I have 12 animals (rabbits and ducks) loose in my barnyard. How many rabbits and ducks are there if I have counted 34 legs?

My answer to a question on Quora: I have 12 animals (rabbits and ducks) loose in my barnyard. How many rabbits and ducks are there if I have counted 34 legs?

There is a classical solution which uses only arithmetic.

  • Observe that rabbits and ducks have different numbers of legs.
  • Make all animals equal in a way that allows counting: at a pet shop, buy sufficient number of boots, \(1\) pair for each duck and \(2\) pairs for each rabbit, ask them put the boots on, and then ask each rabbit to return to you \(2\) of its boots, so that each creature gets exactly \(2\) boots.
  • Ask a question: How many boots are left? It is easy: 12 animals with two boots each have \(12\times 2 = 24\) boots.
  • How many boots were removed? \(34 – 24 = 10\) boots.
  • From how many rabbits boots were removed? 2 boots from a rabbit means \(10 \div 2 = 5\) rabbits.
  • How many ducks are in the barnyard? \(12 – 5 = 7\).
  • As simple as that. But do not forget to return the boots to rabbits.

A comment for a teacher (if by any chance a teacher reads this reply): in this my solution, I am trying to demonstrate Igor Arnold’s characterisation of arithmetic:

The difference between the “arithmetic” approach to solving problems and the algebraic one is, primarily the need to make a concrete and sensible interpretation of all the values which are used and/or which appear at any stage of the discourse.

I was also using the classical old “questions method” for solving word problems; you many find its discussion in my paper A. V. Borovik, Economy of thought: a neglected principle of mathematics education, in Simplicity: Ideals of Practice in Mathematics and the Arts (R. Kossak and Ph. Ording, eds.). Springer, 2017, pp. 241 – 265. DOI 10.1007/978-3-319-53385-8_18. ISBN 978-3-319-53383-4. A pre-publication version (without editorial changes made by publishers): bit.ly/293orpk

After I published the answer, I found a simpler solution:

Give to each animal \(4\) boots, and then ask ducks to return unnecessary (excessive) boots. The calculation now is

\( 4 \times 12 = 48\) boots all together,

\(48–34 = 14\) excessive boots,

\(14 \div 2 =7\) ducks,

\(12 – 7 = 5\) rabbits.

The moral of this story: generosity pays.