26 .

Rule:

To find even numbers the numbers must be exists as x , (x + 2), (x + 4), (x + 5), etc.,