The sum of three consecutive even integers is 180. How do you find the numbers?

3 Answers
Mar 21, 2018

Answer: 58,60,62

Explanation:

Sum of 3 consecutive even integers is 180; find the numbers.

We can start by letting the middle term be 2n (note that we can't simply use n since it would not guarantee even parity)

Since our middle term is 2n, our other two terms are 2n-2 and 2n+2. We can now write an equation for this problem!
(2n-2)+(2n)+(2n+2)=180

Simplifying, we have:
6n=180

So, n=30

But we're not done yet. Since our terms are 2n-2,2n,2n+2, we must substitute back in to find their values:
2n=2*30=60
2n-2=60-2=58
2n+2=60+2=62

Therefore, the three consecutive even integers are 58,60,62.

Mar 21, 2018

58,60,62

Explanation:

let the middle even numbe rbe 2n

the others will then be

2n-2" and "2n+2

:. 2n-2+2n+2n+2=180

=>6n=180

n=30

the numbers are

2n-2=2xx30-2=58

2n=2xx30=60

2n+2=2xx30+2=62

Mar 21, 2018

see a solution process below;

Explanation:

Let the three consecutive even integers be represented as; x+2 , x+4, and x+6

Hence the sum of three consecutive even integers should be; x+2 + x+4 + x+6 = 180

Therefore;

x+2 + x+4 + x+6 = 180

3x + 12 = 180

Subtract 12 from both sides;

3x + 12 - 12 = 180 - 12

3x = 168

Divide both sides by 3

(3x)/3 = 168/3

(cancel3x)/cancel3 = 168/3

x = 56

Hence the three consecutive numbers are;

x + 2 = 56 + 2 = 58

x + 4 = 56 + 4 = 60

x + 6 = 56 + 6 = 62