Puzzles Archive
This is a list of the previous puzzles that have been sent out by E-mail.
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August 12, 1998
A coin game requires:
1. Ten coins in one pile.
2. That each player takes one, two, or four coins from the pile at alternate turns.
3. That the player who takes the last coin loses.
I. When Austin and Brooks play, Austin goes first and Brooks goes second.
II. Each player always makes a move that allows him to win, if possible; if there is no way for him to win, then he always makes a move that allows a tie if possible.
Must one of the two men win? If so, which one?
...this is also from George Summer's "New Puzzles in Logical Deduction".
Well, let's start with a hint for those who are still working on the puzzle.
It must be first decided if drawing from one coin is a winning, losing, or tieing position then whether drawing from two coins is a winning, losing, or tieing position; and so on through drawing from ten coins.

And now the solution...
A number of people dropped me a note (as noted before, one is a number), asking what many have no doubt been thinking... when a game must be played to a conclusion, and the game will have a win/lose outcome, what's all this talk about tieing? We'll see....
From [II], if a player CAN win, he MUST win.
From [2] and [3]:
(a) Drawing from one coin, a player loses.
(b) Drawing from two coins, a player wins by taking only one coin, thus putting the other player in the losing position of drawing from one coin.
(c) Drawing from three coins, a player wins by raking two coins, putting the other player in the same losing position as in (b). If he takes only one coin, the other player may take only one coin and win.
(d) Drawing from four coins, a player loses. if he takes one coin, he gives the other player the winning position of drawing from three coins. If he takes two coins, he gives the other player the winning position of drawing from two coins. If he takes four coins, he loses at once. He cannot win because he cannot leave a number of coins that represents a losing position for the other player.
(e) Drawing from five coins, a player wins if he is able to leave a number of coins that represents a losing position for the other player. So, if he can leave one coin or four coins for the other player to draw from, he wins. Accordingly, he takes four coins, leaving one, or one coin, leaving four.

Reasoning in this manner, one finds that drawings from one, four, seven, and ten coins are losing positions, and that drawings from two, three, five, six, eight, and nine coins are winning positions.
The following tables summarize how these two sets of drawings can be losing and winning positions, respectively:

From A Losing........If A Player......He Leaves A Winning
Position of............Draws.............Position Of
---------------------------------------------------------
..........................1..................3
.....4....................2..................2
..........................4 .................0

..........................1..................6
.....7....................2..................5
..........................4..................3
..........................1..................9
....10....................2.................8
............................4..................6

From A Winning........If A Player......He Leaves A Losing
Position of............Draws.............Position Of
---------------------------------------------------------
.....2....................1..................1
.....3....................2..................1
.....5....................1..................4
...........................4..................1
.....6....................2..................4
.....8....................1..................7
..........................4..................4
.....9....................2..................7

From [1], there are ten coins. Since drawing from ten coins is a losing position, whoever goes first must lose. Since Austin goes first, from [I], Austin must lose. So Brooks MUST win.
---
Mike's comments.... from the above we see that there is no tie. Somebody has to take the last piece and lose.
This is a classic bar bet. You bet the sucker, you tell the sucker the rules, you let the sucker go first because that's the nice thing to do. When the sucker loses, you can either let the sucker escape, or let them go first again to "give them the benefit of going first".
After a time or two, you'll find someone in the bar who wants to take the place of your current sucker cause he's figured it out and he KNOWS he can beat you.
The only dangers come from people who want YOU to go first. If you refuse, you reveal part of the secret. So, either they understand the game in which case you take your lumps for a while and then either fold your tent and leave or think of a different con before things get out of hand. Alternately, maybe they don't understand the game, in which case you might be able to pull it out. While there's no way to recover from a losing position, if your opponent doesn't understand the game, he could give away his winning position.
For instance, you start with 10, take away 1, leaving 9, a losing position.
Your opponent doesn't understand and takes 1 or 4, leaving with 8 or 5, both of which are winning positions you will be shrewd enough to maintain....

August 13, 1998
Everybody Lied
When a psychiatrist was found murdered in his apartment, four of his patients were questioned about his death.
I. The police knew from the testimony of witnesses that each of the four patients had been alone with the psychiatrist in his apartment just once on the day of his death.
II. Before the four patients were questioned they met and agreed that every statement each patient made to the police would be a lie.
Each of the statements made two statements, as follows:
Avery: 1. None of us four killed the psychiatrist.
...........2. The psychiatrist was alive when I left.
Blake: 3. I was the second to arrive.
...........4. The psychiatrist was dead when I arrived.
Crown: 5. I was the third to arrive.
...........6. The psychiatrist was alive when I left.
Davis: 7. The killer did not arrive after I did.
...........8. The psychiatrist was dead when I arrived.
So, who was the killer?
...From [II], negating each of the 8 false statements results in the following 8 true statements:

[1] One of the four killed the psychiatrist.
[2] The psychiatrist was dead when Avery left.
[3] Blake was not the second to arrive.
[4] The psychiatrist was alive when Blake arrived.
[5] Crown was not the third to arrive.
[6] The psychiatrist was dead when Crown left.
[7] The killer arrived after Davis did.
[8] The psychiatrist was alive when Davis arrived.

From [1], [4], [8], [2], and [6], Blake and Davis arrived before Avery and Crown. From [3], Blake must have arrived second; so Blake arrive first. From [5], Avery must have arrived third; so Crown arrived fourth.
The psychiatrist was alive when Davis arrived second, but was dead when Avery left third. So, from [1]m either Avery or Davis killed the psychiatrist.

From [7], Avery is the murderer.

August 14, 1998
The Triangular Pen
A farmer built a triangular pen for his chickens. The pen was made of a wire mesh attached to posts imbedded in the ground.
1. The posts were spaced at equal intervals along each side of the pen.
2. The wire mesh, of uniform width, was attached to the posts at equal heights above the ground.
3. The farmer made the following entry in a notebook:

Cost of wire for side of pen facing barn: $10
Cost of wire for side of pen facing pond: $20
Cost of wire for side of pen facing home: $30

4. He paid for the wire mesh with only ten dollar bills, and received no change.
5. He paid with a different number of ten dollar bills for the wire mesh along each side of the pen.
6. Exactly one of the three costs in his entry was incorrect.
Which one of the three costs was incorrect?
...Here's a hint for those still working on the MindBender. The costs of the sides must be in the same ratio as their lengths. What relative lengths of the sides of the pen are possible?

And now an answer.
From [1], [2], [3], and [6], the lengths of the sides of the triangular pen are in the ration of 1 to 2 to 3, one number of which is incorrect.
From [4], the incorrect number can be replaced only by a whole number.
From [5], the incorrect number must be replaced by a whole number greater than 3. If either 2 or 3 were replaced by a shole number greater than 3, it would be impossible to construct the pen, since the sum lengths of any two sides must be greater than the length of the third side.
So 1 is the incorrect number and the $10 cost for the side of the pen must be incorrect.
If 1 is replaced by a whole number greater than 4 the pen would still be impossible to construct. However, if 1 is replaced by 4, the pen can be constructed. The cost for the side facing the barn must have been $40 instead of $10.

August 17, 1998
Lasciviousness here has its sources, Harlots its use apply.
Without it Lust has never been, and even Love would die.
Now tell me what this wonder is, but pause before you guess it.
If you are mother, maid, or man, I swear you don't possess it.
...the answer is the letter "L".

August 18, 1998
The Jabberwocky Bank Raid
The raid on the bank was very successful - up to a point. The precise point was when the police car came screaming around the corner in response to a phone call from an old lady across the road from the bank. She had got up to make herself a cup of tea, being a poor sleeper, and had seen the raiders' car outside the bank.
There was no real hope of escape, but the five malefactors had piled into their car, anyway, and tried to make a break for it. The police gave chase, and although revolver shots from the car caused the pursuers to crash, a message had already gone out over the police radio.
In the end the raiders crashed their own car, trying to avoid a makeshift road-block. One of them was taken to the hospital with multiple injuries, and the others found themselves in the Jabberwocky Police Station facing a barrage of questions.
There had been shots fired from two revolvers at the police car which had first given chase, but the weapons had been thrown out of the window by the would-be escapers, and none of them would admit to having fired upon the police.
Luckilly for the law, Toves and Slithy were not the best of friends, and they were both only too happy to accuse each other of having been one of the gunmen. In addition, both of them stated that Gyre never carried a gun, and while Slithy also defended himself against the charge, Toves spoke in Gimble's defense.
Neither Gyre or Gimble was prepared to make a statement, and the police felt completely at a loss. The Chief Inspector in charge of the case went along to the hospital, but was allowed only a few minutes with the fifth member of the gang. He, like Gyre and Gimble, had nothing of a constructive nature to say. He simply laughed.
"Well, both Slithy and Toves told the truth for once, even of their other remarks were lies," Wabe chuckled. That was the last thing he said before he died.
Assuming that Wabe's comment was true, being in the nature of a Deathbed Statement, who were the gunmen?
...Toves made three statements, only one of which was true. If the statement "Slithy is a gunman" were true, then Slithy, Gyre, and Gimble must have all been gunmen - which is impossible. Hence, Slithy was innocent while one of the other two was not.
Slithy's statements again contain only one which is true. From Toves, we know that Slithy's claim that he himself is innocent is true - hence Gyre must be a gunmanm while Toves is not.
We also know that only one of Gyre and Gimble is a gunman, which means that Slithy, Toves, and Gimble are all innocent. Therefore, the second gunman must have been Wabe, the dead man.

August 19, 1998
A fishy tale
The Blackheath Piscatorials had very mixed luck when they went out after cod the other week. The weather was ideal, yet Jim and Paul hauled in only ten fish between them, instead of leading the field as theyusually did. Stan did best, witha catch of some fourty-two fish, while his brother Bob managed to land a quarter of the club's total for the day.
Apart from thise already mentioned, the only person who had areasonable day's sport was Arthur, the club's president, whi accounted for 15 percent of the club's catch. Altogether, the only really happy man was Stan. After all, it was not often that one managed to land twice as many finny friends as did the President!
What was the club's total catch for the day?
...This one is pretty straightforward. It's from Nicholas Scripture's book "Puzzles and Teasers".
Since we are told that Stan had a catch of 42 fish, and that he caught twice as many as did the club president who accounted for 15 percent of the total catch), 21 fish = 15 percent, and therefore 100 per cent = 140 fish.

August 20, 1998
Owen Parker, Frank Doyle, Bill Aherne, and Paul Hodges were accustomed to meet on a Thursday evening for a hand of whist.
Knowing that I myself have quite an interest in the game, one of them was describing the play at their last meeting, but for some time I was baffled because of my ignorance of the occupations of the various men involved.
Apparently, the teacher was on Paul's left, while Bill was partnering with the solicitor. Moreover, Owen is neither the teacher nor the civil servant, while Frank's partner was the teacher.
If the man who was telling me about the evening is an engineer, what is his name?
...Owen must have been the engineer. The others, for interest: Frank --solicitor, Bill - teacher, and Paul - civil servant.

August 21, 1998
A factory has ten machines, all making flywheels for racing cars.
The correct weight for a flywheel is known. One machine starts to products faulty (over- or underweight) parts.
How, in two weighings, can the faulty machine be found?
...Take one flywheel from each machine and weigh them as a unit. You can determine the amount of error in this weighing.
Now take one flywheel from machine 1, two from machine 2, and so on until you have ten flywheels from machine 10. Weigh them as a unit. Divide the error in the weight of the total by the error in the weight from the previous weighing. The answer will be the number of machine that is defective.
For instance, if machine 4 is making flywheels 1 ounce to heavy, the first weighing will be one ounce to heavy, the second weighing will be four ounces too heavy.

August 24, 1998
A man is rowing across a river in the gathering dusk. On the approaching bank he sees the dim figures of three men. 'Some' of these men wear black and 'some' white, and the oarsman wishes to know which wear black and which wear white, so he calls out, "What colour are you wearing?"
The reply from the first dim figure is lost in the slight breeze, but the second man calls out, "He's wearing black, but I am wearing white!"
The third figure, referring to the first man, says, "He says he's wearing white, and he is wearing white!"
With this information, and with the knowledge that the men wearing black always lie, and the white dressed men always tell the truth, the man in the boat identified the figures for what they really were.
Can you do the same?
...The first man, who's reply was lost, could only have said one thing. If he was wearing white, he would have said so. If he was wearing black, he would have lied and said he was wearing white.
Therefore, the first man was wearing white, the second was wearing black, and the third was wearing white.

August 25, 1998
"Speaking of consecutive numbers," said a friend, "How many consecutive numbers can you write without a prime number among them?"
"As many as you like," we replied.
"Well, say a dozen", said our friend.
We showed him how it was done.
"And," asked our friend, "How many consecutive numbers can you write down which are all prime?"
How does one write down a series of consecutive non-prime numbers of any desired length?
And what is the greatest sequence of consecutive numbers which are all prime?
...Suppose we want a dozen. Multiply together all the prime numbers less than 12. i.e. 2 x 3 x 5 x 7 x 11 = 2,310
Then add 2, 3, 4, ... 12 to the result. This will be the required series.
Thus the series 2312, 2313, 2314, ...2322
Mr. Kendall says, "Every even number is divisible by 2, therefore a series of consecutive prime numbers cannot be formulated."
However, it is worth noting that 2 is a prime number, as is 3. Thus, there is a series of consecutive prime numbers, 2 and 3. A short series. Some people maintain that 1 is also a prime number in that it is evenly divisible only by itself and 1.

August 26, 1998
A curate is visiting a rector at his rectory one day and they find that their birthdays are on the same day. The vicar remarks that three of his parishioners have their birthdays on the same day.
The ages of the three parishioners have a product of 2,450 years, and added together are equal to twice the curate's age. The vicar asks the curate, "What then are the ages of the three parishioners?"
The curate sat thinking or an hour (for he was not very quick at mental arithmetic) and then he said to the vicar, "You haven't given me enough information!"
So the vicar said, "I'm so sorry, I am older than any of my parishioners, and am the same age as the product of the two youngest."
How old is the vicar?
...The curate knew his own age and therefor he should have been able to work out from the factors of 2450 the ages of the parishioners. He failed, however, because there are two sets of factors which give a correct solution.
i.e. 7 x 7 x 50 and 10 x 5 x 49
10 x 5 = 50 which makes the vicar older than any of his parishioners.

August 27, 1998
Quite often in "western" motion pictures the wheels of stage-coaches appear to be rotating backwards. In a film we recently saw the wheels appeared to be stationary although the horses were galloping "full out".
Fascinated by this, we counted the number of spokes and found there were twelve per wheel. A six-foot hero stood about twice as tall as the wheels, so we estimated that the wheel was about three feet in diameter. If the film was being shown at 24 frames per second, how fast was the stage-coach moving?
...In 1/24th of a second, the wheen must move n/12 revolutions (where n is an integer), i.e., it is revolving at 120n r.p.m.
Hence, the speed of the coach is (120n X 50 X n)/1760mph or 12.85n m.p.h.
The most reasonable stage-coach speed is obtained when n =2, giving a speed of 25.75 m.p.h.


August 28, 1998
Imagine a triangle whose corners are labelled A, B, and C. If you trisect every angle in the triangle, (ABC, BCA, and CAB), six lines will result. Extend the each line so that it intersects the line drawn from the adjacent corner. Connect the resulting three intersections to form a new triangle.
Will that resulting triangle be an equilateral triangle? Will it always be an equilateral triangle?
Here is an image of the triangle:
Note: The pictures below with the math are part of the answer.
...The simple answer here is "Yes".

The complecated math his below




September 1, 1998
A boy looks at a (old fashioned) clock and sees that the hour and minute hand are both together at noon. He realizes that one hand moves 12x faster than the other, and that therefore they will meet again before midnight.
When, to a fraction of a second, will the hour and minute hand again be perfectly aligned?
...An hour after noon, or at 1:00, the hour hand is on the 1, and the minute hand is on the 12. This is consistent with the 12x faster ratio.
As a result, in any given second, the minute hand will move .1 degree, while the hour hand will move .0083333 degree. A bit of playing with numbers reveals that the intersection will occur at 1:05.27.27272
OR explained by Bill -a subscriber to the MindBender
In a second, the minute hand moves 1/(60*60) or 1/3600 of a revolution around the clock face. Thus in N seconds, the minute hand moves N/3600 of a revolution around the clock face.
In a second the hour hand moves 1/(12*60*60) or 1/43200 of a revolution around the clock face. Thus in N seconds, the hour hand moves N/43200 of a revolution around the clock face.
The hands cross some number, N, of seconds past each hour. For any hour, H, the hour hand starts the hour at H/12 of a revolution around the clock face.
To solve, we want the fraction of a revolution for the minute hand and the hour hand to be the same.
For the minute hand, this is just N/3600. For the hour hand, this is the "hour fraction" (H/12) plus the "seconds past the hour" fraction (N/43200).
Thus we have: N/3600 = H/12 + N/43200
Giving us: 12N/43200 = 3600H/43200 + N/43200
Resulting in: 12N = 3600H + N
Or: 11N = 3600H
When H = 1, N = 327 3/11 seconds after 1:00 or 1:05:27 3/11
When H = 2, N = 654 6/11 seconds after 2:00 or 2:10:54 6/11.
. . When H = 10, N = 3272 8/11 seconds after 10:00 or 10:54:32 8/11
When H = 11, N = 3600 seconds after 11:00 or 12:00:00 again.

September 16, 1998
An archeologist comes upon a sandstone carving of a long division problem. However, a number of the digits have worn off, and he wants to know what problem was worked out. Luckily the eight legible digits provide enough information to allow you to supply the missing digits. The author states that while it looks as though there should be many possible solutions, so far only one has been found.

..................X 5 3
..........---------------
X X 9 | 6 X 8 X X X
............X X X 2
............----------
................X 9 X X
................X X 4 X
..............----------
...................X X 4 X
...................X X X X
...This MindBender is from "The Mathematical Puzzles of Sam Lloyd, Vol 2", it is best viewed in a mono-space font, and the answer is:
..................8 5 3
..........---------------
7 4 9 | 6 3 8 8 9 7
............5 9 9 2
............----------
................ 3 9 6 9
................3 7 4 5
..............----------
...................2 2 4 7
...................2 2 4 7


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