The 125 Best Brain Teasers of All Time Page 10
88.Niece. The niece of your mother is your cousin. The mother of your cousin is your aunt (even if only by marriage).
89.FOUR—(1) foul—(2) fool—(3) foot—(4) fort—(5) fore—(6) fire—FIVE.
90.Two out of three. We let B and W-1 stand respectively for the black and white marbles that might be inside the bag at the start, and W-2 for the white marble added to the bag. Removing a white marble from the bag results in three equally likely combinations of two marbles, one inside and one outside the bag:
INSIDE THE BAG OUTSIDE THE BAG
(1) W-1 W-2
(2) W-2 W-1
(3) B W-2
(1) In combination 1, the white marble drawn out is the one added to the bag (W-2), and the white marble inside it (W-1) is the piece originally there.
(2) Combination 2 is the converse of (1): the white marble drawn out is the one originally in the bag (W-1), while the white marble inside it (W-2) is the one that was added.
(3) In combination 3, the white marble drawn out is the one added to the bag (W-2), since there was no white marble originally inside it. The marble that remains in the bag is a black one (B).
In two of the three cases, Carroll observes, a white marble remains in the bag. So, the chance of drawing a white marble on the second draw is two out of three.
91.WHEAT—(1) cheat—(2) cheap—(3) cheep—(4) creep—(5) creed—(6) breed—BREAD.
92.There was never yet an uninteresting life. Each letter has been replaced with the second one before it in the alphabet sequence: T has been replaced by R, H by F, E by C, and so on.
93.3 miles per hour. This classic puzzle has a tendency to mislead solvers, probably because they assume that the average speed is simply calculated by combining the hiker’s two rates (2 mph and 6 mph = 8 mph) and then dividing by 2. They invariably come up with the incorrect answer of 4 mph. Let’s calculate the time Sarah took to hike uphill. For the sake of convenience, assume that the distance uphill (and downhill, of course) is 1 mile. You can actually use any other distance, and the reasoning and result will be the same. So, in this case, you are told that her rate uphill is 2 miles per hour. Remember that Distance = Rate × Time. Therefore, her time up is: 1 (mile) = 2 (miles per hour) × time up, which is ½. Thus, it would take Sarah ½ hour to climb a 1-mile hill. Now, given that her rate downhill is 6 miles per hour, let’s calculate her time down: 1 (mile) = 6 (miles per hour) × time down, which is . So, it would take Sarah hour to descend a 1-mile hill. Now, her total time for the entire trip is, of course, time up + time down, or hour. The total distance she covered is 2 miles—1 mile up + 1 mile down. Therefore, her overall rate is: 2 (miles) = rate ×⅔ (hours), which is 3 miles per hour. As you can see, Sarah’s overall, or average, rate of speed, which has been calculated by taking into account the overall distance she covered (2 miles) and the overall time she took (⅔ hour), is 3 miles per hour, not 4 miles per hour!
94.Grate—Great.
95.Creative—Reactive.
96.Paternal—Parental.
97.31 draws. Again, we cannot count on luck and must assume the worst-case scenario. In this case, you might get all left-foot or right-foot socks first. So, you will have drawn out 30 socks in total. At this point, the next sock you draw out (the 31st) will match one of these, since only right-foot or left-foot socks are inside and one of these will match in color one of the 30 outside the box.
98.. The probability of drawing an ace is and the probability of drawing a king is also . The probability of drawing either one or the other is the sum of these two: .
99.The boy’s father. The boy is an only child, because he says that he has no brothers or sisters. So, when he refers to “my father’s son” he is referring to himself, the only son of his father. Therefore, the son of the man in the photo (“this man’s son”) is himself (“my father’s son”). This also means, of course, that the man in the photo is his father.
100.Saturday, October 13. Set up a time chart for the week of October 1, showing that Jason (= J) works every third day and Alicia (= A) on Saturday:
WEEK OF OCTOBER 1
CLERK MON OCT. 1 TUES OCT. 2 WED OCT. 3 THURS OCT. 4 FRI OCT. 5 SAT OCT. 6 SUN OCT. 7
Jason J J J
Alicia A
Since they will apparently not be working together during that week, set up a chart for the week after:
WEEK OF OCTOBER 8
CLERK MON OCT. 8 TUES OCT. 9 WED OCT. 10 THURS OCT. 11 FRI OCT. 12 SAT OCT. 13 SUN OCT. 14
Jason J J
Alicia A
This chart shows that the two will be working together on Saturday, October 13.
LEVEL 5: GENIUS
101.(22 − 12) − 10 = 0.
102.Gossip is the opiate of the oppressed. The code used for this one is a hard one to crack. Each letter has been replaced by the digits in reverse (backward) order. For example, in the usual order Z would be replaced by 26, since it is the 26th letter of the alphabet; but with this reverse code it is replaced by 1; B would be replaced by 2 in the usual order; now it is replaced by 24, and so on. The complete code (including letters not used in the plaintext) is as follows: A = 26, B = 25, C = 24, D = 23, E = 22, F = 21, G = 20, H = 19, I = 18, J = 17, K = 16, L = 15, M = 14, N = 13, O = 12, P = 11, Q = 10, R = 9, S = 8, T = 7, U = 6, V = 5, W = 4, X = 3, Y = 2, Z = 1.
103.10 trains. Let’s say that you get on a train at the New York station at 12:00 noon. It could be at any other time of course; the reasoning will be the same. As mentioned in the puzzle, five hours later, at 5:00 PM, your train arrives at the Washington station. Now, you must envision the relative positions of the trains on their way from Washington to New York during those five hours. Keep in mind that the Washington-to-New York trains leave on the half-hour. At the Washington station at 5:00 PM, there is a train ready to leave for New York. Call it A. Obviously, the train that had left the Washington station a half hour earlier, at 4:30 PM, will find itself a certain distance from the Washington station when A is about to leave. Call that train B. You can now complete the diagram showing the relative locations of all the trains leaving the Washington station, bound for New York, between 12:00 PM and 5:00 PM as follows:
NEW YORK WASHINGTON
K J I H G F E D C B A
12:00 12:30 1:00 1:30 2:00 2:30 3:00 3:30 4:00 4:30 5:00
Now, when you left the New York station at 12:00 noon, you obviously missed passing the 12:00 K-train that had come from Washington, because it was in the station when your train was leaving. But, as you can see from the diagram above, you passed all the others—the 12:30 J-train (that is, the train that left Washington for New York at 12:30), the 1:00 I-train, the 1:30 H-train, the 2:00 G-train, the 2:30 F-train, the 3:00 E-train, the 3:30 D-train, the 4:00 C -train, the 4:30 B-train, and the 5:00 A-train. That makes 10 trains in all.
104.Option 2. After the first year with Option 1, you would receive just the $4,000. With Option 2 on the other hand, you would receive $2,000 after the first six months, but then you would get an increase of $200. So, for the last six months of that year, you would receive $2,200 dollars. Adding the two semesters up, you would get $4,200 at the end of the first year, whereas with Option 1 you would get just the $4,000. Now, what income do both options generate after year two? Well, with Option 1 you would receive an increase of $800 for the year. So, you would end up earning $4,800. But with Option 2, you would earn $2,400 the first semester—the $2,200 you would have started off with at the beginning of the year (i.e. salary from the previous semester) plus the $200 raise you would have gotten for that semester. Then, in the last six months you would get another increase of $200 on top of this new salary: That is, you would earn another $2,600 ($2,400 + $200). Adding up the two semesters, you would receive $5,000 at the end of the second year. If you continue calculating the incomes generated by the two options in this way, you would see that Option 2 generates more income in the long run, and is therefore the better option.
105.Man = truth-teller, Woman = liar. As in puzzle 49, let’s tra
nslate “Ruma” into English by logical deduction. If the woman were a truth-teller, she would admit it, so her response, “Ruma,” would mean “Yes.” If she were a liar, she would not admit it, and her response, “Ruma,” would still mean “Yes,” but would be a lie. The man clearly told the truth by saying that she said “Yes.” Therefore, he is a truth-teller. He tells Dr. Brown that his partner is a liar, which means that she is indeed a member of the liar clan.
106.Friday. Barb and Charlene give contradictory statements—namely that today is Saturday (Barb) and that today is not Saturday (Charlene). So, one statement is true and the other false. With this deduction, we have located who gave the only true statement—either Barb or Charlene. This means that the other statements were all false. Fanny’s statement implies that today is Thursday, since she says that tomorrow is Friday. That being false, we can eliminate Thursday. Emma’s statement implies that today is Tuesday, since she says that tomorrow is Wednesday. That being false, we can now also eliminate Tuesday. Dina’s statement implies that today is Sunday, since she states that the day after tomorrow (Monday) is Tuesday. So, we can eliminate Sunday, too. Alma’s statement implies that today is Saturday, since she says that yesterday was Friday. That being a lie, as well, we can also eliminate Saturday. We can now see that Barb lied, claiming that today is Saturday, so Charlene told the truth. She allows us to eliminate Monday and Wednesday from the list, as we can now see from her statement. This leaves Friday as the only possibility.
107.Runner 2. Number 1 and Number 5 came in one after the other. Number 1 was not the winner, because he or she did not end up in the spot corresponding to his number. So, he or she could have ended up in second, third, or fourth, followed right after by Number 5. Here are the three possibilities:
(1) Number 1 (2nd)—Number 5 (3rd)
(2) Number 1 (3rd)—Number 5 (4th)
(3) Number 1 (3rd)—Number 5 (5th)
Since Number 5 did not end up fifth, we can discard possibility 3. Now, we are told that Number 3 came in right after 5. So, let’s consider the possibilities now:
(1) Number 1 (2nd)—Number 5 (3rd)—Number 3 (4th)
(2) Number 1 (3rd)—Number 5 (4th)—Number 3 (5th)
Number 4 did not win the race, nor did he or she end up in the fourth spot. This leaves only the fifth spot for Number 4. No other positioning would work for him or her, given the consecutive positioning of Numbers 1, 5, and 3. This leaves Number 2 as the winner. The final order was therefore: Number 2 (1st), Number 1 (2nd), Number 5 (3rd), Number 3 (4th), Number 4 (5th).
108.M. We start by listing the number of ways to throw either a 6 or a 7:
OUTCOME: 6 OUTCOME: 7
FIRST DIE SECOND DIE FIRST DIE SECOND DIE
1 5 1 6
2 4 2 5
3 3 3 4
4 2 4 3
5 1 5 2
6 1
There are 36 possible throws of two dice, because each of the 6 faces of the first die is matched with any of the 6 faces of the second one. Of these 36 possible throws, 11 produce either a 6 or a 7 (as the table above shows). Therefore, the probability of throwing either a 6 or a 7 is .
109.Today. Before its “birth,” today does indeed have a different name—tomorrow. And when it is “laid within the tomb,” that is, when it is over, it assumes a new name—yesterday. Finally, though it lasts only one day, it changes its name three days in a row (“three days together”): from yesterday, to today, to tomorrow.
110.There are several ways to solve the Wolf-Eater puzzle, all consisting of four round trips (nine individual trips in total: four round trips and a final one over). Here’s one. The traveler must start by taking the wolf with him to the other side, leaving the Wolf-Eater with the goat and cabbage on the original side. The Wolf-Eater’s presence ensures that the goat will not eat the cabbage. Upon reaching the other bank, the traveler drops off the wolf and rows back alone. This is his first round trip. Back on the original side, he picks up the cabbage, leaving the Wolf-Eater and goat together, and rows with it to the other side. Once there, he leaves the cabbage safely with the wolf and then rows back alone. This is his second round trip. On the original side, he picks up the Wolf-Eater, leaving the goat there alone, rowing with the monster to the other bank. There he drops off the Wolf-Eater, and picks up the wolf for his trip back (so the Wolf-Eater will not eat the wolf), leaving the Wolf-Eater alone with the cabbage. This is his third round trip. Upon reaching the original side, the traveler drops off the wolf, picking up the goat. Once he reaches the other side, he leaves the goat safely with the cabbage and Wolf-Eater, who are already there. (As mentioned earlier, the Wolf-Eater’s presence ensures that the goat will not eat the cabbage.) He rows back alone, for the completion of his fourth round trip. Back on the original side, the traveler picks up the wolf, and rows back with it to the other side. He gets off the boat with the wolf, and continues his journey with all four.
111.FLOUR—(1) floor—(2) flood—(3) blood—(4) brood—(5) broad—BREAD.
112.BLACK—(1) blank—(2) blink—(3) clink—(4) chink—(5) chine—(6) whine—WHITE.
113.RIVER—(1) rover—(2) cover—(3) coves—(4) cores—(5) corns—(6) coins—(7) chins—(8) shins—(9) shine—(10) shone—SHORE.
114.1009. In Roman numerals, the number is MIX.
115.15. Houses 1–9 = 1 is used once. Houses 10–19 = 1 is used 11 times (10, 11 [twice], 12, 13, 14, 15, 16, 17, 18, 19). Houses 20–29 = 1 is used once (21). Houses 30–39 = 1 is used once (31). Houses 40–49 = 1 is used once (41). House 50 = not used. TOTAL: 1 + 11 + 1 + 1 + 1 = 15.
116.X, Y. All these letters do not change when reflected in a mirror. The letters are listed alphabetically.
117.WINTER—(1) winner—(2) wanner—(3) wander—(4) warder—(5) harder—(6) harper—(7) hamper—(8) damper—(9) damped—(10) dammed—(11) dimmed—(12) dimmer—(13) simmer—SUMMER.
118.The woman herself. Her daughter’s mother is, of course, herself.
119.17 + 13 – 30 = 0.
120.Resistance—Ancestries.
121.Deductions—Discounted.
122.11 + 11 + 1 + 1 = 24.
123.101. First, you add the digits in a number, starting with 28. The sum of its digits is 2 + 8 = 10. Now, you add the result of 10 to 28 to get the next number in the sequence, namely 38. Now, add the digits in 38, 3 + 8 = 11. Add this result to 38—38 + 11 = 49 to get the next number in the sequence. And so on.
124.Jack. Gary and Walter say the same thing—that Hank is the killer—so their statements are either both true or both false. They cannot be true because there was only one true statement in the set, so they are false. Similarly, Sam and Hank say the same thing—that Gary is the killer. Again, both statements cannot be true since there was only one true statement in the set. So, they are both false. We have now identified the four liars, who are all innocent: Gary, Walter, Sam, and Hank. This leaves Jack as the only truth-teller and thus the killer. As we can see, he did indeed tell the truth—Hank didn’t do it—but his statement changes nothing.
125.If any woman would have raised her hand, it would have meant that woman saw at least one red cross on a forehead. This did not happen. So, one of the astute women figured this out.
FURTHER READING
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There are a number of books that I consider to be the “classics.” I simply list them here alphabetically—not by rank or subject matter. Of course, you may want to add others to your personal list.
Alcuin. Problems to Train the Young, around 800 CE. See John Hadley and David Singmaster, “Problems to Sharpen the Young.” The Mathematical Gazette 76 (1992): 102-126.
This contains more than 50 puzzles, including the river-crossing examples you have solved in this book. None of the puzzles require any expert knowledge to solve them. Alcuin wrote it in an era when there was little or no interest in mathematics in Europe. Originally written in Latin, it was clearly intended as an amusement for the well-educated.
Dmitri A. Borgman. Language on Vacation, N
ew York: Scribner, 1965.
Without question, this is one of the greatest books of word puzzles and games.
Claude-Gaspard. Bachet de Mézirac, Pleasant Problems. Lyon: Gauthier-Villars, 1612.
This is the first book to organize puzzles in mathematics according to type—weighing, measuring, and so on. They are the prototypes for many puzzle anthologies today.
Lewis Carroll. Pillow Problems and a Tangled Tale. New York: Dover, 1885.
This may be the number one puzzle book of all time. It contains 72 mathematical posers ranging from those that can be solved by simple mathematical know-how to those that require more advanced or even expert knowledge. The puzzles were originally printed as a monthly magazine serial, and many readers sent in solutions to those posed in it. A number of these puzzles have been included in this book.
Henry E. Dudeney, The Canterbury Puzzles. New York: Dover, Originally 1907.
Published in 1907, this is Dudeney’s masterpiece, in which he designs his 114 puzzles on characters from Chaucer’s Tales.
The book is suitable for young enthusiasts, mathematicians, and veteran puzzlers alike. It is, overall, quite challenging.
Henry E. Dudeney, 536 Puzzles and Curious Problems, edited by Martin Gardner. New York: Dover, 2007.
For two decades, Dudeney wrote a puzzle column, “Perplexities,” for The Strand Magazine. Martin Gardner hailed Dudeney as “England’s greatest maker of puzzles,” unsurpassed in the challenge his inventions continue to provide.
Henry E. Dudeney. Amusements in Mathematics, New York: Dover, 1958.
These 430 puzzles, problems, paradoxes, and brain teasers are particularly inspired. They span all genres—numbers, unicursal and route problems, counter puzzles, speed problems, measuring, weighing, packing, clock puzzles, and many more. Chessboard problems, involving the dissection of the board or the placement or movement of pieces, age and kinship problems are also included here.