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Answer 1

There are 49 dogs signed up to compete in the dog show. There are 36 more small dogs than large dogs signed up to compete. How many small dogs are signed up to compete?

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This is, of course, a bit hoary, but there’s actually a lesson to be learned here.

Many people have, of course, noted that a direct reading of this problem leads to the conclusion that there are 42½ small dogs and thus 6½ large dogs, which is seemingly untenable. However, there are ways to resolve this apparent inconsistency within the constraints of the problem. You may not be able to get an exact answer, but you will be able to establish constraints on the range of possible answers, which is better than simply saying “there is no feasible way to satisfy the requirements of this problem”. Both of these rely on rejecting the unstated assumption that “small dogs” and “large dogs” form a complete partition of the set of “all dogs registered”. There are thus two obvious ways out, as it were, and they can be used separately or in combination to generate a fairly large, but both limited and computable number of, solutions that comply with the constraints given in this problem.

One is to entertain the possibility that at least one dog is registered in both the small and large categories. Thus, the sum of “small dogs” and “large dogs” is greater than 49, because the categories overlap. 42 small-only dogs, 1 small and large dog, and 6 large-only dog satisfies the problem: there are 49 dogs, 43 of which are small and 7 of which are large, satisfying the constraints. Similarly, you can have 41 small-only, 3 small and large, and 5 large-only, and so on up to 37 small-only, 11 small-and-large, and 1 large-only. An even number of small-and-large dogs is prohibited because it leads to hemidoggery, and having more than 11 small-and-large dogs forces the number of large-only dogs to be negative. Thus, this assumption leaves us with six solutions.

The other is to entertain the possibility that there is a third category of dogs, which we will call “dogs of an unusual size”, which are neither large nor small. In this case, there are 7 possible solutions, the first of which is 36 small, 0 large, and 13 dogs of an unusual size and the last of which is 42 small, 6 large, and 1 dog of an unusual size. Again, even numbers of dogs of an unusual size leads to hemidoggery, and having more than 13 dogs of an unusual size leads to a negative number of large dogs. This adds seven more solutions.

Finally, you can entertain both possibilities at the same time. As long as the number of small-and-large dogs is the even and the number of dogs of an unusual size is odd, or vice versa, and neither is too large, it is possible to find a combination that satisfies the constraints. I’m too lazy to work out the exact number of solutions here, especially since you also have to consider the possibility of a dog that is simultaneously small, large, and of unusual size, or is small and of unusual size but not large, or large and of unusual size but not small. And while this is doable, its exceeds the amount of effort I want to put out on writing this answer. And so the total number of possible solutions, and the exact number of dogs of bewildering sizes involved in each, I leave as an exercise for the reader.

My point, of course, is that the problem is solvable if you reject the unstated assumption that every dog is exactly one of “large” and “small”. And any decent mathematician would question this assumption.

Step-by-step explanation:

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Answer 2

Answer:

Here is the answer mate....

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