I looked at this problem and immediately thought of the "least common multiple" concept, so I tried it that way. Don't know if my answer is correct...
Ok, let's state the facts first...
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*Premises*
Starting Positions:
Unknown
Directions of movement:
Either clockwise or counterclockwise (i.e. it doesn't matter)
Starting Time [minutes]:
0
Time for Disks to complete 360° turn [minutes || seconds]: (360° time)
Disk A = 7 || 420
Disk B = 13 || 780
Disk C = 17 || 1020
Disk D = 19 || 1140
Disk E = 23 || 1380
Angular velocity (no acceleration involved/constant velocity) [°/second]:
Disk A = 6/7
Disk B = 6/13
Disk C = 6/17
Disk D = 6/19
Disk E = 6/23
Time that elapsed from the start of each disk rotation until the position of the red dot alignment was reached [minutes || seconds]: (first alignment Time)
Disk A = 2 || 120
Disk B = 3 || 180
Disk C = 4 || 240
Disk D = 7 || 420
Disk E = 9 || 540
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*Solution*
We now know from the premises where the alignment point of every disk is, in regards to time. It doesn't matter whether the disks are spinning CW or CCW, because ultimately each disk reaches the alignment point, due to the fact that the direction of movement is constant (i.e. it doesn't alter from CW to CCW or vice versa) and velocity ≠ 0.
We also know when each disk reaches the alignment point again, for the second time, third time, fourth time etc. We can put it mathematically like this:
Alignment[*i*] := (first alignment time) + (360° time) x (i) Dimension=[minutes]
whereas i is the i-th time the aligned position is reached
again and i is element of the natural numbers:
i (element of) { 1, 2, 3, 4, 5, 6, ...}
In the concrete case of Disk A, the formula can be written as follows:
a[*i*] := 2 + 7 x i
That means for the first alignment point, which we can calculate by setting i = 1, we get:
a[1] = 2 + 7 x 1 = 9 [minutes]
some more examples:
a[2] = 2 + 7 x 2 = 16 [minutes]
a[4] = 2 + 7 x 4 = 30 [minutes] etc.
that means after 9 minutes since the disks have started to spin, disk A reaches it's alignment position for the 1st time again, after 30 minutes for the 4th time again.
If we make this formula for every disk, we get this:
This mathematica program defines the above function for each disk, a[i_] for Disk A, b[i_] for Disk B etc.
Then it makes a list with all solutions of this functions by inserting many numbers from the set of natural numbers for i and saves these results in a list. In the program you see, mathematica has calculated these functions for i between 1 and 1'000'000.
The list for Disk A for the first 30 integers looks like this:
{9, 16, 23, 30, 37, 44, 51, 58, 65, 72, 79, 86, 93, 100, 107, 114, 121, 128, 135, 142, 149, 156, 163, 170, 177, 184, 191, 198, 205, 212}
Now that we have a list of times that represent the reaching of each disk's alignment point, we can look what times all these disks have in common, e.g. if the same number occurs in all the lists of each disk, then it's the time all of the red dots are at the same position.
Example:
a[2] = 16 [minutes]
b[1] = 16 [minutes]
when a reaches the alignment positions for the second time again, b reaches the alignment position for the first time again. If this problem consisted of these two discs only, the solution would be 16 minutes. Now we have to find the times when every Disk (from a to e) function reaches the same solution time.
This is how to do it: The lists can be viewed as normal sets of numbers, and we're basically looking for an intersection of all these lists/sets of numbers, i.e. the numbers these lists have in common.
These are the times when all the discs reach the same positions (Dimension = [minutes]):
{477857, 1153896, 1829935, 2505974, 3182013, 3858052, 4534091, 5210130, 5886169, 6562208}
The first time they reach this position is 477857 minutes, that's 331 days 20 hours 17 minutes .
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And that's my take on this whole problem... i did it quite intuitively, don't know if it's correct...
@suckmymuscle: *the nine minute solution*
Here's how the disks are situated after 9 minutes (in my opinion):
Disk A : Uses the first 2 minutes to reach the alignment position, then reaches it again after another 7 minutes have elapsed. So that means that Disk A is at its alignment position after 9 minutes.
Disk E : Uses the whole 9 minutes to reach its alignment position for the first time. So after 9 minutes, Disk E is at its alignment position, just like Disk A.
Disk B : Uses the first 3 minutes to reach the alignment position for the first time. It reaches the alignment position again after another 13 minutes have elapsed. So after 9 minutes in total, Disk B is about 190° away from it's alignment position.
Same problem with Disk C & D.
Because B,C & D are not at their alignment positions after 9 minutes, 9 minutes can't be the solution to this problem.
