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?

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    $\begingroup$ There is no effort to solve in the question! $\endgroup$ – Santropedro Mar 3 '17 at 16:05

A hypercentaur is a creature with two heads and six legs; an anticentaur is a creature with no head and two negative 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.

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    $\begingroup$ This is officially one of my favorite answers. $\endgroup$ – pjs36 Apr 19 '16 at 16:12
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    $\begingroup$ And a surrealcentaur is a creature with a non-standard number of heads and an imaginary number of legs. :-) $\endgroup$ – Asaf Karagila Apr 19 '16 at 16:13
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    $\begingroup$ @pjs36 I was inspired by a text by Elio Pagliarani, "La merce esclusaā€¯, 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 $\endgroup$ – egreg Apr 19 '16 at 16:22
  • $\begingroup$ math.stackexchange.com/questions/478212/… $\endgroup$ – Artificial Stupidity Aug 19 '17 at 17:03
  • $\begingroup$ I'm going to show your solution to everyone I know, because I love it! It's awesome! You made my day! Thank you! $\endgroup$ – user729424 Dec 6 '19 at 19:15

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.

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    $\begingroup$ Nice, but no fun. ;-) +1 $\endgroup$ – egreg Apr 19 '16 at 16:31
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    $\begingroup$ The classical pre-algebra approach, with a Chinese name that translates roughly as "too much and not enough." Versions of this were taught, under various names, in various civilizations. $\endgroup$ – André Nicolas Apr 19 '16 at 16:36


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.

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    $\begingroup$ Oh yeah thanks....Answer is 50 humans and 24 horses. $\endgroup$ – Prateek Apr 19 '16 at 16:03
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    $\begingroup$ This doesn't account for disabled humans, correct? In which case the system is underdetermined? $\endgroup$ – Roland Apr 19 '16 at 16:03
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    $\begingroup$ @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. $\endgroup$ – Darrel Hoffman Apr 19 '16 at 17:14
  • $\begingroup$ It does indeed. I wouldn't rule out legless spectators in a horse race, but legless participants are indeed rather implausible. $\endgroup$ – Roland Apr 19 '16 at 17:32
  • $\begingroup$ @roland I was legless last time I went to the races $\endgroup$ – WW. May 25 '16 at 11:10

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