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This week, Coach Newton shares a couple of puzzles about adventures in buying and selling, courtesy of Alex Bellos, author of Can You Solve My Problems? Ingenious, Perplexing, and Totally Satisfying Math and Logic Puzzles.

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### Wing Nuts

At an exotic bird market sale, ostriches are \$10 each, emus are \$5 each and puffins go for \$0.50. You cannot buy any fractional birds.

How many ostriches, emus and puffins can be bought with \$100 to make 100 birds in all?

### Seven Eleven

A girl walks into a branch of a familiar chain of convenience stores. She buys four different items; their prices are in dollars and a whole number of cents. The shopkeeper asks her for \$7.11. “Weird!” she says. “I know,” replies the shopkeeper. “All I did was multiply the four prices (expressed in dollars) together.” The girl is now confused: “Shouldn’t you have added them up, and not multiplied them?” she asks. “We can add them up if you insist,” retorts the shopkeeper, “but the answer will be exactly the same.”

What are the four different prices?

## Solutions to week 78

Polychromatic Pigs. Suppose the pen contains a non-yellow pig. Then the pair of that pig with any other pig in the pen will have to include at least one yellow pig, from which we conclude that all of the other pigs in the pen are yellow. So there must be at least 17 yellow pigs in the pen.

Guest Gifts. Suppose that either guest has seven or eight doubloons on his or her dresser. Then that guest knows that nine doubloons must have been distributed, and so can subtract his or her number of doubloons from nine to correctly guess the number of doubloons on the other guest’s dresser on the first morning after they arrive. On the other hand, the other guest, having one or two doubloons, cannot have deduced the total number of doubloons nor the first guest’s total, and so says nothing on the first morning. This scenario violates the premise that both guests solve the puzzle on the same morning, so we conclude that neither guest has seven or eight doubloons. Moreover, both guests know this after the first morning, when neither guest correctly guesses.

Now suppose that either guest has one or two doubloons. After the first morning, once both guests know that neither of them has seven or eight doubloons, the guest with one or two doubloons would then know that there can’t be enough doubloons distributed to make nine (since the other has at most six, plus two is less than nine). So the guest with one or two can subtract from seven to correctly guess the other guest’s number of doubloons on the second morning, whereas the other guest (with five or six) is in the dark as to whether there are seven or nine doubloons in all, and so would not be able to guess on the second morning. So as with the first case, we conclude that not only does neither guest have one or two doubloons, but that both guests know this after neither responds on the second morning.

Continuing in this fashion, now suppose that one guest or the other has five or six doubloons. Then, after the second morning, knowing that neither guest has one or two doubloons, that guest will be sure that nine doubloons have been distributed, and so can guess the other guest’s number of doubloons on the third morning. However, the other guest with three or four doubloons will still not be able to know the total number of doubloons that have been distributed, ad so would not be able to guess on the third morning, once again violating the premise that both guests figure it out on the same morning.

Therefore, we conclude that both guests have three or four doubloons, meaning that Raymond distributed seven doubloons, and that further, both guests realize this on the fourth morning after their arrival.

## Recent Weeks

Links to all of the puzzles and solutions are on the Complete Varsity Math page.

Come back next week for answers and more puzzles.

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