Question

Let’s talk about spinning plates! A juggler wants to perform a trick where he will spin three plates as long as possible. Suppose that a plate of type k starts the plate spinning trick by spinning at k rotations per second. Every plate slows down by one rotation per second each second. Whenever needed, the juggler can increase the spin rate of a plate of type k, by krotations per second. If the plate’s speed is not increased before or at the instant its spin rate reaches zero, it falls off its support and ends the trick. The juggler can perform at most one “boost” of speed to one plate at the end of each second during the trick. Determine the unordered triples (a,b,c) of positive integers such that the juggler can keep three plates, of types a, b, and c, spinning indefinitely.

Answer 1

(Note: rps stands for rotations per second)

First, a specific instance:

1 rps plate : slows down by 1 rps per second => after one second, it would lose 1 rps . Thus after one second, its speed is 1 rps-1rps =0

2 rps plate: two seconds before it falls.

3 rps plate: three seconds before it falls

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n rps plate: n seconds before it falls.

Boosting: The boost returns the 1 rps plate to 1 rps. It returns 2 rps plate to 2 rps....it returns n rps plate to n rps.

The 1 rps plate definitely requires boost every second.

The 2 rps plate definitely requires boost every two seconds but can be boosted after one second also.

The 3 rps plate definitely requires a boost every 3 seconds but can be boosted in the first and 2nd second as well.

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and the n rps plate definitely requires one boost every n seconds but can be boosted in n-1 , n-2, n-3.....1st second as well.

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The man can boost only one plate per second.

If one of those plates is 1 rps, he will have to spin only that again and again. All other types of plates will fall down if he keeps this spinning. So, 1 rps plate is no good.

Let us take (2,3,4) as an example.

If the boost is done as

Time (elapsed)	1	2	3	4	5	6	7	8	9	10	11	12	13	...
Plate boosted	4	2	3	2	4	2 3								...

The above timing works till the sixth second. At the sixth second, there is a clash between the 2 and 3

so, (2,3,4) not good

The whole point was to fill the boxes with the lowest spinning plate first.

Time (elapsed)	1	2	3	4	5	6	7	8	9	10	11	12	13	14	15	16	17	18	19	...
Plate boosted		2	3	2 4		2 3		2 4	3	2		2 3 4		2	3	2 4		2 3		...

The problem at 6 is over come by spinning 3 at the 5th second. The problem at 4 is overcome by spinning 4 at the 1st second.

Now, note that if there were two slots left before 12, both 3 and 4 could have been accomodated.

12 is the common multiple.

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The logic for this problem is that if you over come the toughest situation: when all three need a spin: which occurs at the multiples of the three numbers, the plates can be spun indefinitely.

For three plates, you need atleast two gaps before the common multiple to be able to accomodate the higher spinning plates.

Thus, ANY triplet of integers, excluding 1 and 2 can be used

eg: 3,4,5. The LCM is 60. This is how the table will look:

time	1	2	3	4	5	6	7	8	9	10	11	12	13	14	15	...	55	56	57	58	59	60	61	62	63	64
plate			3		5	3			3	5		3 4			3 5		5		3			3 4 5

Now, at the 60th second, we have a clash. So, cleverly, we spin the 4 and 5 at the 58th and 59th seconds. Thus the table will look like this:

time	1	2	3	4	5	6	7	8	9	10	11	12	13	14	15	...	55	56	57	58	59	60	61	62	63	64
plate			3		5	3			3	5	4	3		5	3		5		3	4	5	3

Thus, three plates require two gaps before the LCM, so the numbers 1 and 2 have to be left out.

Thus, the required triplet is : (n1,n2,n3)

where the lowest n has to be 3.

EXTRA:

Generalizing, 4 plates will require 3 gaps before the LCM.

So, 1,2,3 have to be left. The minimum rps should be 4!

for n plates, n-1 gaps required. Thus, the minimum rps should be n!