Equal Averages

If you work out the mean, median, mode and range of $2, 5, 5, 6, 7$, you'll notice something interesting...

The mean, mode, median, and range of this set are all $5$.
 

Can you find other sets of five positive whole numbers where Mean = Median = Mode = Range?

 

Charlie found that $50, 100, 100, 100, 150$ had Mean = Median = Mode = Range = 100

Alison found that $40, 100, 100, 120, 140$ also had Mean = Median = Mode = Range = 100

How many sets of five positive whole numbers are there with Mean = Median = Mode = Range = 100?

 

Comments

n = any number
first number 2 * n
second number 5 * n
third number 5 * n
fourth number 6 * n
fifth number 7 * n

To get SOME of the answers.

2n, 5n, 5n, 6n, 7n provides a "family" of solutions that satisfy the criteria.

Can you now find some different "families" that also satisfy the criteria?

Case 1: 50, 100, 100, 100, 150

Case 2: X1 < X2 < 100 < 100 < X3

$50 < X1 < X2 < 100$
and $150 < X3 < 200$
and $X3 - X1 = 100$
and $2X3 + X2 = 400$

There are 16 possibilities:

X1 = 51   X2 = 98   X3 = 151
X1 = 52   X2 = 96   X 3= 152
.
.
.
X1 = 66   X2 = 68   X3 = 166


Case 3: X1 < 100 < 100 < X2 < X3

$0 < X1 <50$
and $100 < X2 < X3 <150$
and $X3 - X1 = 100$
and $2X3 + X2 = 400$

There are 16 possibilities:

X1 = 34   X2 = 132   X3 = 134
X1 = 35   X2 = 130   X 3= 135
.
.
.
X1 = 49   X2 = 102   X3 = 149


Therefore there are 33 possibilities altogether.

 

From Penndale Middle School (Lansdale, PA) – Chi Alpha Mu (Math Club)

Knowing that MEAN = MEDIAN = MODE = RANGE = 100,

For any set of five positive whole numbers: A, B, C, D, E

C = 100 (to account for the median = 100)
E = 100 + A (to account for the range = 100)
B or D = 100 (to account for the mode = 100)

Let D = 100, then (A + B + 100 + 100 + (100+A)) / 5 = 100
Simplified, B = 200 - 2A

With this it was found that:

34 ≤ A ≤ 66

Therefore there are 66 – 33 = 33 sets of five positive whole numbers with MEAN = MEDIAN = MODE = RANGE = 100