Let’s have 12 stalls spaced equidistantly, and a cow traveling between them. In the 12th stall a butcher is waiting with a sharp knife. So the farthest stall a cow wants to reach is the 11th stall. In each stall is a trough which contains 4 liters of water. When the cow drinks water from the trough it has to drink all 4 liters. To travel from stall to stall one way the cow requires/″ burns″ 1 liter of water. The forward trip is from the first stall to the eleventh. After the cow has a 2nd drink, she goes back one stall before moving forward. This repeats on the 3rd and 4th drinks, and so on. When the cow reaches the 11th stall, all the water she drank during the trip has to have been burned. Now the cow has to return from the 11th stall back to the first. The same conditions apply for the return trip. The question is: For both trips, from how many stalls did the cow have a drink and how many liters of water did she drink in total?

HINT: One way to approach the unique solution starts with the following. Going forward the cow drinks from the 1st, 2nd, 3rd, and 9th stalls...

  • $\begingroup$ I'm not sure what's the puzzle here. The question seems to have described the exact trip route, and we simply just need to add up the amount of water, which is basically the number of stalls visited times 4? $\endgroup$
    – justhalf
    Oct 23, 2021 at 7:53
  • 2
    $\begingroup$ So the "butcher" who appears in the title does nothing but waiting in the last stall? $\endgroup$
    – WhatsUp
    Oct 24, 2021 at 7:47

1 Answer 1


8 Stalls, 32 liters.

One way to think about this is that the first drink the Cow has propels her 4 stalls, and each drink thereafter propels her 2 stalls. A one way trip of 10 stalls means she has to drink from 4 stalls, so for two trips of 10 stalls she has to drink from 8 stalls. One possible configuration of drinks is:

1,5,7,9 then 11,8,6,4

  • $\begingroup$ Check your arithmetic. $\endgroup$ Oct 22, 2021 at 22:11
  • $\begingroup$ What arithmetic problems are there? The problem statement requires some interpretation, but that answer seems to be a reasonable way to interpret it. >! Assumption: "This repeats on the 3rd and 4th drinks, and so on" means that they each repeat the 2nd (go back one before going forward). So drink from 1, walk to 2 3 4 5; drink from 5, walk to 4 5 6 7; drink from 7, walk to 6 7 8 9; drink from 9, walk to 8 9 10 11. >! Assumption: Now the first walk toward 1 is 100% forward, and each subsequent walk takes one step back toward 11 before continuing forward toward 1. (continued next comment) $\endgroup$
    – Ed Murphy
    Oct 23, 2021 at 1:07
  • $\begingroup$ >! So drink from 11, walk to 10 9 8 (but not to 7 because at this point it's empty); drink from 8, walk to 9 8 7 6; drink from 6, walk to 7 6 5 4; drink from 4, walk to 5 4 3 2. Ah yes, so there is a minor mistake there; the cow needs to drink from 2, walk to 3 2 1, and it's done. $\endgroup$
    – Ed Murphy
    Oct 23, 2021 at 1:13

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