The sum of three consecutive even integers is 114. what are the three integers?

Question

i am not good at math so i need help with math

Answer 1

Yet another answer is realize even numbers are represented by 2*n, n integer. Then 2*n+2*(n+1)+2*(n+3)=114.

Then n = 18 and the numbers are 36 (2*18), 38 (2*19), and 40 (2*20) that are three consecutive even integers that sum to 114. Due to the equation I think these numbers 18,38,40 are unique. No other three consecutive even numbers will sum to 114.

Answer 2

Hi Moe.

 

Since your question asks for "three consecutive even integers" you can call them as follows:

 

x1 = first even integer

x2 = second even integer

x3 = third even integer

 

Since you know they're both consecutive and even, you know that:

 

x2 = x1 + 2

x3 = x2 + 2

 

This is because there's always a difference of 2 between consecutive even integers. Take a look at these examples:

 

6, 8, 10

 

102, 104, 106

 

514, 516, 518

 

No matter what the even integers are, as long as they're consecutive, they're spaced out by 2 each.

 

You know that the sum of your even integers adds up to 114, so:

 

x1 + x2 + x3 = 114

 

But since x3 = x2 + 2

 

x1 + x2 + (x2 + 2) = 114

 

And since x2 = x1 + 2

 

x1 + (x1 + 2) + ((x1 + 2) + 2) = 114

 

So your final equation will be:

 

x1 + x1 + 2 + x1 +2 + 2 = 114

 

3x1 + 6 = 114

 

3x1 = 108

 

x1 = 36

 

x2 = x1 + 2

 

x2 = 36 + 2 = 38

 

x3 = x2 + 2

 

x3 = 38 + 2 = 40

 

So your consecutive even integers are:

 

36, 38, and 40

 

If you add these all together you get:

 

36 + 38 + 40 = 114. So you know you did it correctly.

 

Hope this helps.

Answer 3

Since the three integers are consecutive even integers, the average of the top and bottom integers is the middle integer, which means that 114 can be thought of as 3 times the middle integer. Divide 114 by 3 and fill in the blanks.

Answer 4

So, you know the three numbers must add up to a total of 114. The numbers possible range from 1-114, but you can rule out high values as three consecutive high values will have  a much larger sum (110+111+112=333), and the opposite is true with low values (10+11+12=33) so you could rule those out if you're taking a multiple choice test for example. What you want to do is take 114 and divide it by three. For example, if I were doing the same problem for 210 I would use 210/3 to give me the average value (70) of three numbers that would equate 210. Now you know the average of the three integers must be 70. Combining this with the consecutive requirement means that 69, 70, and 71 are the only integers that fit the criteria. You can use this same logic for your problem.