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While travelling down a road in a socialist country you meet a group of four young men and women sitting in the front porch of their house, with a lively cattle ranch adjoining.

"Our father died, and he willed distinct unit fractions of that ranch's cows to each of us four, none less than $\frac19$," one of them explains. "Initially we were at a loss as to executing the will, particularly because it prohibits slaughtering any cow. But now that you've come, we can divide the cows exactly as per the will, none left over – if you will help us."

"Why me?" you ask in slight confusion.

"Emphasis on the you," another one of the quartet retorts with a sly smile.

You look at the ranch and figure that there are between 80 and 100 cows. How many cows are there and how will they be distributed among the four children?

No cows are harmed or unused in the solution!

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  • $\begingroup$ Are you going to have mercy on us at some point and share the intended solution? $\endgroup$
    – Matthew
    Commented Mar 6 at 19:01
  • $\begingroup$ Right. Are you at some point going to have mercy and let us in? And/or at least accept SWIAV's answer if it's correct? $\endgroup$
    – Matthew
    Commented Mar 7 at 17:07
  • $\begingroup$ @Matthew ugghhh... $\endgroup$
    – Parcly Taxel
    Commented Mar 7 at 18:06

5 Answers 5

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"Initially we were at a loss as to executing the will, particularly because it prohibits slaughtering any cow. But now that you've come, we can divide the cows exactly as per the will, none left over – if you will help us."

This can imply that

the fractions add up to less than one, which is why the ranchers could not split up the cows without slaughtering a few. To solve this, the ranchers can give some cows to you, thereby bypassing the 'no slaughtering' rule.

This part of the problem

none less than 1/9

can imply that

the smallest fraction is 1/9, assuming that the ranchers are not giving you irrelevant information by stating a number that is smaller than the smallest number.

Thus, we can find:

90 cows, with the distribution being 1/2 (45 cows), 1/5 (18 cows), 1/9 (10 cows), and 1/6 (15 cows), leaving 2 cows to you.

But

There is no lower bound, there could be multiple solutions, such as 63 cows, 1/2, 1/6, 1/7, 1/9, and five cows given to you. I've chosen my answer because the number of cows was closest to 100 and the cows given to you was minimal (I would imagine the father wouldn't have wanted to lose a large portion of his cows to some stranger!) I approached this by guess and check and don't have a proof of this being minimal, so perhaps there's a better answer out there.

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    $\begingroup$ Rot13(Gur uvag fgngrf gung gurl jvyy tvir lbh onpx jung gurl qvq abg arrq va gur raq, juvpu vzcyvrf gung lbh tvir gurz fbzrguvat. Vs lbh rqvg gur nafjre gb fnl gurl unir rvtugl rvtug pbjf naq lbh tvir gurz lbhe gjb pbjf, gurl znxr gur fcyvg naq tvir lbh lbhe gjb pbjf onpx, V jvyy hcibgr.) $\endgroup$
    – Weather Vane
    Commented Feb 29 at 17:36
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    $\begingroup$ I totally agree with @WeatherVane. You have the correct fractions; you just need to change the way the cows are being exchanged. $\endgroup$
    – GentlePurpleRain
    Commented Feb 29 at 22:06
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Following spherical-wug-in-a-vacuum's basic approach, what about

$\cfrac{1}{2}$, $\cfrac{1}{4}$, $\cfrac{1}{8}$ and $\cfrac{1}{9}$, which adds to $\cfrac{71}{72}$?

Following the old The Sheik Dies puzzle,

perhaps they had $71$ cows, and if only another cow came walking along then they would have $72$ and could evenly split the original $71$, leaving the extra cow to continue on its way.

That may imply that

"you" from the story text is a cow =).

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  • $\begingroup$ Rot13(Be, lbh unccra gb unir n pbj jvgu lbh). $\endgroup$
    – Weather Vane
    Commented Feb 29 at 12:44
  • $\begingroup$ Is this a pun? Maybe you need to give them an ewe? Regardless I'm not quite getting there. But I'd be interested in seeing the intended answer. $\endgroup$
    – Tyler Seacrest
    Commented Feb 29 at 15:18
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    $\begingroup$ Not quite there? I thought you had solved it. $\endgroup$
    – Weather Vane
    Commented Feb 29 at 17:09
  • $\begingroup$ @TylerSeacrest yeah, there's a reference to a well-known family of jokes in there, which is why the animals in question are cows. $\endgroup$
    – Parcly Taxel
    Commented Mar 1 at 0:39
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    $\begingroup$ @codewarrior0 No cow is harmed in the solution! $\endgroup$
    – Parcly Taxel
    Commented Mar 3 at 2:17
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The ranch has 94 cattle. (Hopefully, or "you" are going to get short-changed.) The fractions to be divided up are 1/4, 1/6, 1/8 and 1/9, adding to 47/72. You loan them 50 (or however many, up to 64) cattle to make 144, they distribute 94 according to the will (36, 24, 18 and 16), and you get the leftover 50. Apparently "you" are someone that has a lot of cattle on hand.

Note that...

...while this would also work with 1/2, 1/4, 1/8 and 1/9, "you" still have to donate 44-64 cattle, and in this case will only get 2 back. (Unless we allow for "your figuring" to be off by 9 and for the ranch to only have 71 cattle.)

Also, I'm curious if we're misusing the word "cow", or if this ranch really has no bulls? (Or are the bulls not part of what's being divided?)

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    $\begingroup$ How can you distribute 1/3 of 100 cows without slaughtering one? $\endgroup$
    – GentlePurpleRain
    Commented Feb 29 at 21:50
  • $\begingroup$ @GentlePurpleRain, corrected! (I was trying to go for as-close-to-equitable-as-possible distribution, but those least-common-multiples are brutal...) $\endgroup$
    – Matthew
    Commented Feb 29 at 22:11
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The puzzle description provides details that I suppose most be red herrings, but it is straightforward to show that no sum of exactly four (or five) distinct unit fractions with denominators between 2 and 9 can sum to one. Therefore, dividing the herd must always lead to a remainder that is either not a unit fraction or equal to one of the four allotted unit fractions. Other answers have touched on 'you' having (or being) cows, and I think this must be the case. I believe the cows are distributed as

84 cows owned by the family and you own 12, for a total of 96 cows.

They are divided among the inheritors in the fractions

1/3 = 32/96, 1/4 = 24/96, 1/6 = 16/96, and 1/8 = 12/96 for a total of 84 cows, and you get your 12 cows back!

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Best I can come up with is there are 90 cows and the fractions are 1/9, 1/6, 1/5, and 1/2. Added up, that makes 88/90. So, there are 90 cows, but two left over, which I guess they will give (as the nice socialists they are and you have none) two cows. You get 2 and the siblings be 10, 15, 16, and 45. (There are other answers, like 1/9, 1/6, 1/4, 1/3, but don't get within the 80-100 cows range. Note: there are only a smallish number of all possibilities, so could try them all.)

If there a pun with the "emphasis on you" (ewe?), I don't get it.

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