A princess is as old as the prince will be when the princess is twice the age that the prince was when the princess’s age was half the sum of their present ages.
This one took a while to figure out and there are numerous valid ways of finding the answer.
Here is the solution I came up with
I created the following table from the riddle:
Current
Future
Past
Princess
x
2z
(x+y)/2
Prince
y
x
z
I then created three equations, since the difference in their age will always be the same.
d = the difference in ages
x – y = d
2z – x = d
x/2 + y/2 – z = d
I then created a matrix and solved it using row reduction.
x
y
z
1
-1
0
d
-1
0
2
d
.5
.5
-1
d
It reduced to:
x
y
z
1
0
0
4d
0
1
0
3d
0
0
1
5d/2
This means that you can pick any difference you want (an even one presumably because you want integer ages).
Princess age: 4d
Prince age: 3d
Ages that work
Princess
Prince
4
3
8
6
16
12
24
18
32
24
40
30
48
36
56
42
64
48
72
54
80
60
To see other solutions check out the comments from when I posted this on my blog.
Kevin, Charles, Larry and Alex are in a room that’s about 110 feet long. In front of them are 5 balls which are exactly 100 ft from the exit. The balls are yellow, purple, green, red and blue, respectively. Each man must carry a ball to the exit. After traveling 20 ft a ball will change color twice. The sequence of color changes is always the same: yellow, purple, green, red, and blue.
At 80 ft Kevin’s ball is red.
At 40 ft Larry’s ball is purple.
At 60 ft Charles’ ball is blue
At 100 ft Alex’s ball is purple.
The remaining ball was blue. Here’s a table of each ball and the color it changes to at 20, 40, 60, 80 and 100 ft. Kevin’s started out yellow, Larry’s was green, Charles’ was red and Alex began with a purple ball, leaving blue as the one nobody picked.