The mean of four consecutive even numbers is 2017. What is the difference between the highest and the lowest digits of the highest even number?

2 Answers
Jun 8, 2017

The answer is 2.
Don't panic, the process is simpler than it looks.

Explanation:

If the average of 4 numbers is 2017, then their sum must be 4 times that (because the last step of finding the average is dividing by the number of data points, we can to this backward to find the sum, the step of finding the mean before that).

20174=8068

Now, we can represent 8068 as the sum of four even numbers. We could set X to any of the four and make it work out, but to keep things simple, let X= the highest number.

(X6)+(X4)+(X2)+X=8068

Because they are consecutive even numbers, we know that each one is 2 greater than the last one, and so we can represent them with X=the largest number, X2=the second largest number, and so on.

Now, just solve this equation algebraically to find X, the highest even integer in the set. First, combine like terms:

4X12=8068

Next, add 12 to both sides.

4X=8080

Finally, divide by 4.

X=2020

If you want to check your work on this part, write out the set of consecutive even numbers with the highest number of 2020. Sure enough, the average of 2014, 2016, 2018, and 2020 is 2017.

And now, the part you've all been waiting for:
The difference between the highest and lowest digits of the highest number is...

20=2

Jun 8, 2017

2

Explanation:

Let the four consecutive even numbers be 2n,2n+2,2n+4,2n+6 where n is an integer.

Given that mean of these four numbers is
2n+(2n+2)+(2n+4)+(2n+6)4=2017
(8n+12)=2017×4
8n=806812

Solving for n we get
n=1007

Highest even number is =2n+6=2×1007+6=2020

Its highest and lowest digits are 2and0

Difference between the two digits=20=2