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A town has 2017 houses. Of these 2017 houses, 1820 have a dog, 1651 have a cat, and 1182 have a turtle. If x is the largest possible number of houses that have a dog, a cat, and a turtle, and y is the smallest possible number of houses that have a dog, a cat, and a turtle, then what is the value of x−y.

Derive Foiler
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  • I already know x is 1182 because there is a Venn diagram that can be made with x = 1182. But I have no idea how to get the minimum. I've seen the solution to PASCAL 2017 QUESTION 24. But their solution is confusing. – Derive Foiler Aug 10 '17 at 20:39
  • @amWhy But the maximum number of turtles is 1182!!! – Math Lover Aug 10 '17 at 20:45
  • Yes @Shuri2060 you are right, and so are you, MathLover! and DeriveFoiler about the max. I must not have been reading carefully. Apologies for any confusion. – amWhy Aug 10 '17 at 20:54

2 Answers2

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You already figured out $x=1182$. Now for computing $y$, we take the following strategy.

Without loss of generality, we presume each of the houses 1 to 1820 has a dog. To minimize the 'overlap', each of the remaining houses (1821 to 2017 i.e. 197 houses) contains both a cat and a turtle. So we are left with $1651-197=1454$ cats and $1182-197=985$ turtles.

Now the question is to find the minimum number of houses out of the first $1820$ houses that have both a cat and a turtle. It is very easy to figure out that it is $1454+985-1820=619$. So $y=619$.

Math Lover
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The trick is to think if the set of houses with all kinds of pets isn't maximal or minimal what we can do to maximize or minimize it and what result that will be.

If there are any houses with only a turtle in it, or only a turtle and one other pet it int we can add the other pets from houses with pets but not turtle. (As there are more dogs and cats than turtles there will always be houses with dogs and houses with cats that don't have turtles). So in order to be maximal all the turtles must be in houses that have all pets. And the maximal number of houses with all pets must be the same as the houses with turtle: $118$2.

If there are any empty houses or houses with only one type of pet, and if there are any houses with all types of pets, we can add a pet from a house with all pets. That way we reduce the number of houses with all types of pets by one. So in order to be minimal either there must be no houses with fewer than one pet , or there are no houses with all types of pets.

If there are no houses with fewer than one pet, let $dc, dt,ct all$ be the number of hoses with dogs and cats, dogs and turtles, cats and turtles, and all three:

Then $dc+ dt+ct+all = 2017; dc+dt +all = 1820; dc + ct + all = 1651; dt + ct + all = 1182$. Solving those we get $ct = 197; DC = 366; dt= 835; all = 619$.

If there are no houses with all types of pets, on the other hand, then $dc + dt \ge 1820; dc + ct \ge 1651; dt+ct \ge 1182$ so $dc + dt + ct \ge \frac {1820+1651+1182}2 > 2017 \ge dc + dt + ct$ so this is clearly not possible.

So the number of houses with all three types of pets is between $619$ and $1182$.

fleablood
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