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I was going through some problems then I arrived at this question which I couldn't solve. Does anyone know the answer to this question?

One day, a person went to a horse racing area. Instead of counting the number of humans and horses, he counted $74$ heads and $196$ legs. How many humans and horses were there?

rubik
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Prateek
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3 Answers3

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A hypercentaur is a creature with $2$ heads and $6$ legs; an anticentaur is a creature with no head and $-2$ legs.

Since $74$ heads make for $37$ hypercentaurs, with $74\cdot 6/2=222$ legs, you have $(222-196)/2=13$ anticentaurs.

Since a hypercentaur is the same as a human on a horse, and an anticentaur is a human deprived of a horse, we have counted $37-13=24$ horses and $37+13=50$ humans.

Parcly Taxel
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egreg
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    This is officially one of my favorite answers. – pjs36 Apr 19 '16 at 16:12
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    And a surrealcentaur is a creature with a non-standard number of heads and an imaginary number of legs. :-) – Asaf Karagila Apr 19 '16 at 16:13
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    @pjs36 I was inspired by a text by Elio Pagliarani, "La merce esclusa”, where a similar problem is solved by introducing the chickenrabbit, with two heads and six legs, and the dechickenized rabbit, with two legs and no head. The number of heads is $18$ and the number of legs is $56$. If you read Italian, you can enjoy the text here – egreg Apr 19 '16 at 16:22
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    https://math.stackexchange.com/questions/478212/is-there-another-simpler-method-to-solve-this-elementary-school-math-problem/478222#478222 – Display Name Aug 19 '17 at 17:03
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    I'm going to show your solution to everyone I know, because I love it! It's awesome! You made my day! Thank you! – user729424 Dec 06 '19 at 19:15
  • I don't know if this will come handy but what is done above is the principle of superposition (Physics) or using $2$ LI basis we went to another basis again by linear transformation (Linear Algebra) – aitfel Aug 30 '20 at 15:04
  • This seems much too comlicated and abstract for such a simple question. And a hypercentaur is not the same as a human OR a horse; it is a human AND a horse. Just let $x$ be the number of horses and $y$ be the number of humans. We know that $x+y=74$ and that $4x+2y=196$. The solution is $x=24$ and $y=50$, – Mark Viola Jan 02 '24 at 17:21
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    @MarkViola I double checked to have written “a human *on* a horse”. Of course I know how to solve this with a linear system, but adding the mythological characters makes it funnier. – egreg Jan 02 '24 at 17:25
  • @MarkViola Your comment (and another answer that you found more acceptable) defines a system of equations and says it's solvable. This answer describes a way to solves it, and defining new species seems less abstract to me than (for example) by Gaussian elimination or Cramer's rule. – peterwhy Jan 02 '24 at 20:53
  • @peterwhy Solving a linear system of two equations in two unkowns does not require introducing general Gaussian elimination or Cramer's Rule. And, if you really beleive that the solution to a 2 x 2 linear system of equations is "abstract," then I would suggest that you pursue another discipline. Happy New Year. ;-). – Mark Viola Jan 02 '24 at 21:05
  • @egreg Ah. My eyes are getting worse. You did indeed write "on," not "or." And I didn't believe that you were unaware of solution of this linear system. I just think that this solution is a bit more difficult to understand than the trivial solution of a 2 x 2 linear system. That is $x$ and $y$ are easier in my mind than tracking these mythological creatures. Happy New Year! ;-) – Mark Viola Jan 02 '24 at 21:09
  • @MarkViola I just meant that to me this answer is less abstract, not that either solutions to a 2 x 2 linear system of equations are "much too comlicated and abstract". But my point was that this answer actually describes how to solve it, not magically writing down "can be solved" or $x=24$ and $y=50$. Happy New Year. ;-) – peterwhy Jan 03 '24 at 04:07
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Just in case you don't like algebra and don't think in terms of centaurs, here's yet another approach (very close to the centaur version but omitting the mythology). Suppose for a moment that all 74 heads belong to humans. Then there would be $2\times 74=148$ legs. That's 48 legs short of the specified number 196. To get that many extra legs, we have to replace some of the humans with horses. Every time we replace a human with a horse, we gain two legs, so we should do $\frac{48}2=24$ such replacements. We started with 74 humans, and we need to replace 24 of them with horses, so at the end we have 24 horses and $74-24=50$ humans.

Andreas Blass
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Hint:

Let there be $x$ humans and $y$ horses.

Each human and horse have one head. Each human has two legs and each horse has four legs.

Then, $$x+y=74$$ $$2x+4y=196$$

which can be solved.

GoodDeeds
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    Oh yeah thanks....Answer is 50 humans and 24 horses. – Prateek Apr 19 '16 at 16:03
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    This doesn't account for disabled humans, correct? In which case the system is underdetermined? – Roland Apr 19 '16 at 16:03
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    @Roland - There could just as easily be disabled horses. Or mutants with extra legs or heads. (Not likely to be any headless humans or horses, at least not if we insist on them all being alive, which also rules out anyone carrying a sack full of severed heads and/or legs.) But if you start going down that route, the problem is completely arbitrary. – Darrel Hoffman Apr 19 '16 at 17:14
  • It does indeed. I wouldn't rule out legless spectators in a horse race, but legless participants are indeed rather implausible. – Roland Apr 19 '16 at 17:32
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    @roland I was legless last time I went to the races – WW. May 25 '16 at 11:10
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    @Roland Aren't you aware of https://en.wikipedia.org/wiki/Para-equestrian? – Anne Bauval Oct 17 '23 at 16:43
  • Not sure why this isn't the accepted answer. The others refer to creatures that are not even mentioned in the OP. – Mark Viola Jan 02 '24 at 17:15