Puzzles Archive
This is a list of the previous puzzles that have been sent out by E-mail.
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September 17, 1998
Mr. Jordan wasn't too bad, as math teachers go. He insisted on getting the work done, but he wasn't above having a bit of a joke with the class occaisionally.
On this particular day, some question about squares had arisen during the geometry lesson, and Mr. Jordan had mentioned that he himself was a bit of "a square", having been born in 1936, which number was a perfect square, being the product of 44 x 44.
'Of course', he said, 'none of you were born in a year which was a perfect square-but you're more likely than I am to be alive in the next such year . . 2025.'
This idea got several of the boys thinking, and when the next math lesson came along, Tom Smith reminded Mr. Jordan of his squareness.
'Never mind, though, sir,' Tomm said. 'My birthday was even more square than yours. The day of the month was a perfect square, the number of the month was a perfect square, and if you add two numbers together, the total is another perfect square. On top of all that, if you add all the figures together which are needed in order to write the date of my birth in full, the result is yet another perfect square!'
What was Tom Smith's birthday?
...This is from "Puzzles and teasers" by Nicholas Scripture.
Tom Smith must have been born on 16th September, 1955.
16 = 4 x 4, September = 9th month, and 9 = 3 x 3.
16+9 = 25 = 5 x 5
1+6+9+1+9+5+5 = 36 = 6 x 6

September 18, 1998
Juggling....
Drinkwell and Company manufactured ornamental pewter jugs in three sizes and in two different finishes. In order to increase sales, the Sales Director suggested that the jugs be sold in sets of siz, and the Board agreed in principle.
However, one member of the Board had reservations about the scheme, and although he had been overruled on the main issue, he did manage to persuade his fellow-directors to agree that certain conditions should be fulfilled when making up the sets of jugs:
1. If both finishes were to be represented in a set, then there must be three jugs of each type of finish, and
2. If more than one size of jug was to be included in a set, then each size must be represented by an equal number of jugs (i.e., if two sizes, then three jugs of each sizel if three sizes, then two jugs of each size).
In how many different ways can the Sales Director arrange sets of jugs, according to these conditions?
...This is from "Puzzles and Teasers" By Nicholas Scripture.

His rather succinct answer is, "Altogether, there are 36 combinations."

September 21, 1998
London has many glories, though most of them are better known by visitors and tourists than by the native Londoners. Speakers' corner, at Hyde Park, is one of the few exeptions, and on any Sunday monring one may see, and hear, the crowds milling around the wooden platforms of the orators.
There is considerable entertainment value here, and the other week I found myself on the fringes of a crowd listening to a man discussing the Breathalyzer test.
His old overcoat and loud checked cap were not particularly prepossesing, but he knew his audience, and his comparisons between Magna Carta and the breathalyzer were both pithy and ear-catching. The main point of his argument seemed to be the popular one that the Government was trying to have its cake and eat it: First, they taxed the beer, then they taxed the car -- then they attempted to raise even more revenue from beer drinking drivers by imposing the Breathalyzer Test!
This line of argument, of course, endeared him to the crowd, but his audience cooled towards him when he went on to tell everyone about his "Safety Formula". According to him, the only way to beat the Test was to allow a certain amount of time-lapse between drinking beer and getting into a car, the time involved being dependent upon the amount of beer drunk. Hence, according to his calculations (please see note below!), one should wait for five minutes after drinking one glass of beer, after two pints the waiting period should be fifteen minutes; after five pints, there should be a cooling off period of an hour and ...
At this point the crowd turned completely, as is the habit of crowds, and the rest of his speech was lost in the catcalls and ribald comments of his sometime adherents.
Time had crept up on me while I had been listening, so I had to dash off while the speaker was trying to regain the audiences sympathy. However, later that night I had a most intriguing time workingout exactly how long the speaker recommended that a person should wait before getting into his car after five pints of beer.
What length of time would you suggest?

(Please note-this formula is as flawed as the Magna Carta argument. As I recall from Drivers Ed classes, one should allow one hour to elapse per ounce of alcohol consumed. This is roughly equivalent to a beer, a glass of wine, or a mixed drink. Not a "super size" drink like the 40 ounce margaritas served in some restaurants..... please drive responsibly.
Or, if you drink five beers in an hour, you've consumed around 5 ounces of alcohol, and have metabolized one, so you should wait 4 more hours before driving.)
...Again, this is from "Puzzles and teasers" by Nicholas Scripture

From the information given:
After 1 pint, wait for 5 minutes
After 2 pints, wait for 15 minutes,
after 3 pints, wait for 25 minutes,
after 4 pints, wait for 35 minutes,
after 5 pints, wait for 45 minutes

might seem to be a reasonable application of the Safety Formula,
However, this is manifestly impossible, since the speaker was about to give a time of over an hour for the five pint delay. Hence, the only possible answer is that after every successive pint the waiting period should be extended by a period of five minutes times the number of pints consumed,
Hence, 5 minutes after 1 pint,
15 minutes after 2 pints,
30 minutes after 3 pints,
50 minutes after 4 pints, and
an hour and a quarter after 5 pints.

(Again, this is a formula for a DWI violation and a possible accident in most jurisdictions today. Try an hour per ounce of alcohol.)

September 22, 1998
By now, with all the weather forecasts giving temperatures in both Centigrade and Fahrenheit, it is fairly well known that 0 Centrigrade is the same as -32 Fahrenheit. What is not quite so well known is that -40 Centrigrade is also -40 Fahrenheit.
At what temperature will the Centigrade reading on a thermometer be just half the figure shown on Fahrenheit scale?
...From "Puzzles and Teasers" by Nicholas Scripture, we learn that 160 Centigrade is 320 Fahrenheit.

September 23, 1998
An Alphametic
A diversion which has managed to achieve a certain measure of popularity in the last few years is the Alphametic. In this one is presented with a sum in which the digits are represented by letters. The task is to replace the letters by digits. For example, one might be given the sum:
......O N E
...+ O N E
-----------------
.....T W O
This is easy, as there are a number of possible solutions:
...2 8 1..........4 0 7
+ 2 8 1.......+ 4 0 7
---------......-----------
...5 6 2.......8 1 4
On the other hand, the following is capable of only one solutiuon, if one accepts that THREE must be divisible by 3, and that FOUR must be divisible by 4, but not by 8:
.........T H R E E
........+ F O U R
--------------------------
.........S E V E N
...Nicholas Scripture suggests that the answer is:

....2 6 8 1 1
+.....4 7 0 8
----------------------
....3 1 5 1 5

September 24, 1998
It was holiday-time and the three young frogs were feeling vored and very fed-up. Finally Honk suggested they should have a race. Tonk didn't like races, but he agreed-on the condition that instead of seeing how fast they could go, they should see how FAR they could go in 20 jumps.
Honk immediately agreed to this condition, because he knew that he was the strongest jumper; Plonk agreed because he always agreed with Honk.
The three started off, and with the first jump, Honk was well in the lead with 16 feet, followed by Plonk with 12 feet, and poor old Tonk way behind with a mere 8 feet.
After the second jumpo the order was still the same. The distances covered this time, however, were very much smaller, the competitors having taken a lot out of themselves with their first tremendous effort. Consequently their total respective distances from the starting point were now 20 feet, 18 feet, and 13 feet.
By the time the third jump had been taken, Honk and Plonk were both 21 feet from the starting position, with Tonk 4 feet 10 1/2 inches behind them.
If they all three continued to increase the distance covered in this way, what would be the result of the race?
...According to Mr. Scripture, From the way the frogs progressed, their jumps decreased according to a well-defined series. For instance, Plonks jumps were 12, 6, 3 .. feet, indicating that the distance was halved with each succeeding jump. In consequence, the final placings were:
Plonk (total distance approx 24 feet), with the other two as joint seconds (with approx 21 feet 4 inches).

September 25, 1998
The Pet Club
The Woolham Secondary School already had over twenty clubs and societies, ranging from archery right through to zoology. As a result, when a pet Club was started there were many who said that it would not be able to attract sufficient support.
The pessimists were proven wrong. At the end of the first term there were 58 members in the club. Admitted, only half actually had pets of their own, and none had more than two pets, but nevertheless the club meetings were extremely well attended. In fact, the support was so considerable that the club was permitted to hold a pet show just before the end of the term.
All the pets came, and the noise (and smell) of six dogs, fourteen cats, and seventeen assorted other pets (mainly budgies (parakeets to Americans), though a grass snake and a couple of tortoises lent further variety) was quite something!
Now then.... how many club members had more than one pet?
...This is again from Nicholas Scripture, and will be the last from his book for a while.
There were 29 pet-owners who were members of the club, and there were 37 pets in the show. Since no one owned more than two pets, 21 members had one pet each, and the remaining 8 members had two pets apiece.

Spetember 28, 1998
What was the profit?
A dealer sold a bicycle for $50, and then bought it back for $40, thereby making a profit of $10 because he had the same bike and $10 besides. Now having bought it for $40, he resold it for $45 and made $5 more, or $15 in all.
"But", says a bookkeeper, "the man starts off with a bike worth $50 and at the end of the second sale has just $55! How then could he make more than $5? You see, the selling if the bike at $50 is a mere exchange which shows neither profit nor loss, but when he buys it at $40 and sells at $45, he makes $5, and that is all there is to it."
"I claim", said an accountant, "that when he sells at $50 and buys back at $40, he has clearly and positively made $10, because he has the same bike and $10, but when he sells at $45 he makes that mere exchange referred to, which shows neither profit nor loss. This does not affect his first profit so he has made exactly $10."
It is a simple transaction which any scholar in elementary schoool should be able to figure out mentally, yet we are confronted with three different answers. Which in your opinion is right?
...This is from "The Mathematical Problems of Sam Lloyd, Volume 1". The problem has no unambiguous answer unless you know what the dealer paid originally for the bike. Since this is not given, the question can not be answered in any satisfactory way.

See... it's like you've been told... there are some questions that just don't have an answer.

September 29, 1998
In describing his experiences at a bargain sale, Smith said that half is money was gone in just thirty minutes, so that he was left with as many pennies as he had dollars before, and but half as many dollars as before he had pennies. Now, how much did he spend?
...In "The Mathematical Puzzles of Sam Loyd, Vol 1" edited by Martin Gardner, we find out that Smith must have started out with $99.98, and spent all but $49.99.

September 30, 1998
Will the cat or dog win the race?
Many years ago, when barnum's Circus was of a truth "the greatest show on earth," the famous showman got me to prepare for him a series of prize riddles for advertising purposes. They became widely known as the Questions of the Sphinx, on account of the large prizes offered to any one who could master them.
Barnum was particularly pleased with the problem of the cat and dog race, letting it be known far and wide that on a certain first day of April he would give the answer and award the prizes, or as he aptly put it, "let the cat out of the bag, for the benefit of those most concerned".
The wording of the puzzle was as follows:
"A trained cat and dog run a race, one hundred feet straight away and return. The dog leaps three feet at each bound, and the cat but two, but then she makes three leaps to his two. Now, under those circumstances, what are the possible outcomes of the race?"
The fact that the answer was to made public on the first of April, and the sly reference to "letting the cat out of the bag", was enough to intimate that the great showman had some funny answer up his sleeve.
...The cat wins, of course. IOt has to make preciesely 100 leaps to complete the distance and retyurn. The dog, on the contrary, is compelled to go to 102 feet and back. It's thirty-third leap takes it to the 99 foot mark, and so another leap, carrying it two feet beyond the mark, becomes necessary. In all, the dog must make 68 leaps to go the distance. But it jumps only two-thirds as as quickly as the cat, so that while the cat is making 100 leaps, the dog can not quite make 67.
But barnum had an April Fool possibilty up his sleeve. Suppose that the cat is named Sir Thomas, and the dog is a female? The phrase "she makes three leaps to his two" would then mean the dog goes 9 feet whilke the cat went 4. Thus when the dog finishes the race in 68 leaps, the cat will have travelled only 90 feet and 8 inches.

Martin Gardner, the editor, added the following comment-"This same puzzle stirred up considerable chagrin in London when Hendy Dudeney published it in the April 1, 1900 issue of "The Weekly Dispatch". Dudeney's version, a race betwen a gardener (a female) and a cook (male), will be found in his "Amusements in Mathematics", problem 428."

October 1, 1998
The three of us made sime bets.
1. First, A won from B as much as A had originally.
2. Next, B won from C as much as B then had left.
3. Finally C won from A as much as C then had left.
4. We ended up having equal amounts of money.
5. I wound up with 50 cents.
Now then-which of the three-A, B, or C is the speaker?
...This is from George J. Summer's "New Puzzles in Logical Deduction". Mr. Summers usually gives a hint so that people who are having troubles can read the hint and not have to be (quite so) embarassed when they read the answer. The hint for this MindBender was: Let a, b, and c be the amounts of money A, B, and C had, respectively, before the betting. Then represent algebraically the amounts of the money each had after the betting. Only one of the three people could have begun with 50 cents.

The answer....Let a be the amount of money A had, and b the amount B had before A and B bet. The from [1], after they bet A had 2a and B had b-a.
Let c be the amount that C had before he bet with B. Then, from [2], after B and C bet. B had (b-a)+(b-a) or 2b-2a, and C had c-(b-a) or c-b+a.
Then from [3], after C and A bet, C had (c-b+a)+(c-b+a) or 2c-2b+2a, and A had 2a-(c -c+a) or a-c+b.
From [4], a-c+b = 2b-1a and a-c+b = 2c-2b+2a. The first equation yields: b = 3a-c, and the second yields: 3b = a+ 3c. Multiplying the first of these latter equations byu 3 and adding the two equations yields: 6b = 10a or b=5/3a Substitution for b yields C = 4/3a.
So A started with a cents, B started wut 5/3a cents, and C started out with 4/3a cents.
From [5], a cannot be 50 cents, because then B and C would have started with ractions of a cent, and 4/3a cannot be 50 cents because then A and B would have started out with fractions of a cent. So 5/3a is 50 cents and B is the speaker.
In summary, A started out with 30 cents, B started out with 50 cents, and C started out with 40 cents.

October 2, 1998
Well.... this isn't math... so some of you might feel happier, but perhaps not much happier.
In a certain card game, one of the hands dealt contains:
1. Exactly thirteen cards.
2. At least one card in each suit.
3. A different number of cards in each suit.
4. A total of five hearts and diamonds.
5. A total of six hearts and spades.
6. Exactly two cards in the "trump" suit.
Which one of the four suits-hearts, spades, diamonds, or clubs -- is the "trump" suit?
...This is from "New Puzzles in Logical Deduction" by George Summers.

From [1], [2], and [3], the distribution of the four suits is either:
(a) 1 2 3 7
or
(b) 1 2 4 6
or
(c) 1 3 4 5
From [6], combination (c) is eliminated because no suit consists of only two cards.
From [5], combination (a) is eliminated because the addition of no two numbers produces a sum of six.
So (b) is the correct combination of suits.
From [5], either there are two hearts and four spaces or there are four hearts and two spades.
From [4], either there are one heart and four diamonds, or there are four hearts and one diamond.
From [4] and [5] together, there must be 4 hearts. Then there must be two spades. So spades is the trump suit
In summary, there are four hearts, two spades, one diamond, and six clubs.

November 9, 1998
In the town of Arlington the supermarket, the department store, and the bank are open together on one day each week.
1. Each of the three places is open 4 days a week,
2. On Sunday all of these places are closed.
3. None of the three places is open on three consecutive days.
4. On six consecutive days:
the department store was closed on the first day,
the supermarket was closed on the second day,
the bank was closed on the third day,
the supermarket was closed on the fourth day,
the department store was closed on the fifth day,
the bank was closed on the sixth day.
On which one of these seven days are all three places in Arlington open?
...If Sunday is the first of the six consecutive days mentioned, then from [1], [2] and [4], the supermarket is closed only on Sunday, Monday, and Wednesday. This is impossible from [3].
If Monday is the first of the six consecutive days mentioned, then from [2] and [4] at least one place is clised each day. This is impossible because all three places are open together on one day each week.
If Tuesday is the first of the six consecutive days mentioned, then from [1], [2] and [4], the department store is closed only Tuesday, Saturday, and Sunday. This is impossible from [3].
If Wednesday is the first of the six consecutive days, then from [1], [2] and [4] the bank is closed only on Sunday, Monday, and Friday, and the supermarket is closed on Sunday. Thursday, and Saturday. This is impossible from [3].
If Thursday is the first of the six consecutive days, then from [1],[2], and [4] the bank is closed only on Tuesday, Saturday, and Sunday. This is impossible from [3].
If Friday is the first of the six consecutive days mentioned, then from [1], [2] and [4] the supermarket is closed on Monday, Saturday, and Sunday. Again, this is impossible from [3].
So, Saturday is the first of the siz consecutive days mentioned; then from [1], [2], and [4] (C stands for Closed, O for Open):
Sun Mon Tue Wed Thur Fri Sat
--------------------------------------
Bank C C O O C O O
Department C O O C O O C
Store
Supermarket C O C
From the above, FRIDAY must be the day when all three places are open.
To complete the table: from [1] and [3], the supermarket cannot be closed on Wednesday or Saturday, so it must be closed on Thursday.
To complete the table:
Sun Mon Tue Wed Thur Fri Sat
--------------------------------------
Bank C C O O C O O
Department C O O C O O C
Store
Supermarket C O C O C O O

November 23, 1998
1. There is a man who lives on the top floor of a very tall building. Everyday he gets the elevator down to the ground floor to leave the building to go to work. Upon returning from work though, he can only travel half way up in the lift and has to walk the rest of the way unless it's raining! Why?
This is probably the best known and most celebrated of all lateral thinking puzzles. It is a true classic. Although there are many possible solutions which fit the initial conditions, only the canonical answer is truly satisfying.
2. A man and his son are in a car accident. The father dies on the scene, but the child is rushed to the hospital. When he arrives the surgeon says, "I can't operate on this boy, he is my son!" How can this be?
3. A man is wearing black. Black shoes, socks, trousers, jumper, gloves and balaclava. He is walking down a black street with all the street lamps off. A black car is coming towards him with its light off but somehow manages to stop in time. How did the driver see the man?
4. One day Kerry celebrated her birthday. Two days later her older twin brother, Terry, celebrated his birthday. How?
5. Why is it better to have round manhole covers than square ones? ...The man is very, very short and can only reach halfway up the elevator buttons. However, if it is raining then he will have his umbrella with him and can press the higher buttons with it.
The surgeon is the child's mother.
It's daytime.
I got a number of excellent answers to this one, but this is the "official" answer.
At the time she went into labor, the mother of the twins was traveling by boat. The older twin, Terry, was born first early on March 1st. The boat then crossed a time zone and Kerry, the younger twin, was born on February the 28th. Therefore, the younger twin celebrates her birthday two days before her older brother.
A square manhole cover can be turned and dropped down the diagonal of the manhole. A round manhole cannot be dropped down the manhole. So for safety and practicality, all manhole covers should be round.

November 24, 1998
A Thanksgiving Discussion
I overheard two friends talking during Thanksgiving dinner. Kris and Terri carpool to work together each day but don't see each other socially very often. The following is a portion of their conversation:
Kris: What are the ages (in whole years) of your three kids?
Terri: The product of their ages is 36.
Kris: That's not enough information.
Terri: The sum of their ages equals your house number.
Kris: That's still not enough information.
Terri: My oldest (at least a year older than either of the others) is the one that absolutely loves turkey and dressing.
Kris: That's enough. Their ages are ...
At this point, I walked away and didn't hear the rest of Kris' last statement. So, how old are Terri's kids?
___________________________

...Answer to A Thanksgiving Discussion
The only sets of three factors that yield 36 are:
Factors
1 1 36
1 2 18
1 3 12
1 4 9
1 6 6
2 2 9
2 3 6
3 3 4
Kris certainly knows Kris' address. Kris would be able to know the correct set of factors when told it had a sum equal to that address --- unless the sum was 13, since two sets of factors have that sum.
Factors Sum
1 1 36 38
1 2 18 21
1 3 12 16
1 4 9 14
1 6 6 13
2 2 9 13
2 3 6 11
3 3 4 10
As soon as Kris knew there was an oldest, the 1 6 6 set was eliminated, meaning the ages of Terri's three kids are 2, 2, and 9.
This MindBender was modified from a puzzle in Martin Gardner's book, "Wheels, Life and Other Mathematical Amusements."
___________________________


2 = 1
The following is a simple proof that 2 = 1.
Find what is wrong with it. (Unless you think that 2 REALLY does equal 1!)
This is suitable and an excellent learning aid for any student that has gotten as far as factoring in basic algebra.
.Multiply each side by A giving- .................. A*=AB
.Substitute A for B giving- ...........................B*=AB
.Multiply by -1 and add A* giving- .............A*-B*=A*-AB
.Factor giving- .............................................(A-B)(A+B)=A(A-B)
.The (A-B)'s cancel giving- ..........................A+B=A
.Substitute B for A giving- ...........................A+A=A
.Simplify giving- ..........................................2A=A
.Divide by A giving- .....................................2=1
Hmmmm .... So what's wrong here?
___________________________

Answer to 2 = 1
Some thought the problem was multiplying by A in step 2 if A were zero, or dividing by zero in step 8 if B were zero. Actually multiplying by zero is not the problem since A and B can be assumed to be non-zero. The problem is somewhat different. It is dividing by zero but not in step 8.
The step going from:
(A+B)*(A-B) = B*(A-B) Factor both sides
to:
A+B = B Reduce
is dividing by zero since it is essentially dividing both sides by (A-B) and it is given that A = B. Once you divide by zero, all bets are off and results are unpredictable.


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