Mathematical Puzzles

Peter Winkler

• Editorial: A K Peters/CRC Press
• Año: 2020
• ISBN:
  9780429557972
• Edición: 1
• Categoría: Matemáticas
• Categoría: Matemáticas general

Research in mathematics is much more than solving puzzles, but most people will agree that solving puzzles is not just fun: it helps focus the mind and increases one's armory of techniques for doing mathematics. Mathematical Puzzles makes this connection explicit by isolating important mathematical methods, then using them to solve puzzles and prove a theorem.

Features

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• A collection of the world's best mathematical puzzles

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• Each chapter features a technique for solving mathematical puzzles, examples, and finally a genuine theorem of mathematics that features that technique in its proof

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• Puzzles that are entertaining, mystifying, paradoxical, and satisfying; they are not just exercises or contest problems.

1. Out for the Count

There’s nothing more basic in mathematics than counting—but at the same time, counting mathematical objects can be dauntingly difficult. We’ll start with some simple puzzles and gradually introduce some of the special tools that we sometimes need for counting.

Half Grown

At what age is the average child half the height that he or she will be as an adult?

Solution: Most people guess too high, maybe thinking that if full height is reached around age 16, then half-height should be around 8.

There are two problems with that reasoning: (1) the rate of growth for a human being is not constant, and (2) babies have a substantial head start of 20 inches or so (when you prop them up).

The right answer: two years old! (For a girl, it’s actually about 214 years, for a boy 212.)

Powers of Two

How many people are “two pairs of twins twice”?

Solution: There are four words in the phrase suggesting the number “2” and most people rightly divine that these 2’s should be multiplied, not added. So the answer is 16, right?

Not so fast—a twin is only one person, thus a pair of twins, only two. So the correct answer is 8.

Watermelons

Yesterday a thousand pounds of watermelons lay in the watermelon patch. They were 99% water, but overnight they lost moisture to evaporation and now they are only 98% water. How much do they weigh now?

Solution: 500 lb. The watermelons contain 10 pounds of solid matter that now comprises 2% of their final weight; divide 10 pounds by 0.02, and there you are. Apparently, quite a bit of evaporation took place!

Bags of Marbles

You have 15 bags. How many marbles do you need so that you can have a different number of marbles in each bag?

Solution: Counting 0 as a number, you might reasonably deduce that you should put no marbles in the first bag, 1 in the second, 2 in the third, etc., and finally 14 in the last. How many marbles is that?

The quick way to answer that is to observe that the average number of marbles in one of your 15 bags is 7. Thus the total number of marbles is 15×7=105.

But there’s a trick: You can put bags inside bags! If you put the empty first bag inside the second along with one marble, then the second inside the third along with another marble, etc., you end with the last bag containing all the marbles. So you only need 14 marbles in all.

Salaries and Raises

Wendy, Monica, and Yancey were hired at the beginning of calendar 2020.

Wendy is paid weekly, $500 per week, but gets a $5 raise each week; Monica is paid monthly, $2500 per month, and gets a $50 raise each month; Yancey is paid yearly, $50,000 per year, and gets a $1500 raise each year.

Who will make the most money in 2030?

Solution: Yancey gains $1500 per year, thus he makes $50,000+10×$1500=$65,000, 10 years later.

Monica gains 50×78(!)=$3900 per year, making $30,000+10×$39,000=$69,000 at the end of the 10th year (which is when 2030 is about to start). So Monica out-earns Yancey in 2030, without even considering the additional raises she is due that year.

Wait, why 78? Because Monica’s monthly raises add up to $50×(1+2+3+⋯+12)=$50×13×12/2=$50×78.

By similar reasoning, figuring 52 weeks in a year, Wendy gains $5×53×52/2=$6890 per year, making $26,000+10×$6890=$94,000 at end of her 10th year. Yay Wendy!

Here’s yet another chance to add up consecutive numbers.

Efficient Pizza-cutting

What’s the maximum number of pieces you can get by cutting a (round) pizza with 10 straight cuts?

Solution: There are a number of ways to tackle this one, but perhaps the easiest is to note that the nth cut can at best cross each of the n−1 previous cuts, and between each pair of crossings, split a previous piece in two. Since you also get a new piece before the first crossing and after the last one, the cut ends up adding n new pieces.

It follows that with n cuts you can create at most 1+2+⋯+n=n⁢(n+1)/2 new pieces, but remember you started with one piece (the whole pizza), so the answer is 1+n⁢(n+1)/2 pieces. For 10 slices, that works out to 56 pieces.

Wait, we haven’t actually shown that you can achieve that many pieces—that would require having every two cuts cross, with never any more than two cuts crossing at the same place. But if we just mark 2⁢n random points along the edge of the pi...