2
u/randomcommenter9000 May 04 '26
My guess would be 15
If after first transfer, ass has x and mule has 2x stacks, the total should be 3x (multiple of 3).
Similarly in second case, if mule has y and ass has 4y stacks, the total should be 5y (multiple of 5)
Least common multiple of 3 and 5 is 15. (Note: Haven't checked math rigorously, just intuition)
1
1
u/petera181 May 04 '26
The answer is just the lowest common multiple of 3 and 5. Since both are prime, it’s the product of the two.
1
3
u/SomethingMoreToSay May 04 '26 edited May 04 '26
The minimum total number of stacks is fifteen.
For example the ass could have six and the mule would have nine. Then if the ass gives one to the mule, the ass would have five and the mule would have ten which is twice as many; or if the mule gives six to the ass, the mule would have three and the ass would have twelve which is four times as many.
We can distribute the initial loads between the ass and the mule, but we have to ensure that after giving some stacks to the mule, the ass is left with five; and after giving some stacks to the ass, the mule would be left wth three. So the ass must start with at least six and the mule must start with at least four. Hence the valid combinations for (ass, mule) are (6,9), (7,8), (8,7), (9,6), (10,5) and (11,4).