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 A certain old Count had ten pets. Some were cats and the rest were dogs, but he could never remember how many of each there were. Each day he fed them biscuits – 56 in all – there it was, written on the container. This meant that each of the dogs got six biscuits, and each of the cats got five. Now, how many cats and how many dogs were there? Because this is such a simple problem, we shall make it harder by insisting that the solution not use any algebra. Trial and error is not allowed either! |
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#2 52. THE COUNT WHO COULDN’T 
| HINT 1
Perhaps try using algebra first to find the correct answer, and then look for a better way! |
| HINT 2
If you followed Hint 1, look at your equation |
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SOLUTION
If we fed five biscuits to each animal, the cats have had their share, and there are six biscuits left over. Hence there are six dogs and four cats. |
| EXTENSIONS
You can obviously produce similar problems to this by changing the numbers. The interesting aspect of this puzzle is the simple solution. Can any set of two linear equations in two variables be solved in this simple way? Or are there some constraints required? |