Puzzles Archive
This is a list of the previous puzzles that have been sent out by E-mail.
Don't forget to signup for "The MindBender" here.
To see the answer, click and hold your mouse button just to the right
of the red arrow 
and drag down. This will highlight
the answer and make it visible.
JULY 13, 1998
As I went over the London bridge,
I met my sister Jenny,
I broke her neck and drank her blood
and left her standing empty.
Tell me, who was my sister?
...This was taken from Larry Niven and Steven Barnes' wonderful book,
"Dream Park". It is in the chapter called "Neck Riddles" where the characters are in a riddle contest with a god, and their lives are in the balance.
The answer was "A bottle of gin". The questioner must have been a sloppy drinker.... I never have to break the neck off a bottle, and I never leave the bottle lying in the street.... tacky!
JULY 14, 1998
What goes through the door without pinching itself?
What sits on the stove without being burned?
What sits on the table and is not ashamed?
...That would be the sun, or more completely, the sun's light.
This was also taken from Larry Niven and Steven Barnes' wonderful book, "Dream Park". It is in the chapter called "Neck Riddles"
JULY 15, 1998
What work is it that the faster you work,
the longer it is before you're done,
and the slower you work,
the sooner you're finished?
...Another riddle from "Dream Park".... and the answer is "Cooking meat
on a spit". If you turn the meat too quickly, it won't heat enough
to cook. If you turn it more slowly, it will cook more quickly.
JULY 16,1998
"I know a word of letters three,
add two and fewer there will be."
So, what's the word? (And for the benefit of aging Beatles fans, a hint -- the word is not "Love".
...The word was, of course, "few".
July 17, 1998
Here is a simple but pretty problem which developed at a recent
election where 5,219 votes were cast for four canidates. The victor exceeded his opponents by 22, 30, and 73 votes, yet not one of them knew how to figure the exact number of votes received by each.
Can you give a simple rule for obtaining the desired information?
...The answer is to add the pluaralities to the total vote and divide by
the number of canidates. The quotient will be the number of votes received by the winning canidate. From that, the number of votes of the other canidates can be ascertained by subtraction. The counts were 1,336, 1,314, 1306, and 1,263.
July 19, 1998
It is told that two men pooled their savings and bought a grindstone. Because they lived some distance apart, they agreed that the elder man should keep the grindstone until he had reduced it's size by one half, then it should be turned over to the other.
The grindstone started out exactly 22 inches in diameter, with a 3 and 1/7 inch hole in the center for the shaft. What would be the diameter of the stone when it is turned over to the second man?
...The "trick" that a few people missed out on is that the use of the grinding wheel is based upon it's volume, which is proportional to the area of the circle that is either side of the grindwheel, not the diameter of that circle.
Of course, the area of the hub doesn't count, any more than the hole in a doughnut provides nutrition.
So, the usable area of the wheel is the area of the whole wheel, less the area of the hub.
Or, (pi * (22/2)^2) - (pi * ((3 1/7)/2) ^2) or 380.1 - 7.8 or 372.3 square inches.
If you use the "easy" answer of dividing the radius in half, even accounting for the hub, you surrender the wheel when it's radius is ((22/2) - ((3 1/7) / 2) + ((3 1/7) /2) or about 6.3 inches, or it's diameter about 12.6 inches. Which gives the wheel an area of (pi * 6.3^2) - (pi * ((3 1/7)/2)^2), or 124.7 - 7.8, or 116.9
square inches. This isn't quite fair....
So, what IS the answer, and how can we find it easilly? "Easilly" has different meanings to different people.
Well, start by getting a pencil, a straight edge, a compass, and a piece of paper.
Now draw the grindstone with a diameter of 22 inches, and the center hole with a diameter of 3 1/7 inches.(You may scale this if you like....) If we draw the largest square we can in the original circle, and then we draw the largest circle we can in that square, the second circle will have 1/2 the area of the original circle. Of course, this doesn't account for the center hole.
Imagine the corners of the square, where they touch the circle, as being labelled A, B, C and D.
Now let's label the inner circle E.
Half the area of the grindstone's hole now needs to be added to circle E. To do this, we inscribe a square in the center hole, and a circle in it, as we did with the outer circle. The new smaller circle will have 1/2 the area of the hub.
Now we duplicate that smaller circle on the edge of the circle E, so that the radius of the smaller circle, G,
is on the larger circle, E.
Now draw a line from one of the intersections of the circles G and E through the radius of the circle E, and extend
it to the other
July 21,1998
Eunice's Marital status
At a party Jack saw Eunice standing alone at a punch bowl.
1. There were nineteen people altogether at the party.
2. Each of seven people came alone; each of the rest came with a member of the opposite sex.
3. The couples who came to the party were either engaged to each other or married to each other.
4. The women who came alone were unattached.
5. No man who came alone was engaged.
6. The number of engaged men present equalled the number of married men present.
7. The number of married men who came alone equalled the number of unattached men who came alone.
8. Of the married women, engaged women, and unattached women present, Eunice belonged to the largest group.
9. Jack, who was unattached, wanted to know which group of women Eunice belonged to.
Which one of the three groups did Eunice belong to?
...From [1] and [2], six couples came to the party. From [3], [4], and
[5], if a equals the number of married present, then 6-a rquals the
number of engaged women present, and 6-a equals the number of engaged men present.
From [6], then, 6-1 equals the number of married men present.
If b equals the number of married men who came alone, then the number of marroed men who came with their wives (a) plus the number of married men who came alone (b) equals the total number of married men
present: a + b = 6 - a. Then the number of married men who came alone (b) equals 6-2a.
From [7], 6 - 2a equals the number of unattached men who came alone.
From [4], the number of unattached women who came alone, then, equals the number of people who came alone (7) minus the number of married men who came alone (6 - 2a) minus the number of unattached men who came alone: 7 - (6 - 2a) - (6 - 2a), or 4a - 5.
So, a equals the number of married women present, 6 - a equals the number of engaged women present, and 4a - 5 equals the number of unattached women present.
Since 4a - 5 equals the number of unattached women present, a cannot equal 0 or 1. From [9], Jack was unattachedl so a cannot be greater
than 2, otherwise the number of unattached men (6 - 2a) would be 0 or less. Therefore, a must equal 2.
So, there were two married women, four engaged women, and three unattached women at the party.
From [8], Eunice was an engaged woman.
July 22, 1998
"Whoever makes it, tells it not.
Whoever takes it, knows it not.
And whoever knows it, wants it not.
Can you tell me what I speak of?"
...Counterfeit money.
July 23, 1998
"I give you a group of three.
One is sitting down and will never get up.
The second eats as much as is given to him and yet is always hungry.
The third goes away and never returns.
Who are they?"
...This puzzle is, again, from "Dream Park", and the answers are,
A stove,
the fire in the stove, and
the smoke created by the fire in the stove.
July 24, 1998
This will be of interest largely to those with some electronic
background, though it shouldn't be inaccessible to anyone with some mathematical background.
Imagine a wire-frame cube, each of edge of which is made with a 1 ohm resistor.
What is the resistance of the cube as a whole, if current is passed from one corner, which we'll call "A", to the opposite corner of the cube, which we'll call "B"?
Hints -
Resistors connected in series, that is, the foot of one connected to the head of another, have a resistance equal to the sum of their individual resistances.
Bad ascii graphic, resistors in series..... O---Resistor 1---O---Resistor 2---O
If resistor 1 has a resistance of 100 ohms, and resistor 2 has a resistance of 200 ohms, together they will have a resistance of 300 ohms.
Resistors in parallel, that is, resistors connected so that their feet are all on a single lead and their heads are all on another lead, can have their resistance calculated by the formula
Rtotal = 1/( (1/R1) + (1/R2) + ..... + (1/RN))
where R1 is the resistance of the first resistor, R2 is the resistance of the second resistor,
and RN is the resistance of the last resistor.
O-------------O----------------0 ---------------------------O
|---------------|------------------|
R1------------R2----------------R3 .....................Total resistance
measuring point
|---------------|------------------|
O------------O----------------O----------------------------O
In this case, if R1 = 100, R2 = 200, and R3 = 1000, the answer would be
1/( (1/100) + (1/200) + (1/1000))
0r
1 / ( .01 + .005 + .001)
or
1 / .016
or 62.5
Note, the total resistance of any parallel circuit is always lower than that of any element in the circuit.
...I found this, most recently, in Martin Gardner's "The Second Scientific American Book of Mathematical Diversions and Puzzles". While
teaching high school, I also saw it in a number of electronics texts. One can make the problem rather difficult - calculations covering sheets of paper are not uncommon. The engineering answer is, of course, to construct a cube of resistors
and to measure the result. Of course, this does leave you prey to the errors of the real world, so use 1% or lower tolerance resiistors.
The mathematical answer is to look at the cube and realize that the three resistors coming out from point A, and the three resistors meeting at point B are in parallel to one another. Each group of these three resistors has a resistance of 1/3 ohm.
The other 6 resistors are also in parallel, so their combined resistance is 1/6 ohm.
The three parallel circuits (the three resistors in parallel at point A, the three resistors in parallel at point B, and the six remaining resistors in parallel) are in series, so the resistance of the cube is 1/3 + 1/3 + 1/6, or 5/6 ohm.
July 27, 1998
Two Cylinders, one of lead and one of titanium, are identical in physical dimensions, and both are painted green, so that you cannot tell which is which. They both weigh the same, the lead cylinder being hollow and the titanium cylinder being solid. Of course, the hollow cylinder, being lead, does
not sound hollow. How can you distinguish between the two without scratching or damaging either cylinder and without using any other object?
...This is from Kendall's "Mathematical Puzzles for the Connoisseur". The answer is the moments of inertia
are different, so if both cylinders are rolled along the ground, the keaden one will roll further.
July 28, 1998
how DO trains turn? Their wheels are on a solid axle, so when one wheel turns a turn, so to
will the other. We know that the inner rail of a curve is shorter in the curve than the outer rail. Some people think that perhaps one wheel slips, but with the weight a train
puts on a wheel, there is no appreciable slippage. So, what's left? How DO trains turn?
...The edges of train wheels are bevelled. So, by sliding slightly from
side to side, the distance covered by one wheel in a single rotation can be different from the distance covered by the other wheel. This does impose a limit on how sharply a train can turn, but that's not usually a real world problem.
July 31, 1998
This is from "Puzzles and Teasers" by Nicholas E. Scripture.
What's in a word?
Maybe the title above is a misquotation -- the fact remains there is often a great deal more in a word than we notice at first glance.
Take the word ESCALATOR, for example: not only can you find an ESCORT, but a CARTER (who is, naturally, COARSE) carrying COAL, and who likes to ORATE while drinking TEA, and so on.
If you consider that you have a fair knowledge of language, just see how many words you can make, using only the letters employed in the
word ESCALATOR. Proper names and plurals do NOT count, and a word must consist of at least three letters.
Scoring - if you found
50 words, you have moderate language skills
75 .., fair
100 .., average
125 .., good
150 .., very good indeed
175 .., excellent
200 .., throw away your dictionary -- you'll never need it!)
...how did you do? (no real answer)
August 3, 1998
The Deadwood Express.
The Deadwood Express arrived at a western mining town with a
consignment of two boxes for a young lady. A lively dispute quickly developed between the expressman and the lady's miner friends.
The difficulty was that the expressman wished to charge for the boxes at the rate of $5.00 per cubic foot, as per his instructions on the freight bill. The miners, however, strenuously objected on the grounds that their custom was invariably to pay so much per running foot -- according to mining laws. They could not see what right an express company had to meddle with the "cubic contents" of a young lady's box, anyway!
The expressman was compelled to accept the proposed terms, so he measured the length of the boxes and charged $5.00 per running foot. Both boxes were perfectly cubical, and one box was exactly 1/2 the height of the other.
The strange part of the problem is that when the expressman placed the two boxes together and measured their combined length it was found that there was not the thousandth part of a cent difference in the ways of charging -- at $5.00 per cubic foot or at $5.00 per running foot.
What were the sizes of the two boxes?
...This was from "Mathematical Puzzles of Sam Loyd, Volume 1", as edited
by Martin Gardner. I am told that the large box must be 13.856
inches on a side, and that the smaller box must be 6.928 inches.
Adding their lengths gives 20.784 inches, or 1.732 feet, which at
$5.00 per running foot would be $8.66. The two boxes together
contain a trifle more than 2,992 cubic inches or 1.732 cubic feet.
At the rate of $5.00 per cubic foot, this would also amount to $8.66.
August 4, 1998
Mr. Scott, his sister, his son, and his daughter are tennis players.
The following facts refer to the people mentioned:
1. The best player's twin and the worst player are of the opposite sex.
2. The best player and the worst player are the same age.
Which one of the four is the best player?
...This was from "New Puzzles In Logical Deduction" by George J.
Summers. The best player and the best player's twin are the same age; the best player and the worst player are the same age, from [2]; and the best player's tin and the worst player are two different people, from [1]. Therefore, three of the four people are the same age. So Mr. Scott's son, daughter, and sister must all be the same age since Mr. Scott must be older than his son and his daughter.
Then Mr. Scott's son and daughter must be the twins indicated in [1].
So either Mr. Scott's son or daughter is the best player and Mr.
Scott's sister is the worst player. Since the best player's twin
must be Mr. Scott's son, from [1], the best player must be Mr.
Scott's daughter.
August 5, 1998
Murder in the Family
Murder occurred one evening in the home of a married couple and their son and daughter. One member of the family murdered another member, the third member witnessed the crime, and the fourth member was an accessory after the fact.
1. The accessor and the witness were of the opposite sex.
2. The oldest member and the witness were of the opposite sex.
3. The youngest member and the victim were of of the opposite sex.
4. The accessort was older than the victim.
5. The father was the oldest member.
6. The killer was not the youngest member.
Which one of the four - father, mother, son, or daughter - was the killer?
...The youngest member of the family was not the victim, from [3]; was not the
accessory, from [4]; and was not the killer, from [6]. So, from [4], there are only three possibilities (A stands for accessory, V for Victim, and W for witness):
..........................................I.....|....II....|....III
-----------------------------------------------
oldest member..................A.....|....A.....|.....K
next-to-oldest member.......V.....|....K.....|.....A
Next-to-yourgest member..K.....|....V.....|.....V
youngest member.............W.....|....W....|.....W
From [5], the father was the oldest member; so the next to-oldest member was the mother. From [2] and the above possibilities, the youngest member was the daughter; so the
next-to-youngest member was the son. Then from the oldest member to the youngest member the three possibilities are:
...........................I.......|........II.........|.........III
-----------------------------------------------
father..................A.....|........A.........|.........K
mother................V.....|........K.........|.........A
son.....................K.....|........V.........|.........V
daughter.............W |........W........|.........W
From [3], I is impossible. From [2], III is impossible. So II becomes the only remaining possibility. Therefore, the mother was the killer.
August 6, 1998
"Who makes it, has no use of it.
Who buys it, has no use for it.
Who uses it can neither see nor feel it."
What is it?
...It's a coffin.
August 7, 1998
There was quite a crowd outside the operating theatre at St. Nick's Hospital. In fact, there were five patients, all ready to be operated upon, lined up on their trolley's.
This itself was not particularly unusual, and normally there was no bother. On this particular day, however, a particularly clumsy nurse bumped into the trolley on which had been placed
all the false teeth of the patients awaiting attention. Everything fell to the floor and the dentureswere hopelessly mixed up.
On rescuing the teeth, it was found that three of the patients needed dentures for both upper and lower jaws, while the other two had top plates only. Since none of the patients was in
a condition to identify their own teeth, what were the chances of the five patients all having their own teeth by their bedsides when they came round after their operations? Also, what were
the chances for each individual that they would wake up and find their own teeth to hand?
...According to Nicholas Scripture's Puzzles and Teasers, there are no
less than 720 different ways of replacing the teeth (allowing for the fact that a top set is easily distinguishable from a bottom set), the chances are 719 to 1 against all the teeth finding their way to their proper homes right away.
And individual who has a top denture only, has a 4 to 1 chance
against being given their own teeth, while a person with a full set
of dentures has the odds loaded against having their proper mouthful to the tune of 14 to 1.
August 10, 1998
In the multiplication problem below, each letter represents a
different digit:
..... A S
x ......A
---------
M A N
Which one of the ten digits does A represent?
...This is from George Summer's "New Puzzles in Logical Deduction", and
is titled "The Three A's". It isn't terribly hard. If you're still
working on it, and don't want the answer, here's a hint.... If there
is a solution (and there is), only the value of A need be found.
And now for the answer...
A cannot equal 0 because then M and N would also equal 0.
A cannot equal 1 because the product is different from AS.
A cannot equal 2 because a three-digit product would not be possible.
A cannot equal 3 because 4 cannot be carried to A x A.
A cannot equal 4 or 7 because 8 cannot be carried to A x A.
A cannot equal 5 or 6 because then S would have to equal 0, making N equal to S, or S would have to equal 1, making N equal to A.
A cannot equal 9 because then 8 would have to be carried, making A equal to S.
So A must be equal to 8.
Though not necessary for the solution of the problem, the numerical values of S, M, and N can now be determined: since 4 must be carried, S equals 5 or 6; but S cannot equal 6 because then A would equal N. So S is equal to 5.
The multiplication is shown below:
8 5
x 8
-------
6 8 0
August 11, 1998
The Student Thief
Professor Dimwit's answer key to a Physics test was stolen one day
during one of his Physics classes. Only three students -- Amos,
Burt, and Cobb -- had the opportunity to steal the answer key.
1. Five Physics classes had been held in the room that day.
2. Amos attended only two of the classes.
3. Burt attended only three of the classes.
4. Cobb attended only four of the classes.
5. The professor conducted only three of the classes.
6. Each of the three students attended only two of the professor's classes.
7. No two of the five classses were attended by the same group of students from the three students under suspicion.
8. Two of the three students, who attended one of the professor's classes that the third student did not attended, were proven innocent of the theft.
Which one of the students stole the answer key?
...This is also from "New Puzzles in Logical Deduction" by George Summers.
If you're stumped and would rather have a hint than a solution,
here's a hint.... How many classes not conducted by the professor did each student attend? To satisfy [6], must each of his classes have been attended by two of his three students?
And now to the solution....
From [6] and [4], Cobb attended two classes not conducted by the professor. From [6] and [3], Burt attended one class not conducted by the professor. From [6] and [2], Amos attended only classes condicted by the Professor. If P represents a class conducted by the professor, and O represents a class not the professor's, then from [1] and [5], the following table can be constructed (an x indicates attendance at a class):
...............AMOS...BURT...COBB
---------------------------------
P
P
P
O.......................X.....X
O..............................X
From [6] and [7] - applying [7] only to the professors classes for now - four records of attendance as possible, as shown below:
I..............AMOS..BURT...COBB
---------------------------------
P...............X.......X
P.......................X......X
P...............X..............X
O.......................X......X
O..............................X
II.............AMOS...BURT...COBB
---------------------------------
P................X
P...........................X......X
P................X.........X......X
O...........................X......X
O...................................X
III............AMOS...BURT...COBB
---------------------------------
P...........................X
P................X......................X
P................X..........X.........X
O............................X.........X
O.......................................X
IV.............AMOS...BURT...COBB
---------------------------------
P.........................................X
P................X...........X
P................X...........X..........X
O..............................X.........X
O.........................................X
Then, applying [7] to all 5 classes -- possibilities I, II, and IV
are eliminated.
Frin III and [8], the two students who were innocent must be Amos and Cob (only one of the professor's three classes was attended by two students).
Therefore, Burt stole the answer key.
Next Page
Other Pages
If you have any riddles or MindBenders to add please
e-mail me or
e-mail Mike Avery
PaintSaint helped make this page. check out his
Riddles page.
