Nitpicker's Guide to Martin Gardner's "aha!" books

Books Nitpicked:

• Aha! Insight (the blue book)
• Aha! Gotcha: Paradoxes To Puzzle and Delight (the red book)

• page 9: Susan doesn't consider the chance of intra-route collision. Even if Stinky doesn't walk with her all the way, he could "run into" her somewhere along the path.
• page 12: If you look closely, the name tags read "Amy, Bob, Clair, Don". These are gender-specific names. Gardner does not mention the different problem of how many ways the tags could be mixed up provided that each baby wears a tag whose name matches the baby's gender. In this situation, that would be a more realistic problem. Gardner could have avoided this by using non-gender-specific names (Pat, Terry, Alex, etc...) or by using last names instead of first names.
• page 21: Quibble could also have had two rats. Gardner's equations are almost correct. The last equation should read: n=x+y+z+w, where 'w' represents the number of pets that are NOT parrots, dogs, and cats. His commentary is also correct, except when he says, "numbers of animals are usually given in positive integers". Professor Quibble (if not Gardner) would correct this to "nonnegative integers". While it may be true no one has fractional pets, plenty of families have "0 dogs" or "0 pets". If Gardner had noted this, his equations would have given the simpler solution as well.
• page 43: "On a ... Mercator projection, [a loxodrome] is ... a straight line". This is incorrect for the standard definition of "straight line". Consider the slope of the line. At the equator, lines of latitude and longitude are approximately equal-- that is, traveling 70 miles east is an increase of 1 degree of longitude, while traveling 70 miles north is an increase of 1 degree of latitude. Thus, the slope of the loxodrome is approximately one. Near the poles, traveling 70 miles north still results in a change of 1 degree of latitude. However, because the longitude lines are closer together, traveling 70 miles east may result in a change of several degrees longitude (let's say 10). The slope of the loxodrome is therefore 1/10 at this point. A curve that changes slope cannot be a line. Perhaps Gardner meant a "continuous differentiable curve", but that's hardly a line. Also, if a loxodrome really was a straight line, we could start at the equator, at 0 degrees longitude, and travel NE until we came to 90 degrees north, 90 degrees east longitude. This path would not "strangle" the north pole as Gardner claims. Thus, the loxodrome cannot be a straight line. [Note: this argument does NOT apply if Mercator maps are drawn such that higher lines of latitude are more "spaced out" than lower ones]
• page 44: "[A]fter the four turtles meet at the square's center, ... they crawl outwards, each moving directly away from the turtle on it's left". This motion is not well-defined. Since the turtles are points, they coincide at the square's center. For one point to move away from a coincident point is like asking one point to move away from itself. Any direction would count as moving away.
• page 75: Bob could also cheat and ask Helen the one question, "Yes or No: If your answer to this question is yes, you will tell me your phone number in the next 30 seconds?". If Helen answers "no", the if-then statement fails, which means the antecedent is true and Helen's answer to the question is "yes", contradiction. Therefore, Helen cannot honestly answer "no", and must answer "yes". Then, since her answer is yes, she must tell Bob her phone number in the next 30 seconds. Of course, Helen could argue that Bob's question was not semantically valid since it was self-referent.
• page 80: If the Scotsman had nested pockets (one pocket inside another), the problem has several solutions.
• page 93: The Paul, John and George puzzle is not particularly hard and does not require a matrix. Note the guitarist and the pianist know each other (they're working together), and the drummer knows them both (he admires them). Therefore, the guitarist and pianist are known by everyone. Statement 3, "George has never heard of John", proves John is the drummer, since he's the only one who could be unknown to anyone. Statement 1 and 2 show the pianist earns more than the drummer (John) while Paul earns less then John. Thus, Paul can't be the pianist, making him the guitarist, and leaving George as the pianist.
• page 95: Jane's uncle "reads" in pitch dark because he is blind and "reading" Braille. The primary dictionary definition of read is "look at and understand written or printed matter". Although the other definitions seem to allow for Braille reading, this is an unusually poor stretch of a word for a puzzle. To me, feeling Braille is not reading.
• page 95: If Henry's aunt is really a midget, and knows she's going to be on an elevator, she would probably carry a walking cane or an umbrella or the like precisely so she won't have this problem. Additionally, she could ask someone else in the elevator to push the button for her. The elevator may have been a problem for her the first time, but she could easily adapt. The only conclusion that can be drawn is that the Aunt enjoys walking up five flights of stairs, possibly because she wants the exercise. Of course, if this is the case, she could just walk up the entire flight of stairs.
• page 97: The manager tied himself to a beam by standing on an ice cube large enough to support him and waiting for it to evaporate overnight. Is overnight really long enough for a large piece of ice to not only melt, but also evaporate? Wouldn't there be tell-tale signs of watery pools on the floor? At the very least, wouldn't there be heavy condensation as the humidity increased?
• page 99: Ach points out you can tie two ropes together using the "loop handles" of a scissors. This assumes, however, we're talking about a traditional pair of scissors. "Scissors" is a functionally-defined word-- anything that can cut paper (or whatever) qualifies as "scissors". There is no guarantee such an object will have "loop handles"-- in fact, I've seen pairs of scissors that do not have these. Ach feel into the trap of "uniform fixedness", assuming all scissors look the same, without realizing "scissors" refers to anything with a cutting function.
• page 102: John "knows" his hat is red because he concludes enough time has passed for Barbara and Mary to reason out the problem if his hat weren't red. This is highly imprecise. It's possible Barbara and/or Mary had come to a similar conclusion, but were waiting to make sure enough time had passed. It's also possible John spoke too quickly because NOT enough time had passed-- the notion of "enough time" is fuzzy.
• page 107: The solution to Problem 8 is to push the cork into the bottle. This only works if the cork is narrower (or equally narrow) at the top as at the bottom. If the cork is thicker at the top (which some are), and the cork is not sufficiently compressible, this trick won't work.
• page 107: Typography makes Problem 2 trivial. Gardner wants us to confuse "under water" with "underwater", but has to be honest and types it the first way, with an obvious space between the two words. Of course, when given verbally, the problem is more challenging.
• page 179: In Gardner's solution to "Square Family: Straight and Equal", his treatment of 1 adjacent to 0 is inconsistent. In one case he splits a 1,0 pair into two separate regions. In other cases, he treats the 1,0 pair as the number ten. If the 1,0 pair is treated as ten, then the 5,5 pair (near the lower left) should be treated as fifty-five and the 4,6 pair (right center) should be treated as forty-six. If the 1,0 pair is treated as the separate numbers one and zero, the sums in two sections add up only to 1+0 or one, instead of ten.

Company paperback) [the red book]
• page 18: Gardner claims that "[s]omewhere on the dull list is the dullest person in the world." This is not necessarily true. It's quite possible that all dull people are equally dull. In fact, if you could differentiate levels of dullness, the least dull person on the dull list might also be interesting. In fact, the boundary between the "dullest interesting person" and the "most interesting dull person" would be hazy. Bechenbach's "proof" succeeds because there is always a *smallest* uninteresting positive integer, not a "dullest". Gardner could have used any distinguishable measurable quantity to make his proof work (tallest, shortest, heaviest, lightest, oldest, youngest, etc..).
• page 47: The curious will is not followed. The eldest son got 6/11, not 1/2 of the cars, for example. Gardner's solution only guarantees that each son got the correct proportion of cars, not the correct amount. Of course, the lawyer's will only summed to 11/12 of his cars, which means the lawyer was either not in "sound mind" when he wrote the will or he intended 1/12 of the cars to go elsewhere.
• page 61: Gardner goes through a lot of trouble to explain about mirrors reversing things. In actuality, mirrors reverse nothing at all. If you put your left hand up to a mirror, you will see it reflected right there, to the left. However, when you meet other people, things are reversed. Their left hand is opposite your left hand; if you face north, they face south; if you can read their T-shirt, they see it reversed (if they look down). In other words, mirrors reverse nothing, while face-to-face meetings reverse everything.
• page 64: "Mr. Randi, the world famous magician", might be a reference to James Randi, noted magician and debunker of paranormal claims.
• page 91: The stated problem is ambiguous. If you consider only three possibilities (0-4 split, 3-1 split and 2-2 split), then it's true that 3-1 is most likely. However, if you consider five possibilities (0girls-4boys, 1girl-3boys, 2girls-2boys, 3girls-1boy, 4girls-0boys), then then 2-2 split is most likely. Until the very end of the problem, it is impossible to know which interpretation Mr. Katz had in mind.
• page 108: Gardner states "a penny is geometrically symmetrical... [and] physically symmetrical ... having ... uniform density". This is false. The design on each side of the coin is different (otherwise, we wouldn't be able to tell 'heads' from 'tails'), making it non-symmetrical. You could easily distinguish one side of an edge-wise cut coin from another. The difference in design may also affect physical symmetry, but it's enough to note that U.S. pennies were sometimes composed of one metal plated by another. Since the plating has different density than the rest of the coin, the coins are not physically symmetrical either.
• page 121: Gardner assumes each birthday is equally likely. This is not quite accurate, but a reasonable approximation. It should be noted, however, that non-uniform distribution of birthdays increases the chance of 2 identical birthdays in a group.
• page 123: The planets are listed in order of average distance from the Sun. From 1979 to 1999, however, Pluto is closer to the Sun than Neptune.
• page 124: Stars do not clump in groups called constellations. The constellation boundaries were determined by the International Astronomical Union, and are fairly arbitrary. Many other stargazers through the years have chosen different constellation boundaries. Further, some constellations are very sparse, not forming a group at all, and some star groupings transcend constellation boundaries. The most dense grouping of stars is the pale-white band of the Milky Way, which meanders through several constellations. Finally, Gardner is discussing random clumping. The stars are not randomly spaced!

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