# A parent wants to give his horses to their three children [duplicate]

A parent leaves their horses to their three children in their will. The oldest gets half of the horses, the middle gets a third, and the youngest gets a ninth. The parent has $$17$$ horses so they couldn't be divided wholly according to their will. A friend of the parent solved this problem by lending one of their horses to make the total $$18$$, so the oldest gets $$9$$ horses, the middle gets $$6$$ horses, and the youngest gets $$2$$ horses, and each gets a little more than they were entitled to. The remaining $$1$$ horse is returned to the parent's friend. What is wrong here?

Solution: $$\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{9}=\dfrac{17}{18}$$, so after giving away the horses to the children, there should be $$\dfrac{1}{18}$$ of the horses left, but after the parent's friend takes their horse back there is none left.

I'm a bit confused by this word problem. Nothing seems wrong to me, and each child actually gets a little more than they were entitled to. Is the solution saying the friend of the parent should leave that last horse because according to the will $$\dfrac{1}{18}$$ of the horses are left untouched? But that horse originally belongs to the friend.

• One contribution to the paradox, the fractions don't add up to a whole. $\frac12 + \frac13 + \frac 16 = \frac{9}{18} + \frac{6}{18} + \frac{3}{18} = \frac{17}{18}$ which is not a whole 1. The only thing which the loaned horse changes is to create even divisibility. Commented May 16 at 2:08
• I feel like whoever wrote that problem (the one with the question "what is wrong here?") did not know how to have fun. The original parable is playing with numbers in a way that is obviously not quite correct, but very satisfying. Commented May 16 at 3:14
• I've seen this problem multiple times over the years (with camels etc. instead of horses), certainly in a book in the 1990s, and would have been very surprised if it had not been asked on this site before. But it has: math.stackexchange.com/q/3773496/96384 Commented May 16 at 4:52
• So @EuroMicelli leads me to the fact the problem has its own Wikipedia page: en.wikipedia.org/wiki/17-animal_inheritance_puzzle Commented May 16 at 16:38
• Just wait 'til you get to the "Three Men in a Hotel Room and the Bellboy" problem Commented May 16 at 19:30

If the parent's will is interpreted literally, then:

• The oldest child gets $$8 \frac{1}{2}$$ horses.
• The middle child gets $$5 \frac{2}{3}$$ horses.
• The youngest child gets $$1 \frac{8}{9}$$ horses.
• There is a leftover $$\frac{17}{18}$$ horse whose new owner is not specified.

There are two problems here:

• At least 3 of the horses need to be divided into fractions. This may not be a problem if the family's intent is to slaughter the horses for meat, but fractional horses are pretty useless for riding, or for pulling wagons or plows.
• What to do about the leftover $$\frac{17}{18}$$ horse. Assuming that the parent doesn't have any other heirs, it will go to the three children anyway, but the question is how to distribute it. There are at least three reasonable interpretations:
• Primogeniture. Everything that's not specifically given to the other children goes to the oldest child. This results in the distribution $$(9\frac{4}{9}, 5\frac{2}{3}, 1\frac{8}{9})$$.
• Equally divide it among the three children, giving an extra $$\frac{17}{54}$$ horse to each. This results in the distribution $$(8\frac{22}{27}, 5\frac{53}{54}, 2\frac{11}{54})$$.
• Use the same $$9:6:2$$ ratio that's specified in the will. This results in a distribution of $$(9, 6, 2)$$.

If you require that each child inherit an integer number of horses, then all three interpretations above happen to round to the same triple of integers: $$(9, 6, 2)$$. So that's the most straightforward solution to this horse apportionment problem.

The introduction of an 18th horse isn't necessary to work out this solution. It's just that the fractions happen to be easier to work with for 18 horses than for 17.

For every $$18$$ horses, $$17$$ whole horses are distributed in a ratio of $$9:6:2$$ to the children, and an extra $$1$$ horse remains unclaimed.

It is true that the children receive more than their share, for $$17$$ horses divided would yield whole horses in ratio $$8:5:1$$ ($$14$$ total alive) plus bits and pieces of $$3$$ remaining (dead) horses to each child.

The "loaned" horse is truly loaned, but becomes the unclaimed horse that allows the children to claim integer horses, while the neighbor lays claim to the unclaimed horse.

• @ronaldchristenkkson I dunno, I kind of like this problem, I think it was made by someone who has actually dealt with wills. Wills in the real world often have poor wording or contentious, unforeseen, or nearly inexplicable terms that defy solution, and this problem at least has a happy ending! Commented May 15 at 17:19

An extension to my comment, Suppose each horse is divisible in 18 parts, 17 horses would make up 306 pieces, According to the parent, the children will get 153, 102,34 pieces respectively, What's left is 17 pieces. Now these 17 pieces are left, Now what the Extra loaned horse does is Redistribute these 17 remaining piece without using the pieces of horses loaned by increasing the amount of pieces each person gets by 9,6,2 respectively, This is essentially adding a loaned horse to increase amount of horse each person gets so that the remaining 17 pieces are consumed only. The Loaned horse acts as catalyst in chemical reaction

This is a very strange way of dealing with this question - by handling each $$18$$ one a time, yielding a remainder of $$1$$ out of the $$18,$$ and putting that remainder back in the pool. This ultimately is dealing with $$17$$ at a time, because after $$18$$ are handled, then one is put back, so if $$f_i(N)$$ is the number of horses given to child $$i$$, you'd get:

$$f_1(N)=9+f_1(N-17)\\f_2(N)=6+f_2(N-17)\\f_3(N)=6+f_3(N-17).$$

This would be true for any $$N>17,$$ but when $$N=17,$$ this is no longer true - because we can't take out $$18$$ and then put back $$1.$$ But we can if we borrow one horse, giving us $$18$$, and then put one back.

So we always can reduce it to $$N<17$$ if we allow a borrow-and-put-back at $$17$$.

A more natural approach would be to give each child $$\left(\frac N2,\frac N3,\frac N9\right),$$ including fractional horses, which we can think of as debts, and then divy up the remainder $$N/18$$ the same way. You end up with, if $$f(N)=(f_1(N),f_2(N),f_3(N)),$$ that $$f(N)=N(1/2,1/3,1/9)+f(N/18)$$, which gives:

$$f(N)=N(1/2,1/3,1/9)\sum_{k=0}^{\infty}\frac1{18^k}=N(1/2,1/3,1/9)\frac{18}{17}=N(9/17,6/17,2/17)$$

Of course, if $$N$$ is not divisible by $$17,$$ this will give some choices for which child gets the the ceiling of their values. Luckily, the three resulting fractional parts will be different, so the simplest apprroach is to give the floor of the values and then, depend on whether there are one or two horses remaining, give an additional horse to the one with the top fractional part or the children with the top two fractional parts.

But I'd sue the lawyer who drew up this will.

• I think the natural approach is 18 at a time, if you put the remainder back in the pool, you are not reserving the 1/18 that is unallocated, you are "loaning" that over and over with the effect that you would only ever have at most a handful of horses left over, even if you had a herd of millions to begin with Commented May 15 at 20:06
• But putting it back in the pool is exactly the same as putting it back in the pull when you get fewer than 18 elements left. Basically, if you have functions of three variables, then $f(N,R)$ where $R$ the remainders accrued will be in terms of $f(N-18,R+1),$ and then $f(N,R)=f(N+R,0)$ when $N<18.$ So $N+R$ is all that matters, and that will always $N_0-17m$ for where $m$ is the number of divisions we do. @RobinSparrow In other words, keeping a seperate pile of "left out in prior rounds" is irrelevant - they get merged back into the remainder all the time. Commented May 15 at 20:34
• So if $N=18q+r$ with $0\leq r<18,$ then: $F(18q+r,0)=(9q,6q,2q)+f(r,q)$, and $f(r,q)=f(r+q,0)=f(N-17q,0).$ Whether we put the remaindered back in the main pile immediately or after we are down less than $18$ in the main pile, the result is the same. Commented May 15 at 20:52
• What I'm saying is that if you take seriously the idea that 1/18 should not go to the sons, then propose to borrow and put back remainders internally within the herd, that does not honor the will. If you have 17 million horses, a kind (and rich) friend could gift and then reclaim 1 million unclaimed horses, but they would be unhappy to receive 1 or a few pieces of horses in return. I see what you're doing and saying, and I definitely don't want to beat this hypothetical horse to death, I'm just responding to you saying it's strange to view this as an 18n situation. Commented May 15 at 20:53
• @RobinSparrow Oh, I agree that the will is bollocks and doesn't make sense unless it says what to do with the remainder. I'm just explaining why the given answer really acts as processing $17$ at a time, not eighteen, and that "borrowing one and returning it" when it is read this way makes the last step more uniform with the other steps. But the whole interpretation of the will doesn't make sense, since the will doesn't make sense, hence the last line of my answer. Commented May 15 at 20:54

The important first point here is that the sum of the children's allocated fractions does not add up to one. Convert all the fractions to the lowest common denominator, $$18$$, and you can immediately see that only $$\frac{17}{18}$$ of the herd is bequeathed in the will.

In fact, the children receive all the horses, so there's an extra $$\frac{1}{18}$$ of the herd (nearly a full horse) distributed. It is therefore unsurprising that the children all end up with more than they are originally promised. Perhaps this is what the question is looking for when it asks what is "wrong": all the horses are distributed, even though a small fraction of the herd is not technically part of the will.

The chicanery with the borrowed and returned horse tweaks the allocation so each share now fits a whole numbers of horses. The result of having $$18$$ horses to divide is, of course, that the fractions can be distributed exactly into each of the bequeathed amounts and that this results in one left over, because of what was noted in the first paragraph.

The borrowed horse is smoke and mirrors: what is actually going on is that the children are getting their allotted share plus a share of what's left over (the $$\frac{1}{18}$$ of the herd not actually bequeathed), in similar proportion to the original allocation. Due to the clever way in which the numbers in the question are designed, this extra share added to the fractional horses they originally receive gives each a whole number of horses.