The range, mean and median of four distinct positive integers 10,x,y and z are the same. What is the greatest possible value of one of those numbers?

Question

I guessed the answer for half an hour and still couldn't find it.

Answer 1

Let z be the smallest and x be the largest

Then

0 < z < y < 10 < x

OR

0 < z < 10 < y < x


THe range is x- z

The median is (y+10)/2

The mean is (x+y+z+10)/4

They are all equal to each other, so there are 3 equations and 3 unknowns.


x - z = (y+10)/2

x - z = (x+y+z+10)/4

(y+10)/2 = (x+y+z+10)4

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Multiplies the first equation by 2 and the second and third equations by 4.
The system becomes:

2x - 2z = y + 10

4x - 4z = x + y + z+ 10

2y + 20 = x + y + z + 10


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They simplify to:

2x - y - 2z = 10

3x - y - 5z = 10

x - y + z = 10


Eliminates Y....

Subtracts the first equation from the second:
x - 3z = 0


Subtracts the third equation from the second:
2x - 6z = 0
x = 3z


So both equations are x = 3z.

Plugging into the third equation:
3z - y + z = 10

4z - y = 10

4z - 10 = y


In terms of free variable z,
x = 3z
and
y = 4z - 10


Since all variables are positive integers,
z must be 3,4,5,.....


For z = 3, y = 4(3)-10 = 12 - 10 = 2 < z=3 AND
x = 3(3) = 9 < 10 which are both contradicitons

Z=4 ---> y = 4(4)-10 = 16 - 10 = 6 and x = 3(4) = 12, so
0 < z=4 < y=6 < 10 < x = 12

As a check, the range is 12-4 = 8
the mean is (4+6+10+12)/4 = 32/4 = 8
the median is (6+10)/2 = 8