Personal Accounts and Folded Paper
Old challenge
Send in a true personal account with an estimate of its mathematical probability of occurring. The least likely will be the winner.
Answers
In the second-best answer, Steve Gluck reports that, "In a seemingly irresolvable dispute between my son and daughter, I proposed we flip a coin. My son said coin flipping was unfair because his sister always won. [After he chose tails,] 13 consecutive flips were heads. The probability of such an event was 1 in 213 or 8,192." Actually, since such opportunities might have arisen, say, on 8 occasions, Math Chat puts the probability at about 1 in 1,000.
In the best answer, Fred Wedemeier reports that his grandfather lost his wedding ring working in the field on the family farm. One day Fred's father got down off the tractor and found the ring stuck in a small crack in one of the tractor's tires.
Fred estimated that the probability that the crack on the tire would hit the ring in the field and that his father would get off at that time and find it to be 1 in 1022. Math Chat estimates at least a 1 in 1,000 probability that the tire would catch the ring some time over the years and a 1 in 10 probability someone would notice, for an overall probability of about 1 in 10,000.
Charles Sullivan writes humorously, "I just flipped a coin 30 times, and got this sequence of heads and tails: hthhhthhhtthttthhthhhtththhhth. Amazing! The chances of getting that particular sequence are less than 1 in a billion! And yet it happened to me, sitting right here in my office, with an ordinary penny (1966). And on the first try!" The humor, of course, lies in the fact that there is nothing special about this sequence. And the probability of getting some sequence is 100 percent.
Folding paper
Brian Ford wants to fold a large sheet of paper 1 millimeter (about 1/25 inch) thick in half 50 times, and asks whether the height of the folded stack is (1) less than a foot, (2) about 21 feet, or (3) higher than the Sears Tower. The answer is (3)! Indeed, its height would be 250 millimeters, or about 10 million miles, about a 10th of the way to the sun.
'Flatland'
Tarah Smith Nellis writes: "Math Chat has been an intriguing, thought-provoking addition. I borrowed a copy [of the book award 'Flatland'] from a library and was initially delighted by the author's cleverness. However, about 20 or so pages into the novel, I was dismayed to encounter a severe put-down of the intelligence of women."
Math Chat replies: Actually, as the second of our award books, "The Mathematical Tourist" by Ivars Peterson, explains, "Flatland is a sharply delineated satire, which reflected widely debated social issues in Victorian Britain. Abbott [the author] was a strong advocate of women's rights, and he couldn't resist taking a satirical swipe at his society's attitudes toward women." Indeed, one edition of "Flatland" announces on the front cover: "Humour, satire, logic, all combined in a brilliantly entertaining classic of the fourth dimension," and on the back cover: "In a world where women [are mere] needles ... and men are various polygons who recognize each other by their angles (from the wedge-like soldiers to the perfect circle), social satire of Victorian mores is hilariously sharp and potent."
I like to use "Flatland" as a lead-in to another award book, "The Boy who Reversed Himself," by William Slater, a modern children's book that gives a fascinating speculative account of all higher dimensions, and in which the starring character is a girl. Mathematics is for everyone.
Old challenge on towers of 5s (from June 27)
We sometimes use the symbol ^ to denote powers, so that 5^999,000 denotes 5x5x...x5 (999,000 times), 5^5 = 55 = 3125, and 5^5^5 denotes 5^ (5^5) = 53125 (not 31255). William Foster asks for the remainder after dividing by 7 of 5^5^ ... ^5 (999,000 times).
Answer
Michael Stern used the computer program Mathematica to compute that 5^5, 5^5^5, and 5^5^5^5 all leave remainder 3 when divided by 7, and surmised correctly that the simple pattern continues. Neal Castagnoli, Martin Hildebrand, and Howard Sheldon explained why. Consider the example 5^5^5^5.
Since as discussed June 27, powers of 5 modulo 7 repeat in cycles of 6, we need to consider 5^5^5 modulo 6. In a similar fashion, powers of 5 modulo 6 repeat in cycles of 2. Since any power of 5 is odd, therefore, modulo 6, 5^(5^5) = 5^1 = 5. Finally, modulo 7, 5^(5^5^5) = 5^5 = 3125 = 3.
New challenge
A recent National Public Radio Weekend Edition puzzle asked to replace one letter in HOMEGAME with another letter to get the name of a famous philosopher. Bruce Cobi asks how many different letter sequences you can obtain this way. What if you are also allowed to rearrange the letters?
(The NPR answer was none of these, but to replace the Greek letter OMEGA with U to get HUME. )
* To be eligible for 'Flatland' and other book awards, send answers and new questions to:
Math Chat
Fine Hall, Washington Road Princeton, NJ 08540
or by e-mail to:
Frank.Morgan@williams.edu
