Frank Morgan's Math Chat


August 17, 2000

Magician's Solution From Math Olympiad

MathFest 2000 of our Mathematical Association of America attracted over 970 registrants to Los Angeles August 2-5. In one invited address, Jeffrey Lagarias of AT&T described the notorious 3x+1 Conjecture. Start with any positive integer. If it is odd, triple and add 1. If it is even, divide by two. Continue. The Conjecture says you will eventually reach 1, no matter what integer you start with. Sometimes it takes a while (try 27). The conjecture has been verified up to about 1017.

Questionable Mathematics. Eric Brahinsky sends in this clipping from the San Antonio Express-News (June 1):

"All of the air around the earth weighs a breathtaking 5 3/4 quadrillion tons, or

11,500,000,000,000,000,000 pounds.

But the earth, itself, weighs ... 6.59 sextillion tons, or

13,180,000,000,000,000,000,000,000 pounds!! ...

That makes the combined weight of our atmosphere and earth

13,180,011,500,000,000,000,000,000 pounds...."

Brahinsky continues: "The author did his arithmetic correctly but betrayed a basic misunderstanding of the science of measurement by mishandling significant digits. All those 0s after the initial 1318 in the earth's weight are not precisely determined, so one should not consider them as true 0s when figuring a sum. The combined weight of the earth and air should just be given as

13,180,000,000,000,000,000,000,000 pounds.

This also reminds me of an old joke: A guide at a science museum asks her tour group, 'Do any of you know how old these dinosaur bones are?' One man calls out, 'Seventy-five million and seven years!' The puzzled guide asks, 'How did you come up with a number like that?' He explains, 'Well, I was here seven years ago, and the guide then said they were 75 million years old.'"

Math Chat invites readers to submit further examples of questionable mathematics.

Old Challenge (from the 2000 International Math Olympiad).

A magician has one hundred cards numbered 1 to 100. He puts them into three boxes, a red one, a white one and a blue one, so that each box contains at least one card. A member of the audience selects two of the three boxes, chooses one card from each and announces the sum of the numbers on the chosen cards. Given this sum, the magician identifies the box from which no card has been chosen. How many ways are there to put all the cards into the boxes so that this trick always works? (Two ways are considered different if at least one card is put into a different box.)

Answer (Al Zimmermann). The answer is 12. There are two basic approaches.

One approach is to put card number 1 into one box, cards 2-99 in a second box, and card 100 into a third box. If the sum is 100 or less, the cards came from the first two boxes. If the sum is 101, the cards came from boxes 1 and 3. If the sum is greater than 101, the cards came from the last two boxes.

The second approach is to put all the cards divisible by 3 into the first box, those with remainder 1 (modulo 3) into another box, and the rest into the third box. When the sum is announced, the remainder of that sum when divided by 3 indicates which box was not used.

Since there are three boxes, each of these two approaches can be implemented in 3! = 6 different ways (3 choices for the color of the first box times 2 choices for the color of the second box). So the total number of ways is 12.

Correct solutions also by the Brahinskys, Joseph DeVincentis, Elliot Kearsley, and Arthur Pasternak.

New Challenge (Al Zimmermann). For some reason you need a standard fair die (with the numbers from 1 to 6 on the six faces). All you have is an unfair "loaded" die. How can you use the loaded die to fairly choose random numbers from 1 to 6?

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Copyright 2000, Frank Morgan.