An innovative algebraic approach

Usually, when faced with a word problem, I take the most obvious approach and call it done. But then, sometimes, I read of an alternative approach that makes me go "Whoa." This is one of those times.

Here's the problem:

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?

The standard approach is to note that (most) humans have one head and two legs, while (most) horses have one head and four legs. You can set up two equations, using $h$ for humans and $e$ for equines:

$h + e = 74$
$2h + 4e = 196$

And solve them simultaneously -- dividing the second by two gives $h + 2e = 98$, and it's clear that $e = 24$ and $h=50$.

User egreg on Math.StackExchange offered a different method, the one that made me go "Whoa.". I'll quote it here:

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 $\frac{74 \times 6}{2}=222$ legs, you have $\frac{222−196}{2}=13$ 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.

What a bizarre, yet brilliant insight! By making up mythical animals, some of which have a negative number of legs, the answer pretty much drops out straight away! But how did egreg come up with it?

Well, I'm not party to the inner workings of other people's brains, or else I'd be a killer poker player. However, I can have a good stab at explaining a plausible thought process: supposing there are equal numbers of horses and humans (37 each), how many legs would we be out by? We'd have 26 legs too many. That means we've got 13 horses too many - so there are 24 horses and 50 humans.

But I think we can all agree, the idea of an anticentaur is a much defter touch than simply halving the error in the number of legs.


Colin is a Weymouth maths tutor, author of several Maths For Dummies books and A-level maths guides. He started Flying Colours Maths in 2008. He lives with an espresso pot and nothing to prove.


6 comments on “An innovative algebraic approach

  • Fred Terracina

    At this point it it is not obvious to me that it is clear that the solution to

    h+2e = 98

    works out to h=50 and e=24

    I hope reading Dummies for Mathsand algebra helps me to correct this deficiency.

    I am 74 yrs old and have decided re acquainting myself with maths will stimulate or refire some of my neurons Thank you for your so far ( Only read one chapter of D4Maths. Great fun so fat.

    • Colin

      Hi, Fred, good name! (My younger son is called Fred, too.) Glad you’re enjoying the book!

      There are two equations here: $h+e=74$ and $h+2e=98$. The difference between those on the left is $e$, and on the right is $24$, so $e=24$. From there, $h$ has to be 50.

      I hope that helps!

      • fred terracina⚡

        Ah you fired a neuron thanks

  • Toby Bailey

    Hypercentaurs seem a bit complicated. But you can easily do it in your head as follows.

    If all the creatures were humans you would have 2×74=148 legs. We have 196-148=48 extra legs here and that means 24 of the creatures must be horses.

    • Colin

      Also nice :o) I agree that the hypercentaur is a less-than-obvious approach, but I rather like the weirdness of it!

  • Stephen Cavadino

    A brilliantly bizarre solution. Thanks for sharing Colin.

Leave a Reply

Your email address will not be published. Required fields are marked *

This site uses Akismet to reduce spam. Learn how your comment data is processed.

Sign up for the Sum Comfort newsletter and get a free e-book of mathematical quotations.

No spam ever, obviously.

Where do you teach?

I teach in my home in Abbotsbury Road, Weymouth.

It's a 15-minute walk from Weymouth station, and it's on bus routes 3, 8 and X53. On-road parking is available nearby.

On twitter