1000 Play Thinks Errata

I proofread and test solved the drafts for all the puzzles in this book. Ed Pegg Jr. slyly suggested that I maintain an errata page for the book, and that seemed like a good idea considering the number of errors that slipped through the cracks.
If you find any sort of error that is not listed here, please mail to and I will add appropriate comments to this page. All comments will be given to the publishers for consideration in future reprints. Minor editorial issues are welcome, but will not be listed here.
Please note that these comments reference the first printing, October 2001.
Thanks go to Dick Hess, Ed Pegg Jr., Joe DeVincentis, Brooke Allen, and especially Ken Duisenberg for numerous comments.
Nick Baxter
21  The second paragraph incorrectly inventories the colors of the disks; ignore this and rely on the diagram.
24  Ignore the first sentence.
82  The J at the end of the first group of letters should be a T.
162  The challenge to create 4 squares by changing 2 matches cannot be done; you must move 3 matches.
176  Rotations and reflections should be disregarded.
177  The final question is truncated; find a path the the largest number of legs, following the given rules.
188  The problem should ask for as many lines as possible.
189  The problem should ask for as many lines as possible.
Martin Gardner published a general formula from a reader that predicts the maximum number of paths for kpartite graphs. Although it worked for bipartite graphs and complete graphs, it failed for most other cases, including these three problems. I've made a stab at this, and would welcome any other suggestions of such a general formula.
219  Proceed at your own risk; the solution contains errors, and I don't know of any valid solution.
256  The rosette cannot have any "gaps". If you look at the single point in common for all the circles and draw all the radii from that point, then the angle between consecutive radii cannot be greater that 180 degrees.
301  The largest triangle should have sides of 5 units.
366  I don't believe that rotations should be ignored in this case (see solution).
399  The blank red circle in the diagram should contain the digit 4.
433  You should rearrange all seven (not six) of the hexagons, retaining the same overall arrangement. The given solution is incorrect; I believe there is no solution.
445  There are more than two solutions. I count 16; let me know if you find more than this.
552  The problem diagram implies that the consecutive digits must be increasing (since 12 is omitted from the artwork, but 10 remains). This condition should be explicit; also it's assumed that "90" are consecutive. The solution given is also incorrect (see below).
553  The digit 1 should be included in the diagram.
580  Only the distances from 1 to 6 are intended.
581  Only the distances from 1 to 12 are intended.
582  Only the distances from 1 to 21 are intended.
622  The policeman is the first to move.
664  The winner of a game is to have the marble that is closest to the given point.
669  A game not counted if either die has in illegible side face up.
706  The last word should be "rectangles".
714  The problem is misleading, and intends to pose the general question: what is the minimum number of colors needed to color any 2pire map? The diagram is also misleading, since it is not a 2pire map: the pairing of regions into 2pires is not given. Hint: at least one of the 316,234,143,225 ways of pairing the regions into 2pires will lead to the intended solution!
797  The density of the balls is constant. The volume of each set of balls is not 1 cubic meter; the intention of the question is that each set be packed into a cube that is 1 meter on edge. (Otherwise, it should be obvious that a cubic meter of steel weighs the same regardless of its shape!)
809  You must use a balance scale (and just the packages) to solve the problem. The parcels are of equal weightexcept for the single difference caused by the ring.
811  The new balls can be placed either on the left or in the middle space.
829  The last sentence is truncated. If each jogger runs at a constant 5 kph and the fly travels at 10 kph, how far does the fly travel?
831  You should bounce the ball off at least four of the six sides.
832  The reference grid is missing from the table. The full table is 3 high and 8 wide. The blocked space on the table is 2 high and three wide, and is 3 units from the left cushion.
The third problem should ask for six sides, not five.
The target pocket is not indicated in any of the problems. The problem can be solved without knowing this information; read on if you would want to know this extra information.
For the first problem, the target is the upperleft pocket; for the third problem, the target is the lowerright. For the second problem, the target pocket isn't even shownit's in the corner two units to the left of the upperright pocket. The second problem can also be solved using the upperleft pocket.
864  The question is really to find the shortest path from the left side to the right side that does not go along the top or bottom?
945  Remove "isosceles".
946  The class size should be 15, and you need to find the minimum number of boys that have all four characteristics.
954  Many of the white circles are supposed to be touching each other. Imagine that the background of the puzzle is not black, so you could actually see the black borders that are part of each white circle.
959  The orientation of the fruit on the plates does not matter, just the color.
999  The last sentence should say that cars cannot separate while the train is moving. It is impossible to exchange places and also retain the order of the cars for both trains. Do the exchange and retain the order of only the green train, or allow one train to proceed from the switch and retain the order of cars in both trains.
2  The given proof contains little detail.
Let r be the radius of R, the original circle; let a be the radius of A, the first circle. Draw circle B (with radius b) tangent to circles R and A, and with its center on the given vertical line (and distance d from the given diameter of R). We will show that the right triangle with hypotenuse d and one leg b is similar to one half of the given isosceles triangle, and thus the tangent to B is the same line as the left edge of the isosceles triangle.
Two right triangles with leg d give the following:
(a+b)^2 = d^2 + b^2
(rb)^2 = d^2 + (2ar)^2
Giving:
b=2a(ra)/(r+a), and
d=2a/(r+a) * sqrt(2r(ra))
b/d = sqrt((ra)/2r)The altitude of the isosceles triangle is sqrt(r^2a^2) and base is 2(ra). After computing the length of the other sides, we find that the sine of the base angle is the same as b/d, and the proof is complete. (Sorry for the lack of diagram.)
13  The chances of winning in the first case is 1/10. The chance of winning in the second case is 1/10  (99/100)^10, which is slightly less. If the drawing in the latter case is without replacement, then the odds are the same.
18  Brooke Allen submits the following: As the original solution points out, the gloves have "handedness". So even in darkness, you should be able to feel the difference and separate the gloves into two piles: lefthanded and righthanded. Taking 10 from one pile, you are guaranteed at least one of each color; add 1 from the other pile, you are sure to have exactly one pair.
21  Joe DeVicentis found a 3move solution: vertcw1, horcw5, vertcw3
152  First of all, there is a typo in the given solution. The last sentence should end with "...perpendicular to line AD."
For the diagram given, the proof works, but omits a few assumptions. The proof of the general problem appears in other publications, and occasionally contains the same flaws. When the line segment connecting the centers of the circles, OO', does not intersect AD, then it is not possible for both BD and CD to be diameters at the same time, thus destroying the proof's construction. Additionally, when the radius of either circle is larger than segment OO' (when the center of one circle is inside of the other), then the longest segment will be the diameter of the largest circle going through A.
161  The top red line (from leftmost point to upper rightmost point) should be cyan (it is the same length as the cyan line to the lower rightmost point) and the cyan line from lower left to upper right should be red.
177  The solution indicates that there is a seventeen leg path, but the diagram shows a solution with only sixteen. Remove leg16, and replace leg2 with three legs starting with the reverse of leg16 (then to the column between leg9 and leg10, then to the column that starts leg3).
181  The red line is correct, but the numbers don't indicate the proper order. The problem should also state that each pair of points is connected by a straight path.
187  In the solution, line8 should connect the yellow and red shells, and add line12 connecting the green fish to the red shell.
188  The solution has at least 20 lines, not 16 as indicated by the drawing.
189  Line9 is missing and should connect the dog with the other cat.
203  The second and fourth diagrams in the second row are the same. One of them should look like a regular Z.
219  The colors do not follow the advertised pattern.
331  This 19 rod solution (including the four used for the square) shown to the right was discovered by the late Andrei Khodulyov. Erich Friedman had discovered the previous unpublished best with 21 rods.
366  The problems does not state that rotations of the same arrangement should be ignored. Since a dining room table is not necessarily rotational symmetric, nor is it easily rotated, it is reasonable to assume that rotations should be counted separately. Thus the solution should be 8!, or 40320.
433  The given solution includes a number of mismatched edges. There is no solution for the given pieces.
464  The solution text does not agree with the drawing; there are four pieces.
552  The solution erroneously includes numbers with leading zeros. The correct solution is 81, with 9^n as the general solution. Personally, I'd prefer that consecutive digits could not appear in either order, and that 90 were not considered consecutive digits; this relies on more natural definitions, and is a much more interesting puzzle.
622  The policeman can catch the thief in no more than 7 moves.
629  There should be 11 glasses, not 8.
649  At a minimum, the solution should indicate that p = 1  1/2 + 1/6  1/24 + 1/120  1/720. This implies the general solution for any number of hats, and that the limit rapidly converges to 1/e.
663  There are two solutions in addition to the given diagram.
707  The simplest solution can be completed with only six "moves". Labeling the three spaces AC, and the disks 19, then the six "moves" will be: 7C, 4B, 1A, 1B, 5A, 1C.
714  The given diagram is quite clever, with each 2pire touching each of the other 11 2pires. Thus the minimum number of colors is obviously at least 12. However, as with the 4Color Map problem, it is quite a different matter to prove that this is the minimum for all possible 2pire maps.
797  The solution glosses over the issue of the relative sizes of the balls. Optimally packed balls of any size will occupy the same fraction of infinite space. However, when confined to a 1 meter cube, this no longer applies. If the balls are small compared to the cube, then the difference, if any, will be negligible. But in general, the smaller balls will pack more efficiently and thus will weigh more. And in some special cases (such as when the number of small and large balls are the same), the large balls will weigh more.
811  The solution is reversed for parts 1 and 3; the solution for 2 is show for 3; the solution for 4 is the same as for 5. For problem 9, the solution should note that there is no nonsymmetrical solution.
832  For problem 2 as stated (without specifying a pocket), Ken Duisenberg found a way to sink the ball in the upperleft pocket with seven bounces: aim the ball with an "up 1, over 5.6" angle, hitting first near the bottomright corner of the added barrier.
854  In the first line of the second paragraph, change "wheel" to "stool".
946  There must be at least 2 such boys. In general, the minimum is the sum of the number of boys which each of the individual characteristics (14+12+11+10) less 3 times the class size. If this value is negative then the minimum is zero.
961  The values for Feb and Mar are swapped.
999  The given solution has uses an illegal move (the train in move 7 stops on a diagonal track). Leaving the red car where it was on the top siding after move 6, the following are the new moves 1013:
10. red engine to top siding
11. blue engine (blue, blue, red) to right
12. blue engine (blue, blue) to left
13. red engine (red) to right.To retain the car order of both trains:
8. Blue engine (blue 1, blue 2) leave switch to the left
9. Red engine (red 1) to bottom siding
10. Red engine to top siding
11. Red engine (red 2) to left
12. Red engine to bottom siding
13. Red engine (red 1) to left
14. Red engine (red 1, red 2) leave switch to the right.
95  The third color used in the problem in the lowerleft is confusing.
115  Some shading for the cross is missing.
150  Problem should clarify that the distance is measured between trunks and that the trees are not touching.
279  The wording seems to indicate that the slack should be taken out by pulling at a single point. The diagram indicates the proper interpretation. Earth should be capitalized.
382  The pictures are of chimpanzees, not monkeys.
564  The addition diagram on the right should be 30+x=z.
683  A simpler explanation is that whatever shows on the first die, the second die has a 50% chance of matching its parity.
688  The answer should be presented first, then followed by the lengthy discussion.
811  For some problems balls are placed on the left side even though the text indicated that this should not be. Problems 4 and 5 are the same. The solution diagrams for 2 and 9 show extraneous balls below the balance.
839  nexttolast word in nexttolast paragraph should be "same".
853  "tries" should be removed.
906  Note that exactly 10 mirrors should be moved.
954  Redraw problem diagram to match the style of the example diagram or the solution diagram.
959  The blue fruit is printed as black.
987  The drawing for configuration 168 is missing, and should appear in the fifth position of the third row (replacing the duplicate 136). The fourth row are the three nonright triangles.
995  For day 2, replace bird 2 with bird 4.
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