2) By accident, a computer that was handling certain calculations dropped all exponents to the base line, so that – for example – the number 7 (to the power 5) appeared as 75. Is there a calculation or value that isn’t affected by such a computer glitch?

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

Here are the answers to this month’s puzzles:

1) The number describes itself when you read it: one zero, two ones, three twos, two threes.

2) 2 (power 5) x 9 (power 2) = 2,592 There isn’t a formula I’m aware of for this answer; you just have to stumble across it. [Adapted from Posamentier and Lehmann’s Mathematical Amazements and Surprises]

2) This is typical of the count-carefully genre of puzzle. We’re going to cross the Atlantic by boat from New York to London. A ship leaves London every day at four pm (GMT) bound for New York and arrives exactly seven days later. Similarly, at four pm GMT every day, a ship (including ours) leaves New York for London and arrives exactly seven days later. All boats follow the same route and avoid collision by trivial deviations. How many ships from London does our vessel encounter during its voyage? Don’t include ships that arrive just as ours leaves New York, or those that leave London just as ours arrives?

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

Here are the answers to this month’s puzzles:

1) The definition from elementary science is: power = work/time. But time is money (one cliché), which means that power = work/money. If knowledge is power (another cliché), then knowledge = work/money. Expressed differently, for a fixed amount of work, knowledge and money are inversely related; the more you know, the less you’re paid. [Adapted from Ian Stewart’s Hoard of Mathematical Treasures]

2) 13. Assume that our ship leaves New York on January 10. It arrives January 17. The ship that left London on January 3 arrives in New York just as we depart, so we don’t count that one. While on the ocean, we meet the vessels that left London on January 4 through to January 16 inclusive. 13 vessels. [Adapted from Ian Stewart’s Hoard of Mathematical Treasures]

2) A mathematician’s office was interested in creating fractions based on the frequency that digits occur. The office prepared a form on which appeared the ten digits from 0 through 9 in a row like this: 0 1 2 3 4 5 6 7 8 9. When the office considered the number 2,372,208, it noted that this number has one 0, so on the form, the staff placed the denominator 1 below the printed 0. There are no ones in 2,372,208, so on the form below the printed 1 the staff placed the denominator 0. There are three 2s in 2,372,208, so below the printed 2 on the form the staff placed the denominator 3. And so on. The form for the number 2,372,208 gave rise to the denominators 1 0 3 1 0 0 0 1 1 0. One day, a staff person was startled to notice that a number consisting of multiple digits was exactly the same as the row of denominators it created on the form. What was the number?

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

Here are the answers to this month’s puzzles:

1) This just takes a little thought. On the straight segments, we could draw rectangles showing that Robert travels the same distance as Winona. At the corners of the large triangle, however, Robert has to move quickly along the arc of a circle to stay 10 meters to her right. Robert’s three arcs amount to a full circle, and therefore – at the corners – he runs the circumference of a circle whose radius is 10 meters. He therefore runs 20 x 3.14 meters, or 62.8 meters farther than Winona. If Winona runs a triangle that is 15 meters on each side, Robert would again run 62.8 meters more than she does. His extra distance is independent of how far she runs. It depends only on how far apart they are. [From Paul J. Nahin’s Dr. Euler’s Fabulous Formula]

2) 6210001000

2) Consider a manic bird, which flies straight from point A to point B and back again. He does it repeatedly and never diverts from his path. A and B are distance d apart. The bird’s speed, relative to the ground, is always V when there is no wind. The time to make the round trip, when there is no wind, is obviously T = 2d/V. So far, so good. Now suppose that there is a steady breeze blowing directly from A to B at speed W. The breeze is always slower than the bird’s independent speed V. When flying with the wind from A to B the bird’s ground speed is V + W, right? And when flying against the wind from B back to A, the bird’s ground speed is V – W. But here is the question: what is the impact of the wind on the bird’s total flight time?

The answers to last month’s puzzles were supplied last month. Here are the answers to this month’s puzzles:

1) The cumulative product of the square of even integers divided by the product of adjacent odd integers, all rational, produces the transcendent number pi/2. The formula is known today as Wallis’s product formula for pi. [From Paul J. Nahin’s Dr.Euler’s Fabulous Formula, Princeton Univ. Press, 2006)]

2) There’s no effect, because the influence of the wind balances out. Is that your thought? But consider that the total flight time is given by d/(V+W) + d/(V–W). By elementary algebra, this reduces to 2Vd/(Vsquared–Wsquared). Divide numerator and denominator by Vsquared and the result is (2d/V)/1 – (W/V)squared], which is equal to T/[1–(W/V)squared]. This shows that the time for the round trip is always increased when there’s a steady wind. [From Paul J. Nahin’s Dr.Euler’s Fabulous Formula, Princeton Univ. Press, 2006)]

2) A truel is a duel fought by three persons, in this case Mssrs. Black, Gray and White, who have congregated at dawn, armed with pistols that fire three shots each. They are determined to kill each other. Mr White is deadly; he always hits his mark. Mr Gray strikes the target two times in three. Mr White is successful one shot in three. Any shot that hits a person is fatal. They agree that Mr Black will fire first, then Mr Gray, and finally Mr White, rotating until only one survives. Despite this situation, all are fully rational and know the facts recited above. Mr Black considers where to aim. What should he do?

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

Here are the answers to this month’s puzzles:

1) A cloud. [The riddles of the 7th century scholar, St Aldhelm, survive in a 10th century manuscript at Canterbury Cathedral. With thanks to Lacey & Danziger’s The Year 1000]

2) If Mr Black tries to hit Mr Gray and succeeds, Mr White shoots next. Mr Black is his only target and therefore Mr Black is a dead man. It’s better for Mr Black to aim at Mr White. If Mr Black is successful, Mr Gray fires next. But Mr Gray hits the target only two times in three, so Mr Black may survive and return fire with a chance at winning. But there’s a superior option. Mr Black should fire into the air. Mr Gray’s turn is next, and he’ll aim at Mr White because Mr White is a fatal antagonist. If Mr White survives Mr Gray’s shot, Mr White will target Mr Gray because the latter is more dangerous than Mr Black. This allows Mr Black the chance to eliminate Mr White. The best strategy is for Mr Black to arrange things so that instead of the first shot in a truel, he has first shot in a duel: Mr Gray or Mr White will have died, and Mr Black will aim at whoever survives.

2) Create a sequence of 100 numbers that aren’t prime. Sequence here means a list of counting numbers (whole numbers) in which each successive number is one unit higher than the preceding number.

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

Here are the answers to this month’s puzzles:

1) 135 and 176. The series is based on the ways in which a given number of objects can be divided. For example, consider how to divide three objects into separate piles. They can be divided 1-1-1, 2-1, or 3-0. Three is therefore said to have 3 partitions, and the series 1,2,3,5,7 etc (above) might be called the partitions of the counting numbers from 1 to 15. There’s more than trivial interest at stake here, because partitions crop up in the real world almost as often as the Fibonacci series. A modern illustration of their usefulness lies in the density of energy levels in certain quantum systems. If you didn’t find this puzzle a challenge, work out the partitions of 200. How many ways can 200 objects be divided? I’ll place the solution at the end of the next answer.]

2) There’s an easy way to do this. Take all the numbers from one through 101 and multiply them together to find factorial 101. Factorial 101 is divisible by all the numbers from 1 to 101. If N is any number from one through 101, then 101! + N will also be divisible by N. So all the numbers from 101! + 2 through 101! + 101 are not prime. Hey presto, you have 100 numbers that aren’t prime. You don’t have to start with one through 101, of course. Any similar sequence of successive counting numbers (a,b,c,…) is good enough. Just create the factorial then add a, b, c, …successively to it. As to the partitions of 200, there are 3,972,999,029,388 ways to divide 200 objects.

[Both puzzles this month come from Marcus du Sautoy’s The Music of the Primes]

2) While we’re on the subject of friendly numbers, can you find a pair between 1,000 and 1,300?

The answers to last month’s puzzles were supplied last month. Here are the answers to this month’s puzzles:

1) These are numbers whose factors add up to the other. The factors of 220 are 1, 2, 4, 5, 10, 11, 20, 22, 44, 55, and 110. Add them up and the result is 284. The factors of 284 are 1, 2, 4, 71, and 142, which makes 220. This was an intimate and secret bond that appealed to Europe during the middle ages, reflecting a general belief in magic and numerology. The numbers 220 and 284 were written on charms to foster love. At other times and in other places, 220 would be carved on one fruit and 284 on another. It was thought that eating one and offering the other to a lover would act as an aphrodisiac. No other friendly numbers (as they are called) were found until 1636 when Fermat identified the pair 17,296 and 18,416. [from Fermat’s Enigma by Simon Singh]

2) The answer is 1,184 and 1,210, discovered in 1866. Other friendly numbers include 9,363,584 and 9,437,056 (found by Descartes); if you have an idle moment, try to identify the factors in one that point to the other. [from Fermat’s Enigma by Simon Singh]

2) While we’re pondering numbers that end with 25, let’s consider 425. Can you find the distinct positive integers whose sum equals 425 and whose reciprocals add up to 1? Fool around with a few numbers and see how close you can come.

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

Here are the answers to this month’s puzzles:

1) The question contains a clue. There are only two such three-digit numbers: 376 and 625. The square of the former is 141,376. The square of the latter is 390,625. [from Mathematical Amazements and Surprises by Posamentier and Lehmann]

2) The answer is of course: 3,5,7,9,15,21,27,35,63,105,135. [from Ian Stewart’s Hoard of Mathematical Treasures]

2) Speaking of unusual numbers, the series 1-1+1-1+1-1+…seems conventional, yet has something strange about it. How many answers are there if we add and subtract as indicated?

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

Here are the answers to this month’s puzzles:

1) This is the smallest number that can be expressed as the sum of the cubes of two positive integers (ie as x cubed plus y cubed where both x and y are positive) in two different ways. The first way is 1 (1 cubed) plus 1,728 (12 cubed). The second way is 1,000 (10 cubed) plus 729 (9 cubed). 1729 is known as the Hardy-Ramanujan number. The famous English mathematician Godfrey Harold Hardy (1877-1947) was visiting his colleague Srinivasa Ramanujan (1887-1920) in hospital and remarked that the number of his taxi, 1729, seemed dull. Ramanujan replied to the contrary that it was very interesting, giving the explanation above. By the way, consider carefully: if we accept the cubes of negative integers, what is the smallest number that can be expressed as the sum of two cubes in two different ways? the answer is 91. 91 is the sum of 6 cubed plus –5 cubed, as well as the sum of 4 cubed plus 3 cubed. And 91 is a factor of 1729! [borrowed with thanks from Alfred S. Posamentier and Ingmar Lehmann’s Mathematical Amazements and Surprises]

2) The result depends on how you group the series. Bracketed as (1-1)+(1-1)+(1-1)+… , it reduces to 0+0+0+… which is 0. But grouped differently, the result changes. For example, 1+(-1+1)+(-1+1)+(-1+1)+… is 1+0+0+0…, which surely equals 1. On the other hand, let’s make the sum = s. Then s = 1-(1-1+1-1+1-1+…) = 1-s. Shift s over to the left side of the equation, and we find that 2s = 1, which makes s = ½. There’s an element of sleight of hand here, by virtue of the fact that infinite series cannot be added and subtracted according to commonplace rules, which shows that infinite care must be taken with them. [from Ian Stewart’s Hoard of Mathematical Treasures]

2) Here’s an old conundrum. Where on earth can you travel one mile south, then a mile east, then a mile north, and end up where you began? Add a Mensan sophistication to your answer.

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

Here are the answers to this month’s puzzles:

1) Multiply the number by 2. Or by 3. By 4, or in fact by any number from 2 through 16. The result is the same sequence of digits but starting at a different place. For example, multiplied by 3, you get 1764705882352941. What happens after 16?. Well, for a start, multiplied by 17, the result is 9999999999999999. The number (0588235294117647) is the decimal expansion of 1/17, repeated ad infinitum. [borrowed with thanks from Ian Stewart’s Hoard of Mathematical Treasures]

2) The conventional answer is the North Pole. You voyage a mile south, then head east. The eastward journey takes you along a latitudinal line (ie a line on the globe that is equidistant from a Pole). Go one mile north, and you’re back where you started. But there’s actually an infinite number of solutions. As preparation for a sample alternative, consider the latitudinal circle, which is one mile in circumference and located near the South Pole. Call this circle A. Walk one mile north along any great circle route, and find latitudinal circle B. Now we’re ready to begin. If you take any point on circle B and walk a mile south, you’re back at circle A. Walk a mile east and you’ve circumnavigated circle A. Going north a mile takes you back where you started. If circle A were half a mile in circumference, you could walk around it twice to make a mile. By extension, you can see the infinite number of answers to the puzzle. [borrowed with thanks from Alfred S. Posamentier and Ingmar Lehmann’s Mathematical Amazements and Surprises]

2) Consider the moon. Between each new moon, there are seven stages, which have names in English. What are they? At the one-quarter and three-quarters points in a lunar month, how much of the moon’s face is visible?

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

Here are the answers to this month’s puzzles:

1) 496 is next. And the one after is 8,128.

2) Crescent, first quarter, waxing gibbous, full moon, waning gibbous, third quarter, crescent. At the one-quarter and three-quarter points, half the moon’s face is visible.

[Both puzzles this month come from Ian Stewart’s Hoard of Mathematical Treasures]

2) From the numbers one through ten inclusive, can you choose the sides of five rectangles – each number chosen only once – such that you can assemble the rectangles to make an 11 x 11 square without overlapping the rectangles.

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

Here are the answers to this month’s puzzles:

1) Fibonacci showed in 1202 that the answer is yes. His method is called the greedy algorithm. The algorithm begins by finding the largest unit fraction that is less than or equal to the fraction you want to represent. Subtract the fraction from the original fraction. Now repeat the process for the remainder, looking for the largest unit fraction each time that is different from the ones you found earlier. Suppose, for example, that you want to represent 6/7. The largest unit fraction that is less than or equal to 6/7 is ½. Subtract this from 6/7. The result is 5/14. What is the largest unit fraction different from ½ that is less than or equal to 5/14? The answer is 1/3. Subtract 1/3 from 5/14. The result is 1/42. Obviously, the largest unit fraction different from ½ and 1/3 that is less than or equal to 1/42 is 1/42 itself. So the algorithm stops. Putting the pieces together, 6/7 equals ½ plus 1/3 plus 1/42. Amazingly enough, every fraction less than one is also a sum of unit fractions with distinct even denominators. And it looks like the same is true for distinct odd denominators, but we don’t yet have a proof.

2) Ignoring rotations and reflections, we’re aware of two solutions. The first was found by M. den Hertog, the second by Bertie Smith. They are respectively rectangles with sides: 1 x 6, 2 x 10, 3 x 9, 4 x 7, and 5 x 8; and 1 x 9, 2 x 8, 3 x 6, 4 x 7, and 5 x 10. Interestingly (and as a hint), in assembling the 11 x 11 square, neither case has any rectangle butted like the base of the letter T.]

[Both puzzles this month come from Ian Stewart’s Hoard of Mathematical Treasures]

1) First, appropriate to the holiday season just past: what sound does a drowning mathematician make?

2) Next, we mull over a small community of intelligent monks with embarrassment issues: each blushes dramatically if he has a blob painted on his forehead, but none can say anything that might embarrass anyone. The community is in straitened circumstances. Only three monks are left, namely Alfred, Benedict and Cyril. Their novice Dominic creeps into their cells one night and paints a blob on each of their foreheads. He then leaves the monastery. The three monks wake up in the morning and are together all day. They wonder about their foreheads, but no one can be sure, so no one blushes and no one mentions the possibility of blobs. A visitor, who is known always to tell the truth, arrives in the evening and says, “At least one of you has a blob on his forehead.” (a) Does that remark make a difference to the monks? To close off the range of possibilities, the monks aren’t visually impaired and there are no reflective surfaces in the monastery. (b) Assume that the community has 100 monks instead of three. The visitor makes his comment and continues, “I’m going to ring a bell every ten seconds until someone blushes.” After how many rings does someone blush and why?

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

Here are the answers to this month’s puzzles:

1) Log, log, log, log, log, log, log, log, log, log.

2 (a) Of course it makes a difference. Alfred thinks: if I don’t have a blob, then Benedict sees a blob on Cyril but not on me. Benedict will ask himself whether he has a blob and will think, “If I, Benedict, don’t have a blob, then Cyril sees that neither Alfred nor I have blobs, so he will deduce that he has a blob and will blush. But Cyril hasn’t blushed. Therefore I, Benedict, must have a blob.” But Benedict hasn’t blushed, therefore Alfred knows that he, Alfred, has a blob on his head. Each monk reasons in similar vein, so they all blush after the visitor’s remark. 2 (b) Monk number 100 sees that the other 99 all have blobs. He reasons, “If I don’t have a blob, then the other 99 all know this. I’m therefore out of the reckoning. The remaining 99 will make similar deductions, and after 99 rings they will all blush.” But they don’t blush after the 99th ring. Monk 100 then knows that his assumption is wrong and that he must have a blob. He blushes. All the other monks reason similarly and blush at the same time.

[Both puzzles come from Ian Stewart’s Hoard of Mathematical Treasures]

2) There’s a class of puzzle in which you begin with a particular word and, changing one letter at a time, arrive at another concluding word. I call them CLOTS, ie Change Letters One at a Time. Yes, each intervening word must also be a word. The challenge is to arrive at the specified result in the shortest number of steps, when it’s possible to get there at all. Example: start at CATS and arrive at CAST. The shortest process is CATS-CARS-CART-CAST. Now, try to transform SHIP into DOCK, and ORDER into CHAOS? What is the least number of steps?

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

Here are the answers to this month’s puzzles:

1) 370, 371 and 407

2) We can’t be sure that these answers are the shortest paths, but they are the best we can do. Anyone have better answers? SHIP-SHOP-SHOT-SOOT-ROOT-ROOK-ROCK-DOCK. ORDER-OLDER-ELDER-EIDER-CIDER-CODER-CODES-CORES-SORES-SORTS-SOOTS-SPOTS-SHOTS-SHOPS-SHIPS-CHIPS-CHAPS-CHAOS. SOOTS is formed from the verb “to soot” which means what it seems. Both puzzles this month derive from Ian Stewart’s Hoard of Mathematical Treasures.

2) Anton takes his ten hungry children to the baker, but there’s only one loaf of bread left. The bread is perfectly spherical. The baker has a slicer, which cuts the loaf into pieces of equal thickness. When the baker has finished, there are ten slices. The end pieces seem to have more crust than the middle ones, but which child in fact receives the most crust and why?

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

Here are the answers to this month’s puzzles:

1) The core of this game/puzzle is that each curve reduces the potential rays by two, while adding one potential ray (the new dot). It therefore matters whether we start with an even or odd number of dots and players. The arrangement of the dots doesn’t matter, because curves may vary in shape. But it’s important whether we’re allowed to draw lines that cross other lines; if yes, we avoid the Bridges of Konisberg problem. In our example which begins with three dots, eight curves can be drawn before we run out of options. There are two players, and Barbara went first. She will therefore lose.

2) All slices have the same amount of crust. The easiest way to see this is to compare the sphere with a cylinder whose height and diameter match the sphere. The surface area of the corresponding slices of sphere and cylinder are the same.

2) And an old favorite, albeit not much of a challenge: ten dogs break into a garden and dig ten holes. Assume that five of them broke into a garden last month and dug five holes in five days. If all dogs dig at the same rate all the time and all holes are the same size, how long did the ten dogs take to dig the ten holes?

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

Here are the answers to this month’s puzzles:

1) Yes. Round wheels keep a bicycle level on a flat road. For square wheels, the road must consist of a series of inverted arcs, concave down, that abut each other. More exactly, the surface must be shaped like downward facing catenaries. A catenary is the u-shape formed by a hanging chain. If the inverted catenaries all meet at right angles, square wheels of the right size will fit snugly and the centre of the wheels will remain level as the bicycle rolls along.

2) Five days. It takes a dog five days to dig a hole.

2) Mathematical Jones was intrigued by prime numbers. He wanted to prove the Riemann hypothesis, but began with an easier problem. Of all choices, he landed on the following: find a method that will result in a sequence of consecutive numbers, as long as you like, that do not include a prime. He thought for five minutes, then smiled. He’d solved the problem. What method did he use?

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

Here are the answers to this month’s puzzles:

1) The answer is one of the mysteries of our universe. We don’t know why “pi” (3.14159) should be pervasive in our number system, along with “e” and other irrational constants. Curvature of our space can’t be the cause. Nor can it be that cardinality isn’t a quality of our local region of space. But sure it is that the sum of the reciprocals of the squares of the positive integers is pi squared divided by 6.

2) Mathematical Jones tried to find a sequence of 200 consecutive numbers without a prime. To locate such a sequence, he started with 1 and multiplied all the numbers from 1 through 201, resulting in factorial 201. Call this product X. X isn’t prime, because it’s divisible by all the numbers from 1 through 201. Jones then added 2 to X, giving the next number of his sequence. Both X and 2 are divisible by 2 and therefore this number isn’t prime. He added 3 to X; (X + 3) is divisible by 3 and therefore this also isn’t prime. And so on consecutively through 201. None of the numbers (including X) is prime. Jones realized that he could make the series of consecutive numbers as long as he liked by the selection of his starting sequence.

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]

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]

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

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.

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.

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.

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.

2)

    9567
+ 1085
 10652

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.

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.

2)

    9567
+ 1085
 10652

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.