mensacalgary.org » Puzzles Etc

PUZZLES

editor — Wed, 01 Sep 2010 00:34:29 +0000

1) Find the next number in the sequence: 10, 25, 39, 77, 679, 6788, …

2) Create a sequence starting with any two random numbers. Add them together to create the third number in the sequence. Add the third and second to create the fourth. Add the fourth and third to create the fifth. And so on. The ratio of consecutive numbers in the sequence (ie the second to the first – whatever they may be – the third to the second, fourth to the third, etc) converges; how quickly can you determine the number to which the ratio converges?

The answers to last month’s puzzles were supplied last month.

Here are the answers to this month’s puzzles:

1) Start with any number X containing two or more digits. Multiply all of X’s digits together. Take the result and multiply its digits together. Continue until your result is a single digit. The amount of times you have to multiply X’s digits to get down to a single digit is the “persistence” of X. Our question this month touches on persistence. The key to our sequence is that 10 is the smallest number with persistence one, while 25 is the smallest with persistence two, and so forth. Ten is the first term, because it’s the smallest number (of two digits or more), which reduces to one digit in one step. 25 is the second term, because it reduces to 10, which reduces to 0. Therefore 25 is the smallest number that reduces to a single digit in two steps. 39 is the third term, because it’s the smallest number that reduces to one digit in three steps (39 becomes 27 which becomes 14 which becomes 4). 77 reduces to one digit in four steps. And so on. The next number after 6788 is 68889 (it’s the seventh term and smallest number that reduces to one digit in seven steps). [from Here’s Looking at Euclid, by Alex Bellos, based on work by Neil Sloane]

2) They converge on the golden mean, phi, which is 1.618…[ from Here’s Looking at Euclid, by Alex Bellos]

PUZZLES

editor — Sun, 01 Aug 2010 03:34:31 +0000

1) To stretch your visualization skills, imagine any bizarre quadrilateral on a flat surface. The quadrilateral must, of course, consist of four straight sides that completely enclose a space. Join the midpoints of adjacent sides. What new shape have you created? Now return to the original quadrilateral in your mind and bisect its angles. Use pen and paper if you wish. Connect the points where the bisectors meet. What shape have you drawn? Finally, bisect the angles of this last shape and connect the points where the bisectors meet. What is this last shape you’ve created? [from Mathematical Amazements and Surprises, by Alfred Posamentier and Ingmar Lehmann]

2) Starting with a cube, consider how to create an object that is invisible and has infinite surface.

The answers to last month’s puzzles were supplied last month.

Here are the answers to this month’s puzzles:

1) The first shape is a parallelogram. The second is a rectangle. The third is a square.

2) One answer is the Menger sponge. Imagine that our cube has been notionally divided into 27 subcubes of equal size. Remove the subcube at the centre of each face and the subcube at the heart of the original cube. We’re left with a cube that has three square holes right through it. There are 20 subcubes remaining. Treat each of the remaining subcubes as we treated the original, and repeat the process again and again. The surface area keeps multiplying as the volume shrinks. At the limit of infinity, the surface area approaches infinity while the volume approaches zero. [from Here’s Looking at Euclid, by Alex Bellos]

PUZZLES

editor — Thu, 01 Jul 2010 00:34:56 +0000

1) Pick a random number between 0 and 1. Use a table of random numbers if you wish. Write the number down. Now pick another random number between 0 and 1. Add the two random numbers. Repeat the process. On average, how many random numbers are needed to make the total greater than one?

2) This is an appropriate place to repeat an old challenge. Start with a needle. Draw lines on a sheet of paper, which are exactly the length of the needle apart. Drop needles on the paper at random. If you multiply the number of drops by two, and divide by the number of times a needle touches or straddles a line, what is the result?

The answers to last month’s puzzles were supplied last month.

Here are the answers to this month’s puzzles:

1) The answer raises the bizarre nature of the number which we traditionally designate “e”, namely 2.71828… It crops up in the strangest places. The answer to the question is indeed “e”.

2) The answer, of course, is our old friend pi (3.14159…). For one analysis of the problem, see http://mste.illinois.edu/reese/buffon/buffon.html#intro

PUZZLES

editor — Tue, 01 Jun 2010 00:34:43 +0000

1) What is the next fraction in this sequence: 1/1, 3/2, 7/5, 17/12, 41/29, 99/70, 239/169, 577/408, 1393/985, …. ?

2) Does the sequence converge on any number, and if so what is that number?

The answers to last month’s puzzles were supplied last month.

Here are the answers to this month’s puzzles:

1) The next fraction is 3363/2378. The rule is add the numerator and denominator to form the new denominator, and add the numerator plus twice the denominator to form the new numerator.

2) The square root of two.

PUZZLES

editor — Sat, 01 May 2010 00:34:22 +0000

1) What is the smallest number that is divisible by 13, which – if divided by any of the numbers from 2 to 12 inclusive – leaves a remainder of 1? [from Jocoby and Benson’s Intriguing Mathematical Problems]

2) Two cyclists are headed directly towards each other at constant speed. At the beginning of Interval X, they are a quarter of a mile apart. One is traveling at 8 mph, the other at 12 mph. A mutant fly flits between the cyclists at a constant 30 mph, directly back and forth, reversing its direction each time without pause. At the beginning of Interval X, the mutant fly is just leaving the 8 mph cyclist, headed towards the other. The fly continues back and forth until it is crushed when the cyclists collide. Its death marks the end of Interval X. If we ignore the length of the bicycles, forward incline of the cyclists’ bodies and similar issues, how far did the fly travel during the period we are considering?

The answers to last month’s puzzles were supplied last month.

Here are the answers to this month’s puzzles:

1) We start by finding the smallest multiple of the numbers from 2 to 12. The answer is 27,720. The number we’re looking for must therefore be 27,720x + 1, where this total is divisible by 13 and where x is an integer. The number is therefore 2132x + (4x + 1)/13. If this total is divisible by 13, then (4x + 1) must be a multiple of 13. It is obvious that x = 3 provides the smallest integral value of x that satisfies the condition. The result is (27,720 x 3) + 1, which is 83,161.

2) There’s a lot of sleight of hand in this puzzle. All we have to know is that the riders are approaching each other at a combined 20 mph and are ¼ mile apart. Interval X will end when the cyclists meet, which will be in 1/80 of an hour. The fly moves at 30 mph. Distance = speed x time, so the fly travels (30 x 1/80) which is 3/8 of a mile during Interval X.

PUZZLES

editor — Fri, 02 Apr 2010 00:34:01 +0000

1) A woman with three daughters passes her neighbour’s house. The neighbour asks the daughters’ ages. The woman answers that their ages multiplied together is 36, and their ages added together is the same number as his address. The neighbour stares at his address. The woman then says she forgot to mention an essential piece of information. The information is that her eldest daughter’s name is Jenny. The neighbour now is able to determine the daughters’ ages. How does the neighbour do it? (NB We’re dealing only with whole integer ages.)

2) Here’s a classic alphametric from 1924. What numbers do the letters stand for?

SEND
+ MORE
MONEY

The answers to last month’s puzzles were supplied last month.

Here are the answers to this month’s puzzles:

1) The man makes a chart of the possible ages, consisting of the three numbers whose combined product is 36.

Daughter A’s Age	Daughter B’s Age	Daughter C’s Age	Sum of Ages
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

There is only one case in which the neighbour needs additional information, and that is if the sum of the ages is 13. The neighbour concludes that when the woman gave the essential information it was to differentiate between the two cases where the sum is 13. The statement that the woman’s eldest daughter’s name is Jenny means that there is only one eldest daughter. This eliminates the possibility of twins age 6. The three ages are therefore 2, 2, and 9.

9567
+ 1085
10652

PUZZLES

editor — Mon, 01 Mar 2010 00:34:14 +0000

2) Here’s a classic alphametric from 1924. What numbers do the letters stand for?

SEND
+ MORE
MONEY

The answers to last month’s puzzles were supplied last month.

Here are the answers to this month’s puzzles:

1) The man makes a chart of the possible ages, consisting of the three numbers whose combined product is 36.

Daughter A’s Age	Daughter B’s Age	Daughter C’s Age	Sum of Ages
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

9567
+ 1085
10652

PUZZLES

editor — Mon, 01 Feb 2010 00:34:29 +0000

1) Create a square consisting of four rows and four columns. Each “box” in the square contains a different number from one through 16. Every number appears only once. The sum of every row is identical; likewise the sum of every column, quadrant, the diagonals, and the sum of the central four boxes.

2) Why is a raven like a writing desk? Or, if Lewis Carroll’s famous conundrum doesn’t appeal to you, then answer this gentle query: you’re at a party. Can everyone at the party have a different number of friends present? For greater certainty, a person can’t be his or her own friend, and someone may have no friends.

The answers to January’s puzzles were supplied in the January issue.

Here are the answers to this month’s puzzles:

1) Hint: Albrecht Durer engraved a picture of this square, and the date of his engraving lies in the bottom row, central two boxes.

16 3 2 13
5 10 11 8
9 6 7 12
4 15 14 1

2) As to Lewis Carroll, he didn’t supply an answer, perhaps because there’s a “b” in both and an “n” in neither. As to friends and gatherings, at least two people will have the same number of friends at the party. The reason is that a friend is defined as someone else, not one’s self. In math terms, we can identify each person with a different number from A through (say) J. Assume person A has no friends in the gathering, person B has one friend, and so forth, through J who has J – 1 friends. But there is a contradiction between J, who counts A as a friend, and A, who claims to have none. To avoid the contradiction, J must have the same number of friends as one of the other people (J – 2 friends, or J – 3 friends, for example). Lest the issue of zero seem relevant, eliminate it and the result is the same. A has one friend, B has two, and so forth through J, who has ten. But this ten includes him or herself, which is excluded. J must therefore have the same number of friends as one of the other partygoers.

PUZZLES

editor — Fri, 01 Jan 2010 00:34:44 +0000

1) Here’s the scene. You and a math teacher face three wooden doors. Behind one is a valuable diamond bracelet. Behind the other doors are mud puddles. You don’t know what’s behind which door; the teacher does. You select a door, and the teacher – before opening the door you selected – opens another to reveal a puddle. The teacher offers you the opportunity to change your selection and win what lies behind the second unopened door, or stick to your first choice and win what lies behind that door. What should you do?

2) Every whole number can be obtained by multiplying a certain number of primes. For example, 48 requires 2×2x2×2x3, which is an odd number of primes. 49 requires 7×7, which is an even number. Every whole number is therefore odd or even in the number of primes required. By convention, 1 is considered even in this typology. As we rise through the whole numbers, do we encounter more even types, odd types, or are they equally balanced?

The answers to December’s puzzles were supplied in the December issue.

Here are the answers to this month’s puzzles:

1) The question is whether the odds are fifty-fifty that the bracelet lies behind one of the two unopened doors, and the answer is no. Work it through slowly. One time in three, the bracelet is behind the door you chose earlier. Two times in three, the bracelet lies behind the other unopened door.

2) The answer is peculiar. Except for #1, you find that you encounter either more odd types or an equal quantity of odd and even types. Until you reach 906,150,257. When for the first time you find that there have been more even types than odd. (We hope you didn’t test the numbers one by one.

PUZZLES

editor — Tue, 01 Dec 2009 00:34:43 +0000

1) Here is a mental exercise. First, imagine an equilateral triangle on a flat surface. Label the triangle’s corners A, B and C. Place one point of a compass on A and the other on B, and draw an arc from B to C. Draw similar arcs from C to A, and A to B. The result is an equilateral triangle whose corners are joined by arcs. What formula gives you the area of this curve-sided triangle? You may use paper to calculate.

2) What shape is produced if you rotate this curved triangle on a flat surface?

The answers to November’s puzzles were supplied in the November issue.

Here are the answers to this month’s puzzles:

1) The curved triangle you’ve imagined is called a Reuleaux triangle. If the straight length of its equilateral parent is x, we find the Reuleaux area by taking pi minus root 3, multiplying the result by x squared, then dividing by 2.

2) The area covered by rotating a Reuleaux triangle (as craftsmen will know) is a square! Well, not precisely a square, because of very slightly rounded corners. The actual area produced is 0.9877… of the area of a square. Close enough.

PUZZLES

editor — Sun, 01 Nov 2009 00:34:37 +0000

1) Mobius raised this classical problem in about 1840. As amended it reads: a king with five argumentative arrogant sons and five bright reasonable daughters died. His Will gave the kingdom to his five sons provided they, within thirty days, divided it into five parts such that each part had a common boundary with the other four. If they couldn’t do this, the kingdom as a single unit would go to his daughters to rule together. Of course, the sons quarreled and one after the other died mysteriously until just one remained. But even the best mathematicians in the realm couldn’t find a way to divide the kingdom as required, and after the expiry of the thirty days the kingdom passed to the princesses who governed it peacefully and happily. Question: was there a mathematical way for the princes to do what the Will asked?

2) If the answer to question 1 above is yes, what is the answer to the question: are four colours enough to make any map on a plane surface unambiguous?

The answers to October’s puzzles were supplied in the October issue.

Here are the answers to this month’s puzzles:

1) Try it. The answer is no.

2) If the answer to question one is yes, then more colours are needed. But the answer to question one is no and only four are required.

PUZZLES

editor — Thu, 01 Oct 2009 00:34:52 +0000

1) We are in an insane asylum where the only people we meet are doctors and patients. There’s a profound truth to this which we shall, for the moment, ignore. In the asylum, doctors and patients may be either sane (in which case they believe only what is true and know it as true, believing that what is false is false) or insane. The insane are entirely inaccurate in their beliefs (what they believe is true is false, and the propositions which they believe are false are true). Everyone says what he/she actually believes. In our first encounter, we hear a statement, which makes us believe that the speaker is a sane patient. We take immediate steps to have the person set free. Question: what is the simplest such statement?

2) We visit another asylum with identical inhabitants and interview four of them: Adel, Barbara, Cameron, and Denise. Adel believes that Barbara and Cameron are alike as to sanity. Barbara believes that Adel and Denise are alike as to sanity. We scratch our heads and ask Cameron whether he and Denise are both doctors, to which Cameron replies in the negative. Question: is there something wrong with this asylum?

The answers to this month’s questions appear below.

Answers
1) “I’m not a sane doctor.” There are other solutions, but unlikely more elementary than this. The explanation of the answer is that an insane doctor wouldn’t hold the stated belief if it were true, and a sane doctor wouldn’t hold the belief because it would be false. An insane patient couldn’t hold the belief, because the belief would be true. But if the speaker is a sane patient, the statement would be true and consistent.

2) If Adel and Barbara are both sane, then Barbara and Cameron are alike as to sanity, and so are Adel and Denise. This means that all four would be sane, in which case Cameron and Denise would be sane and thus alike. But suppose Adel and Barbara are both insane. This would mean that B and C are different from each other, and likewise A and D aren’t alike as to sanity. Hence C and D are both again sane and alike. Now suppose A is sane while B is insane. B and C would be alike, meaning that C is insane, but A and D would be different, meaning that D is likewise insane. Finally, if A is insane and B sane, then B and C are different from each other, which means that C is insane, while A and D are alike, which makes D also insane. Where have we got so far? If A and B are alike, then C and D are both sane. If A and B are different, then C and D are both insane. The logical possibilities mean that C and D are either both sane or insane. Whew! Now suppose they are both sane. This makes C’s statement true, the one in which C says that he and D are not both doctors. This makes either C or D a sane patient. If, on the other hand, C and D are both insane, then C’s statement is false, which means C and D are in fact both doctors, which makes them insane doctors. This asylum therefore contains at least one sane patient or two insane doctors.

PUZZLES

editor — Tue, 08 Sep 2009 00:34:44 +0000

1) What is the simplest way to make a hypercube (aka a tesseract)?

[The answers to August’s puzzles were supplied in the August issue. Here is the answer to this month’s puzzle.]

1) Start with the simplest object, ie a point. Connect two points to make a line, two lines to make a square and two squares to make a cube. Now comes the tricky part: place one cube above another and connect the cubes by adding six additional cubes, connecting the faces of the two original cubes. Yes this is difficult to draw, but it’s simple and real. [from Lisa Randall’s description in Warped Passages]

PUZZLES

editor — Sat, 01 Aug 2009 00:34:46 +0000

1. The following letters comprise the entire alphabet. Well, in fact, a few letters are missing, and those letters are the focus of this puzzle. Without writing anything on paper, can you rearrange the missing letters and find the word that is spelled? D F G I J K M N P Q T U V W X Y Z

2. This is an easy one. There are eight coins and seven weigh the same while one is lighter. You have a pair of balance scales. Identify the light coin in two weighings. And one of the possibilities has an alternative solution. What is it?

The answers to July’s puzzles were supplied in the July issue.

Here are the answers to this month’s puzzles:

Answers:

1. Bachelors

2. Place three coins on each side of the balance. If they are equal, the light coin is one of the remaining two. Place one of the remaining two on each side of the balance and the light coin is quickly found. If the initial weighing isn’t equal, the higher side contains the lighter coin. Take two of the coins from the higher side. Place one on each side of the balance. If they’re equal, the third coin is the light one. If unequal, the lighter is the odd one out. In the initial weighing, however, if the three coins on each side of the balance are equal to the other three, there’s an alternative next stage. Instead of weighing the other two coins, you could weigh one of them against one of the known equal coins. If equal, you know that the final coin is the odd one. If unequal, you also know the answer.

PUZZLES

editor — Wed, 01 Jul 2009 00:34:12 +0000

1) Four Mensans have sowed confusion in their city. They are Allan, Beatrice, Carlo and Denise. How have they sowed confusion? They have formed a SIG, that’s how. Well, not quite. The real source of confusion is that each of them has a cat named after one of the other three people, and a dog named after yet another of the three. You guessed correctly that no two dogs and no two cats have the same name. I can tell you that Denise’s dog is named after the person who owns the cat Carlo, and Carlo’s cat is named after the same person. Beatrice’s cat is named after the person who owns the cat in turn named after the person who owns the dog Allan. So…. who owns the dog Denise?

2) In Mensa Calgary’s ping pong tournament for 2008, there were 96 entrants. We devised a complex formula for the first round, which resulted in 80 entrants remaining after 32 games. Thereafter, we simplified: a person was eliminated after one loss. No matches were tied. How many games in all were played to determine the winner?

The answers to June’s puzzles were supplied in the June issue.

Here are the answers to this month’s puzzles:

1) If the names are given corresponding letters of the alphabet, this is the ownership chart:

Owner A B C D

Dog C A D B

Cat D C B A

Therefore Carlo owns Denise the dog.

2) This is the simplest and most straightforward puzzle in our series. We start with 32 games. Eighty players remained, and one of these were eliminated after each subsequent game. After the 32 games, 79 were required to determine the champion, meaning 111 games in all.

PUZZLES

editor — Mon, 01 Jun 2009 00:34:02 +0000

1) Using only the following numbers and signs, create an equation: 2,3,5,7,11,12,-,x,x,/,(),(),=

2) What are four words, each containing nine letters, that begin with “c” and have “ten” in their exact centre?

The answers to May’s puzzles were supplied in the May issue.

Here are the answers to this month’s puzzles:

Answers

1) ((7 x 11) – 5) / 12 = 2 x 3

2) contented, contender, contended, centenary

PUZZLES

editor — Fri, 01 May 2009 00:34:41 +0000

1) A little girl can take either one or two steps at a time. She can therefore climb a three-step staircase in three different ways: 1+1+1, 1+2, and 2+1. In how many ways can she climb an eleven-step staircase.?

2) Another simple classic: a man realizes he is dying. He calls for his children and asks them to divide his gold coins in the following manner. To the eldest child, he gives one gold coin and a 7^th of the remaining coins. To the second child, he gives two gold coins and a 7^th of what remains. To the third child, he gives three gold coins and a 7^th of what remains. And so on, giving coins and a 7^th of the dwindling remainder to each successive child. The last child receives all that is left. The children find, to their surprise, that each receives the same amount from their father’s estate. So…how many children were there, and how large was the estate?

The answers to April’s puzzles were supplied in the April issue.

Here are the answers to this month’s puzzles:

1) 144. We don’t have to count, of course, because the child’s possibilities form a Fibonacci sequence. As the sequence increases, the ratio of successive numbers approaches closer and closer to 1.6180339887…

2) The question appeared in Fibonacci’s Liber Abaci of 1202, and the answer is straightforward. If the estate is E and the first child receives x, the first child in fact receives 1+1/7 (E – 1). The second child receives 2+1/7 (E – 2 – x). The two shares are equal, so the two sums can be equated. The result allows us to cancel out E, showing that x = 6. Each child therefore receives six gold coins. If we equate 6 to 1+1/7 (E – 1), we find that the estate was 36 gold coins. The number of children is therefore 36/6 or six.

PUZZLES

editor — Wed, 01 Apr 2009 00:34:24 +0000

1) This is a simple classic. Peter is cleaning up after a marathon poker party and feels the need to smoke. He finds 29 cigarette butts in various receptacles. He recalls that four butts yield enough tobacco for one cigarette and begins to build cigarettes from the butts. He smokes as many cigarettes as he can make. How many butts will remain at the end of the clean-up?

2) Another simple classic. Five men, namely Anderson, Baker, Campbell, Draper and Emory dined with their wives at a circular table. Men and women alternated, and husbands and wives sat three places apart (ie two other diners sat between husband and wife in every case). Mrs Campbell was on Mr Anderson’s right. Mr Campbell sat two seats to the right of Mr Emory (ie one intervening diner). Mrs Baker was two places to the left of Mrs Emory. Who sat on Mrs Anderson’s left?

The answers to March’s puzzles were supplied in the March issue.

Here are the answers to this month’s puzzles:

1) 29 butts produce seven cigarettes with one butt left over. The seven cigarettes produce seven butts, which makes eight in all. These in turn yield two cigarettes, which produce two butts to be disposed of.

2) Mr Draper.

PUZZLES

editor — Sun, 01 Mar 2009 00:34:15 +0000

1) Viewed from the side, two ladders span a narrow alley like a lop-sided ‘X’. The first ladder (AB) runs from the base of one wall straight to (and part way up) the opposite wall. The second ladder (CD) runs from the base of this second building straight back to (and part way up) the first wall. The ladders are different lengths. The ladders cross at point E, which is directly above point F on the floor of the alley. What are the shortest lengths of each of these distances, measured in inches, such that all measurements are an exact number of inches?

2) The manager of a ranch asked a stable hand how many men were tending the horses in the corral. The stable hand said: “I saw 82 feet and 26 heads.” How many men and horses were there?

The answers to February’s puzzles were supplied in the February issue.

Here are the answers to this month’s puzzles:

1) The key is that EF (the height of the cross-point) must be the shorter leg of one 3-4-5 right-angle triangle and the longer leg of another such triangle at the base of the alley. EF therefore must be a multiple of 12. The shortest triangles (AEF and CEF), however, do not produce an exact number of inches for the other measurements. For this we take the values and simply multiply them all by 12. The cross point becomes 144 inches. The long ladder is 500 inches. The short ladder is 375 inches. The alley is 300 inches wide.

2) The most intriguing solution seems to start with the men. If the 26 heads all belong to men, that accounts for only 52 feet. But there were 82 feet in all. This leaves 30 feet to explain by horses. Each horse has two feet more than a man, which means that there were 15 horses. Returning to the heads, the total of 26 less the 15 horses leaves 11 men.

PUZZLES

editor — Sun, 01 Feb 2009 00:34:35 +0000

1) What two numbers add up to 10, but when multiplied give 40?

2) A village has baskets, all of equal size. When the community harvests its grain, it finds that three baskets of wheat, two of barley and one of flax total 39 kilos. Two baskets of wheat, three of barley and one of flax total 34 kilos. One basket of wheat, two of barley and three of flax total 26 kilos. How many kilos does one basket of each grain weigh?

The answers to January’s puzzles were supplied in the January issue.

Here are the answers to this month’s puzzles:

1) 5 plus the square root of minus 15. And 5 minus the square root of minus 15.

2) Wheat weighs 37/4 kilos, barley 17/4 and flax 11/4.

PUZZLES

editor — Thu, 01 Jan 2009 00:34:17 +0000

1) Jack wanted everything at his birthday party to go well so he prepared thoroughly. He started by placing a band of narrow red crepe paper, two feet long, on a table. Atop this band, he placed a blue one the same length. Lifting the right and left ends, he taped each together and examined what he’d done. In front of him was a blue ring, its ends taped together, sitting inside a red ring, also with its ends taped together. No, he thought to himself, that’s not sophisticated enough. He tried again, placing a piece of blue paper atop a red one as before, but this time before taping the ends together, he twisted the right ends through 180 degrees. He sat back. Now the taped ends were blue to red, and red to blue. Could he move a pencil all the way around between the inner and outer bands of paper? When Jack gently pulled the nested bands apart, how many pieces did he have?

2) Jack always tells the truth. He has rolled a pair of dice and says, “I didn’t roll any fours.” What are the odds that he has rolled a pair of sixes?

The answers to December’s puzzles were supplied in the December issue.

Here are the answers to this month’s puzzles:

1) Yes, he could move a pencil all the way around, separating the inner and outer bands. But when he gently separated the bands he found there was only a single long one, though it had four half-twists.

2) Two dice have 36 possible combinations. Counting them off, eleven of them include at least one four. This leaves 25 possible throws, of which one is the double six we’re looking for. The odds are therefore 1/25.

editor — Mon, 01 Dec 2008 00:34:25 +0000

1) Your task is to create an iron chain that consists of 15 links. You start with 5 chains that have 3 links each. It costs a dollar to cut a link and 50 cents to reweld a cut link. What is the cheapest cost to create the 15 link chain?

2) A wealthy benefactor in 2001 gave each member of Calgary’s Flying Saucer Club a few sacks of gold coins. She gave the same number of sacks to each member, but then added extra sacks for each female member. The benefactor increased the amount of these gifts year by year over the next three years. The benefactor in 2004 gave a total of 400 sacks of gold coins. That year, there were 21 male members and each received 11 sacks of gold coins. How many female members were there in 2004?

The answers to November’s puzzles were supplied in the November issue.

Here are the answers to this month’s puzzles:

1) Cut each link in one of the 3-link chains. Join four of the 3-link chains with the 3 cut links. That’s 3 cuts and 3 welds, ie $4.50.

2) If 21 males received 11 sacks each, the benefactor gave them 231 in all. This means 169 sacks were given to the women. 169’s factors are 1, 13 and 169, which means that the women members received 13 sacks each and there must have been 13 women. QED. Kudos and admiration if anyone can predict the value of these coins on December 31, 2009.

editor — Sat, 01 Nov 2008 00:34:49 +0000

1) A mouse has built a home inside a perfect circle of straw. The diameter is 5 feet. The door is at the point furthest north. His armchair is one foot due south of the door. The mouse gets out of the chair and walks west till he reaches his wall. Then he walks south for a foot and a half to his fridge. How far is the fridge from the chair?

2) A mouse stores up grain for winter. On Monday, she stores x bags full. On Tuesday, twice that number. On Wednesday, twice the total so far filled, and so each day doubling the total of all previous days. By the end of the seventh day, one-third of the bags were filled. How long in all did the job take?

The answers to October’s puzzles were supplied in the October issue.

Here are the answers to this month’s puzzles:
1) The walk from chair to fridge places the mouse on the east-west diameter of his house. His walk forms two sides of a rectangle, and the diagonal is the circle’s radius, hence two and a half feet.

2) Each day, the mouse doubles the number of all previous days. On day eight, all previous days amounted to one-third. The mouse therefore filled two-thirds on day eight, and the job accordingly took eight days.

editor — Wed, 01 Oct 2008 00:14:17 +0000

1) Three bright children are in a Mensa classroom. The teacher says, "You’re all obedient and very intelligent. Here’s a little IQ test that is fair to each of you. Close your eyes. I’m going to place a red or blue hat on each of your heads. When I say ‘open’, you will open your eyes. If either of the other children is wearing a red hat, clap your hands once. When you know the colour of the hat on your own head, clap twice more. The children, being Mensa members, understood the instructions clearly. They closed their eyes, and the teacher put red hats on each of their heads. The teacher said ‘open’. The children opened their eyes. They all clapped once immediately. After a few moments, the third child clapped twice more and said his hat was red. How did the child know?

2) The same classroom situation as before. This time, after the children opened their eyes, the third child clapped three times right away and said all the hats were red. How did the child know??

The answers to September’s puzzles were supplied in the September issue.

Here are the answers to this month’s puzzles:
1). The third child reasoned that since all three clapped, there had to be at least two red hats. If the third child had a blue hat, child #1 or #2, being very intelligent, would know the other two hats were red and his/hers was therefore red. But neither of them clapped. Therefore, #3’s hat was red.

2) While the teacher talked, the third child reasoned: the test is fair. But if the teacher gives us hats that aren’t the same colour, the test won’t be fair. So all the hats will be either red or blue. As soon as the child saw that the other hats were red, the child knew his/hers was red too.

editor — Mon, 01 Sep 2008 00:09:40 +0000

1) A fly lands in a room that measures 30 feet (length) by 12 (width) by 12 (height). When his IQ rises, he finds himself on the end wall, 1 foot from the ceiling, 6 feet from each side. He spots a female fly on the opposite end wall, 1 foot from the floor and 6 feet from each side. To impress the lady, he wants to walk to her by the shortest path. How far must he walk?

2) Take a chessboard and remove two squares from opposite corners. Divide the rest of the board into oblongs that are 2 squares long and one wide. Or can you?

The answers to August’s puzzles were supplied in the August issue.

Here are the answers to this month’s puzzles:
1) Picture the scene in 2 dimensions. We have 4 rectangles, each immediately above the other, and each 30 by 12. To the left of the second from the top, is a square that is 12 by 12. To the right of the bottom rectangle is another square the same size, ie 12 by 12. This is the room reduced to 2 dimensions. We can now draw the straight line which the fly must walk. We can also construct the right-angled sides of which the straight line is the hypotenuse. The triangle’s sides turn out to be 24 by 32 by 40, which is the familiar 3-4-5 right-angled triangle. Therefore, 40 feet.

2) The board’s 64 squares alternate white and black. This means that opposite corners are the same colour. When you remove two white or two black squares, you leave an unbalanced set of pairs behind. Oblongs (2 squares by 1) always contain two colours and therefore there will always be two white or two black squares left over, the opposite colour from the one you took.