I apologize for the irregular posting over the past few months. I've had to deal with some personal issues (don't worry, everything is fine!). The good news is that, with this entry, everything should start returning to normal.
Having said that, let's dive into August's snippets!
• We can't ignore Katie Steckles' game video after all that! Katie teaches 2 games (or does she?). The first one involves numbered fishes, and the second one involves cards with stars and moons on them:
• There's usually more than one way to use your knowledge. In my tutorial about mental division, I teach a simple method for mentally dividing the numbers 1 through 6 by 7. Presenting it as an exacting feat of mental division is one thing. How else could you present it? Take a look at how Scam School presents the same feat:
If you watch the full explanation, you'll notice another difference between the way I teach it and the way Brian teaches it. He puts emphasis on the last digit, which works well for performing the feat this way. In my version, I teach how to work out the first few digits, as you'll need those first when giving the answer verbally. This is a good lesson in the benefits of changing your point of view!
Back in January of 2012, I wrote about the Chinese Remainder Theorem. Also, Martin Gardner taught the basics well in his book Aha!: Insight, including a trick where you can determine someone's secretly chosen number between 1 and 100 just from hearing the remainders when divided by 3, 5 and 7. Going over that post again, I've developed a few improvements to this trick that make it seem much more impressive, and maybe even easier to do.
The first problem is getting the remainders. With a standard calculator, it's not easy to do. The answer here is to simply have them divide the number by 3, 5 and 7, and have them tell you ONLY the number after the decimal point. Using the amount after the decimal point, you can work out the remainder. When dividing by 3, there are 3 possibilities for numbers after the decimal point:
Nothing after the decimal point: Remainder = 0
Number ends in .3333...: Remainder = 1
Number ends in .6666...: Remainder = 2
To find the remainders when dividing by 5:
Nothing after the decimal point: Remainder = 0
Number ends in .2: Remainder = 1
Number ends in .4: Remainder = 2
Number ends in .6: Remainder = 3
Number ends in .8: Remainder = 4
If you read my Mental Division: Decimal Accuracy tutorial, you'll get familiar with the 7s pattern. It's trickier than 3 or 5, but easily mastered. You only need to pay attention to the first 2 digits after the decimal point:
Nothing after the decimal point: Remainder = 0
Number ends in .14: Remainder = 1
Number ends in .28: Remainder = 2
Number ends in .42: Remainder = 3
Number ends in .57: Remainder = 4
Number ends in .71: Remainder = 5
Number ends in .85: Remainder = 6
Using the decimals makes the trick seem more difficult from the audience's point of view, but it only requires a little practice to recognize which numbers represent which remainders.
The other improvement involves the process itself. Have them start by dividing their number by 7 and telling you the numbers after the decimal point. Using the steps above, you can quickly determine the remainder from 0 to 6. If the remainder is between 0 and 5, you can remember them by touching that many fingers of your left hand to your pant leg (or table, if present). For a remainder of 6, just touch 1 finger from your right hand to your pant leg or table.
What ever the remainder, imagine a sequence starting with this number, and adding 7 until you get to a number no larger than 34. For example, if the person told you their number ended in .42, you know the remainder is 3, and the sequence you'd think of is 3, 10, 17, 24 and 31 (we can't add anymore without exceeding 34). If the remainder was determined to be 4, instead, the sequence you'd think of would be 4, 11, 18, 25 and 32.
Next, ask the person to divide by 5, and tell you the part of the answer after the decimal point. Once you get this number, recall your earlier sequence (which can be recalled via the number of fingers resting on your pant leg or table), and subtract the new remainder from each number, until you find a multiple of 5. For example, let's say that when dividing by 7, their remainder was 4, so the sequence was 4, 11, 18, 25 and 32. Let's say that when dividing by 5, their remainder was 3. Which number from your initial sequence, when it has 3 subtracted from it, is a multiple of 5? is 4-3 a multiple of 5? No. Is 11-3 a multiple of 5? No. Is 18-3 a multiple of 5? YES! Now you have the number 18.
Whatever number you have at this point, is the smallest of 3 possible numbers. The other 2 possibilities are 35 more than that number and 70 more than that number. At this point, you can know that the chosen number is either 18, 53 (35+18) or 88 (70+18). The remainder when dividing by 3 will determine which one of these is the correct answer. For example, if they say their number, when divided by 3, ends in .3333..., you know the remainder is 1. So, run through all 3 numbers quickly and ask yourself which one is 1 greater than a multiple of 3. Is 18 - 1 a multiple of 3? No. Is 53 - 1 a multiple of 3? No. Is 88 - 1 a multiple of 3? YES! Therefore, their number must be 88.
There are certainly other approaches. In fact, a magician's magazine called Pallbearer's Review once presented this trick as a challenge, asking their readers to supply their methods. They received a wide variety of entries and many approaches. Most of them involved far more difficult methods than the above approach, which I prefer for actual performance.
As you've no doubt guessed, this puzzle involved the yin (dark) yang (light) symbol. For this puzzle, I've drawn it in a very mathematically precise way over at Desmos. The outside is a unit circle (so, the radius is 1 unit), the main semicircular divisions of the design have a radius of 1⁄2 unit, and the opposite-color dots have a radius of 1⁄6 unit. Here's the challenge: What's the equation of the line that divides the design so that each side of the line contains exactly equal amounts of the dark and light portions?
As with many puzzles, this one seems hard, until you break it down into simpler steps. Let's start with a much easier puzzle: If the top half of the puzzle were dark, and the bottom half light, as in this rendering, where would you draw the line? The answer is easy. It should be a vertical line, so the equation would be x=0.
Next, we change the design a bit, so as to get closer to the yin yang symbol. Starting with the previous design, we cut a half circle (1⁄2 unit radius, remember?) of the dark portion from the right side, and add that to the left side, giving us the design below. The vertical line obviously won't work anymore, and we'll need to rotate that line by some amount to compensate, but how much?
Again, the secret is to take steps slowly. If you remember your high school geometry, you remember that the formula for a circle is A=πr2, and that our design as a whole, being a unit circle, has an area equal to π.
The formula for the semicircle we've moved, then, is A=1⁄2πr2. Plugging in 1⁄2 for the radius, we get π⁄8 units. So, to compensate for π⁄8 units out of a full circle with an area of π-units, we simply rotate the formerly-vertical line counterclockwise by 1⁄8 of the circle, or 45°! The upper left quadrant completely dark, so that makes this adjustment simple. The blue line is the dividing line for this design:
This, in fact, is the answer to the original puzzle as posed by Dudeney and Gardner. This is NOT the answer to the problem I posed above. When I first ran across this puzzle, it annoyed me that it wasn't done with the full yin yang symbol. The dots are a symbol of how, in nature, nothing is purely one thing or the other, and are a very important part of the design.
It's time to go back to the full design. Compared to the previous step, we're going to be removing some of the dark area from the right side to the left side. This means that we'll end up rotating our dividing line some distance clockwise this time, and we need to figure out by how much. Yes, once again, we'll be using our area formulas to work out the adjustment. We even know that the result should be easy to interpret, since the result will π over something, and this comes out of a π-unit circle.
The dots, of course, are full circles, so we use the formula for the area of a full circle once again. The dots have a radius of 1⁄6 of a unit, and plugging that into the formula, we get π⁄36. In other words, the line needs to be moved back clockwise 1⁄36 of a full circle, or 10°. That brings the line to being 35° off of the original vertical line.
A 35° line must be our answer, right? Wrong. Go back and look at the original question: Here's the challenge: What's the equation of the line that divides the design so that each side of the line contains exactly equal amounts of the dark and light portions? We need to work out enough details for the line equation y = mx + b, where m is the slope, b is where the line crosses the y-axis, and y and x remain as variables. The line, of course, crosses the y-axis at 0, so b = 0. That reduces the equation to just y = mx, so we need to figure out the slope.
Ever since I started this blog, I've been waiting for this day. I started Grey Matters on 3/14/05, specifically with the goal of having its 10th blogiversary on the ultimate Pi Day: 3/14/15!
Yes, it's also Einstein's birthday, but since it's a special blogiversary for me, this post will be all about my favorite posts from over the past 10 years. Quick side note: This also happens to be my 1,000th published post on the Grey Matters blog!
Keep in mind that the web is always changing, so if you go back and find a link that no longer works, you might be able to find it by either searching for a new place, or at least copying the link and finding whether it's archived over at The Wayback Machine.
This year, I began connecting my posts with the interest of the reader, and it worked well. My series of “Visualizing” posts, Visualizing Pi, Visualizing Math, and Visualizing Scale were the biggest collectively-read posts of the year.
My own personal favorite series of posts in 2011, however, was the Iteration, Feedback, and Change series of posts: Artificial Life, Real Life, Prisoner's Dilemma, Fractals, and Chaos Theory. These posts really gave me the chance to think about an analyze some of the disparate concepts I'd learned over the years when dealing with various math concepts.
After we lost Neil Armstrong in 2012, I was inspired to add the new Moon Phase For Any Date tutorial to the Mental Gym. A completely different type of nostalgia, though, drove me to post about how to program mazes. Admittedly, this was a weird way to kick off 2013.
With 999 posts before this one, this barely even scratches the surface of what's available at this blog, so if you'd made it this far, I encourage you to explore on your own. If you find some of your own favorites, I'd love to hear what you enjoyed at this blog over the years in the comments below!
Here at Grey Matters, I've written about Martin Gardner quite a few times myself, as I have great respect for him. Enjoy exploring the resources, and take some time to remember a man who has brought joy, wonder, and mystery to the world over the past 100 years.
It's time for October's snippets, and all our favorite mathematical masters are here to challenge your brains!
• I'm always looking for a good mathematical shortcut, in order to make math easier to learn. More generally, I'm always looking for better ways to improve my ability to learn. I was thrilled with BetterExplained.com's newest post, Learn Difficult Concepts with the ADEPT Method.
ADEPT stands for Analogy (Tell me what it's like), Diagram (Help me visualize it), Example (Allow me to experience it), Plain English (Let me describe it in my own words), and Technical Definition (Discuss the formal details). This is a great model for anyone struggling to understand anything challenging. This is one of those posts I really enjoy, and want to share with as many of you as I can.
• If you enjoyed Math Awareness Month: Mathematics, Magic & Mystery back in April, you'll love the 31 Tricks and Treats for October 2014 in honor of the 100th anniversary of Martin Gardner's birth! Similar to Math Awareness Month, there's a new mathematical surprise revealed each day. It's fun to explore the new mathematical goodies, and get your brain juices flowing in a fun way!
• If that's not enough, Scam School's latest episode (YouTube link) at this writing also involves three equations. If you have a good eye for detail, you may be able to spot the catch in each one before they're revealed:
That's all for this October's snippets, but it's more than enough to keep your brain puzzled through the rest of the month!
Generally, The College Mathematics Journal isn't available online for free, but they have generously posted the full contents of their January 2012 Martin Gardner issue online for free! It's full of the kind of recreational mathematics which Martin Gardner loved and Grey Matters readers are sure to appreciate and enjoy. There are too many articles to single any one out for special attention, so I suggest jumping in and seeing what catches your eye first!
Scam School has been scammed! Somebody slipped Brain Brushwood some counterfeit coins, and he needs your help to separate the counterfeit coins from the real ones.
OK, this is really just the start of a puzzle, but it's a rather fascinating puzzle. Just when you think you've got the hang of the puzzle, another version can come along and make things tougher.
All in all, not a bad pair of puzzles. For the second puzzle, I would make sure to keep the weighed coins in separate piles, of course, so I can make sure to round up all of the counterfeit coins.
Let's add a new dimension to that second puzzle, just to challenge your thinking. What if, instead of 1 bag holding all counterfeit coins, there were an unknown number of bags? As in the original puzzle, each bag holds either all real coins, each weighing exactly 1 gram, or all counterfeit coins, each weighing 1.1 grams. You still have only one weighing to find out which bags, if any, contain counterfeit coins.
If you want to try and work this out for yourself, stop reading here, as I discuss the solution below.
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If you think about it, what you really need is a series of yes-or-no answers for each bag in a way that allows you to get this information in a single weighing. How do you achieve this?
Our old friend the binary number system comes to the rescue! If you need a quick understanding of binary numbers, watch Binary Numbers in 60 Seconds.
The solution is take 1 coin (20) out of the first bag, 2 coins (21) out of the second bag, 4 coins (22) out of the third bag, and so on. For each bag, you double the number of coins taken from the previous bag, all the way up to 512 (29) coins taken from the 10th bag.
You'll probably note that it's easier to denote the bags as 0 through 9, instead of 1 through 10. With the bags numbered 0 through 9, we can just remember that we take 2x coins out of bag number x.
How does taking coins out in powers of 2 help? First, consider the weight we'd get if all the coins were real. We'd have 1,023 coins weighing a total of 1,023 grams. Any weight over 1,023 grams, then, can be attributed to the counterfeit coins.
Let's say we try this out, and find that we have a total weight of 1,044.7 grams. Take away the weight of the real coins, and we're left with 21.7 grams extra. At 0.1 grams extra for each of the counterfeit coins, we now know there are 217 counterfeit coins among the 1,023 coins.
That's great, you might say, but we still don't know which bags are counterfeit. If you stop and think for a minute, you may have more information than you think. First, since you took 512 (29) coins out of bag 9, the coins in that bag couldn't be counterfeit. If they were, you'd have a minimum of 512 counterfeit coins in the total.
The same argument could be made for bag 8, from which you removed 256 coins (28). Still, isn't it difficult to work out all the possibilities for the remaining bags?
No, and I can explain why in a very simple way. In our everyday decimal system, how many ways are there to write the number 217? There's only one way, of course, and that's by writing a 2 in the hundreds place, a 1 in the tens place, and 7 in the ones place. The same is true for any other base, include base 2 (binary).
Done either way, the binary equivalent of 217 is 11011001, but what does this tell us? Each of these numbers represents one of the bags. To be fair and include all 10 bags, we should write it as a 10-digit binary number, 0011011001, and arrange each number under its corresponding bag like this:
9 8 7 6 5 4 3 2 1 0
0 0 1 1 0 1 1 0 0 1
Are you getting the idea now? The only way for there to be 217 counterfeit coins in the group is if we'd take 128 counterfeit coins (27) from bag 7, 64 counterfeit coins (26) from bag 6, 16 counterfeit coins (24) from bag 4, 8 counterfeit coins (23) from bag 3, and 1 counterfeit coin (20) from bag0.
So, in our example with 217 counterfeit coins, the binary tells us that bags 7, 6, 4, 3, and 0 all contain counterfeit coins, and the rest are real. The decimal equivalents of the number from those bags, 128 + 64 + 16 + 8 + 1 = 217, confirms this answer.
Hopefully, you understand the concepts well enough at this point to figure out which bags are counterfeit if the total weight was, say, 1,062.2 grams (answer after the book excerpt below).
Martin Gardner covered this classic puzzle in a version with medicine (shown below), in his book Aha! Insight, which covers an amazing variety of perplexing situations which are solved with simple insights. They're all presented in the same friendly manner as the Medicine Mix-Up puzzles below.
I hope you enjoyed this look at a classic puzzle. There are many more versions out there, as well. Search the internet for the terms counterfeit coins, weighing, and puzzle to discover more ingenious approaches and ideas.
This time around, we've got a round up of math designed to amaze and surprise you!
• @LucasVB is the designer behind some of the most amazing math-related graphics I've ever seen. You can see some of his amazing work at his tumblr site, and even more at his Wikimedia Commons gallery. Even if you don't understand the mathematics or physics behind any given diagram, they're still enjoyable, and may even prompt your curiosity.
• Our old friend @CardColm is back with more math-based playing-card sneakiness! In his newest Postage Stamp Issue post, he presents a sneaky puzzle that you can almost always win. After shuffling cards, the challenge is to cut off a portion of cards, and see how many of the numbers from 1 to 30 you can make using just the values of those cards. It seems very fair and above-board, but the math behind it allows you to win almost every time!
Even if you're not familiar with the name, you've probably seen it. It's consists of several pieces, each made up of 3-4 cubic units, and the basic challenge is to put these pieces together to form a cube.
I haven't yet discussed solving the classic Soma Cube challenge, so if you want to solve it for yourself without learning how to solve it, stop reading NOW.
You're still here? OK, then I assume you want to learn how to solve it, and more importantly, how to remember the solution. YouTube user MisterCorzi has posted probably the best Soma Cube solving video I've seen, as it emphasizes the points that make the solution easy to recall:
Once you have the basic idea, here's a diagram of MisterCorzi's approach, for quick and easy reference:
I should emphasize that this is only 1 of 240 possible ways (not including rotations or reflections) to solve the Soma Cube, so you can still find plenty of challenge by trying to find the other solutions. There are also many other shapes you can try to make, so create, have fun, and explore!
Granted, the single unmoving camera angle could make things hard to follow, but I've gathered numerous links which I hope will make everything clearer.
First, Ivars Peterson mentions that April is Mathematics Awareness Month, with the 2014 theme being Mathematics, Magic and Mystery, after the Martin Gardner book of the same name (Amazon.com link).
Dr. Benjamin starts out by squaring 2-digit numbers in his head. This feat is relatively easy to learn, and the Mental Gym even features a 2-digit squaring tutorial and quiz. The later explanation features some excellent advice on working up to squaring 3- and 4-digit numbers.
This is followed by the missing digit feat, which is explained much later in the video, so I'll come back to it.
Next up is a magic square feat. The explanation can be tricky to follow. Fortunately, Dr. Benjamin has posted the instructions for his Double Birthday Magic Square online for free. There are several essential tips in the video that make the performance of this far better than if you'd just learned from the PDF alone.
When he talks about how he developed the magic square routine in the first place, he mentions a 2003 magic square article in a magic magazine. This seems to be Harry Lorayne's article, 4×4 Magic Square Breakthrough??. The original magazine article isn't easy to find, but the entire article was reprinted in Harry Lorayne's book, Mathematical Wizardry (Amazon.com link), which I reviewed here back in 2006.
The calendar feat, as many Grey Matters readers already know, is a favorite of mine. You can follow along Dr. Benjamin's somewhat brief explanation of the feat with the help of the Day of the Week For Any Date tutorial and quiz here. I have done my own work simplifying the calendar feat in my Day One ebook.
Impressively, Dr. Benjamin even fields a question about mentally determining whether a 3-, 4-, or 5-digit number is prime or not, despite not performing any feats related to this. If you're wondering why he's using this particular approach, my prime number testing post from earlier this year may make things clearer.
Dr. Benjamin also discusses here what to do when you're not sure whether the missing digit is a 0 or a 9. My preferred approach here would be to say, “I'm not getting anything. It wasn't a zero, was it?” Note that by making this a negative question, you can follow up their answer with “I thought so” or “I didn't think so”, which makes you sound like you knew all along, even though you're just asking a question.
The lecture is wrapped up with the mental multiplication of 2 five-digit numbers. This isn't done as quickly as the other squaring feats. Instead, this is done with lots of verbal calculation and what seems to be some nonsensical words thrown in. First, as he explains after getting the number 37,947 to square, he points out that he's going multiply 37,000 by 947, double that number, square 37,000, square 947, and add all those results together.
Don't worry, there's no complicated sleight-of-hand in these tricks. Not only does math make them easy, but you don't even have to do any math during the routines, since all the math involved has been worked out ahead of time.
I'll start with the simpler of the tricks. In this first one, you have someone think of any hour of the day, and you tap numbers on an analog watch while they silently count up to 20. When they reach 20, they say “Stop!”, and your finger is on the hour they secretly chose!
At first, the workings may confuse you, but a little experimentation with different numbers will help you understand it. Obviously, this is also true for anyone for whom you perform it, so don't treat this as a big mystery, but rather as a simple and interesting experience.
The next trick, courtesy of Card Colm, is a little more involved. You have someone name any card suit, have a regular deck of cards shuffled, and then the number cards (Ace through 9) are removed in the order in which they're found in the deck. You then make an unusual bet based on divisibility of various numbers formed by those cards.
This trick is called the $36 Gamble, and the method is found in Card Colm's post, The Sequence I Desire. Magic: When Divided, No Remainder. Beyond just the mathematical method, there's plenty to explore under the hood of this routine, including Arthur Benjamin's method for determining divisibility by 7, and a very deceptive shuffling method, which appears fair.
If you enjoy the deceptive shuffles discussed in the above post and its links, you also might enjoy Lew Brooks' book Stack Attack, which features the False False Shuffle. The false shuffle and the routines in Stack Attack mix well with the principles behind the $36 Gamble. In my 2006 review of the DVD of the same name by the same author, you can get a better idea of the contents.
Even though I've only linked to 2 tricks here, practicing them, understanding them, and digging in to the variations I've mentioned is more than enough to get your mental gears turning, so have fun exploring them!
The most commonly-used source for this routine is Martin Gardner's book, Mathematics, Magic & Mystery, in which it has the simple title of “Predicting A Sum”. Interestingly, the example number used in the book is 3,845, very close to James Grime's chosen example number, 3,485.
There are many ways to make this routine more impressive. You could employ a secret help, who provides the needed numbers at each step. This takes the heat off of your choices, and makes the whole routine seem fairer, and more impressive.
Another idea to consider is that you don't have to do this as a prediction. Instead, you can do it as a feat of mental arithmetic. Have two people, one of whom is your secret helper, write the numbers down out of your view, and then show the numbers to you briefly. You apparently quickly memorize and calculate the numbers in your head faster than they can do it with a calculator!
With a little creativity, you can think of numerous ways to perform this. Want to seem psychic? Divine the total without ever seeing the numbers (thanks to your secret helper again). You could also perform a routine similar to Scam School's first Pi Day Magic Trick, in which they circle a number, read off the remaining digits, and you can determine the digit or digits they didn't name. Since you know in advance what the total will be, it's all recall with no calculation!
There's a routine with a similar basis called “Alberti's Game”, in which you and two other people randomly choose 3-digit numbers to create 2 multiplication problems, and you're able to predict the sum of the answers to those multiplication problems. Because there's several steps involved, and the answers tend to be large 6-digit numbers, this version seems even more impressive.
When you're just starting out in mathematics, infinity is little more than a neat concept.
Infinity, at that point, is simply the idea that numbers go on forever, and once you accept that, you seem to be fine with the concept of infinity. As you learn more about math, though, you start running into more and more problems relating to infinity, and the concept starts to get weird.
David Hilbert came up with a wonderful example that helps people begin to grasp unusual concepts concerning infinity. His example is known as Hotel Infinity, and is explained here by Martin Gardner, with some clear and amusing illustrations.
Imagine a hotel with an infinite number of rooms (especially easy if, like me, you live in Las Vegas). Also, imagine that an infinite number of people are staying there, so every room is occupied. What happens when 1 person, a UFO pilot in Martin Gardner's version, wants a room? Everybody can be moved to a room number that's 1 higher than their current room, so the first room is now available for the UFO pilot.
Similarly, if 5 couples show up, everyone can be moved to a room number that's 5 higher than their current room, so five rooms are now available for the new couples.
Let's make this more challenging: What happens if an infinite number of people now want a room? You can't simply have everybody move to a room number that's infinitely higher than their current room, so how would you solve this problem?
The answer is surprisingly simple. Have everybody move to a room number that's twice as big as their current room number! Now, the infinite number of previous guests are all staying in even numbered rooms, and the infinite number of new guests can now move into the odd-numbered rooms! Since there are an infinite number of even numbers and odd numbers, this works.
In the late 19th and early 20th centuries, Georg Cantor started talking about different sizes of infinities in a manner similar to this, and even the great mathematical minds of the time scoffed. Eventually, however, mathematicians did come to accept this idea. How exactly can there be different sizes of infinities? You can learn more about this unusual concept in a basic way via Martin Gardner's Ladder of Alephs article. Videos from TED-Ed and Numberphile examine this concept in more detail.
Even though such discoveries about infinity are relatively new, even the ancient Greeks understood the importance of analyzing infinity. Zeno of Elea developed several paradoxes involving infinity which still challenge mathematicians today. TED-Ed's video below explains the Dichotomy paradox:
Not that Hilbert's Hotel Infinity thought experiment even makes this clear. It's even a little startling to realize that it can help you reduce this to a simple algebra problem.
In BetterExplained.com's newest post, An Intuitive Introduction to Limits, these odd ideas about infinity help you understand the concept of limits in calculus. The introduction sums up the challenge perfectly: “Limits, the Foundations Of Calculus, seem so artificial and weasely: “Let x approach 0, but not get there, yet we’ll act like it’s there… ” Ugh. Here’s how I learned to enjoy them:” Concrete examples, including a buffering soccer video, make even this odd concept clear.
If you're confused by the infinite series video, take some time and go back through the earlier concepts of infinity to make sure you understand them. Start by reading the first half of this post, followed by the next quarter of this post, then the next eighth of the post, then the next sixteenth...
In honor of the holiday, I thought I'd share some classic Irish-themed puzzles for you to ponder.
I'll start with a simple puzzle, which you can simply enjoy without choosing to solve it. Developed by Canadian puzzler and magician Mel Stover, this first one is called The Vanishing Leprechaun:
If you'd like to get a closer look at this page, there are several sites, such as this one, which feature the artwork in detail. On that linked page, you can click the top illustration to switch between the two modes yourself, of look at the 2 individual stages in the illustrations below that. The solution to the puzzle is also available on that page, but spend some time trying to figure it out for yourself first.
It seems the Irish have a knack for vanishing in tricky and amusing ways. Take the story or Casey, for example, who marched in many a St. Patrick's day parade. Sam Loyd's puzzle How Many Men Were In The Parade?, which comes to us courtesy of Martin Gardner, and can be read online (Part 1, Part 2), or as follows:
During a recent St. Patrick's Day parade, an interesting and curious puzzle developed. The Grand Marshall issued the usual notice setting forth that “the members of the Ancient and Honorable Order of Hibernians will march in the afternoon if it rains in the morning, but will parade in the morning if it rains in the afternoon.” This gave rise to the popular impression that rain is to be counted as a sure thing on St. Patrick's Day. Casey boasted that he “had marched for a quarter of a century in every St. Patrick's day parade since he had become a boy.”
I will pass over the curious interpretations which may be made of the above remark, and say that old age and pneumonia having overtaken Casey at last, he had marched on with the immortal procession. When the boys met again to do honor to themselves and St. Patrick on the 17th of March, they found that there was a vacany in their ranks which it was difficult to fill. In fact, it was such an embarrassing vacancy that it broke up the parade and converted it into a panic-stricken funeral procession.
The lads, according to custom, arranged themselves ten abreast, and did march a block or two in that order with but nine men in the last row where Casey used to walk on account of an impediment in his left foot. The music of the Hibernian band was so completely drowned out by spectators shouting to ask what had become of the “the little fellow with the limp,” that it was deemed best to reorganize on the basis of nine men to each row, as eleven would not do.
But again Casey was missed and the procession was halted when it was discovered that the last row came out with but eight men. There was a hurried attempt to form with eight men in each row; again with seven, and then with five, four, three, and even two, but it was found that each and every formation always came out with a vacant space for Casey in the last line. Then, although it strikes us as silly superstition, it became whispered through the lines that Casey's “dot and carry one” step could be heard. The boys were so firmly convinced that Casey's ghost was marching that no one was bold enough to bring up the rear.
The Grand Marshal, however, was a quick-witted fellow who speedily laid out that ghost by ordering the men to march in single file; so, if Casey did march in spirit, he brought up the rear of the longest procession that ever did honor to his patron saint.
Assuming the number of men in the parade did not exceed 7,000, can you determine just how many men marched in the procession?
I'll keep you guessing until Thursday, when I'll reveal the answer to this puzzle.
That's all for now. I simply wished to share some quick puzzles for the holiday. If you have any favorite St. Patrick's Day puzzles you'd like to share, let me know about it in the comments!
Disclaimer: I was provided with the book by the author without charge for the purposes of review. The thoughts below are purely mine as a result of going over the ebook on my own time for the purpose of informing Grey Matters readers.
Certainly, the first thing that catches your eye is the long and unusual name. The title comes from 3 puzzles in the ebook that share a common basis. Presh Talwalkar presents these puzzles and explains them in his post announcing the ebook release.
That post, in fact, is a good introduction to the style of the book. Generally, each puzzle is introduced on 1-2 pages, and the answers are provided on following pages, so as to give you an opportunity to stop and think about it.
Yes, you could just flip to the answers on the next page right away, but this will rob you of one of the true values of the ebook. Each puzzle offers the thinker what Martin Gardner dubbed an aha! moment, a moment where insight reveals a clarity that simplifies either the problem itself, or the approach of solving the problem. Helping you experience many aha! moments for yourself is the real value.
You might expect such a book to be organized, perhaps by theme, such as, say, coin puzzles vs. sequence problems there, or by puzzle type, such as arithmetic puzzles vs. probability puzzles. Instead, the problems are roughly arranged in order of complexity of solution. In other words, only the simplest insights are needed to solve the first ones, but you need to be wary of mental traps in the latter puzzles.
Some might complain that chapters would be better organized, but even if you forget the location of a puzzle, a search can find it easily enough. With the puzzles being in order of complexity, a simple glance at the page number can give you a sort of rating, as in, "Oh, I've done pretty well with these puzzles, and I'm halfway through the book. That's a good sign!"
Incuded at the end of the ebook is bonus, but it's not a bonus puzzle. It's an article on the friendship paradox that's also available on Presh's blog here. If you're not already familiar with Mind Your Decisions, this is a good introduction to it. As with many posts there, it gets you pondering, and there's more information available online, including at TED.com and phys.org.
This post picks up from the previous post, in which I explored an integer lattice using Desmos.
Today, we'll keep exploring the integer lattice, and see what other surprises it offers!
As before, the 20 by 20 integer lattice is here (opens in new window), and the 10 by 10 integer lattice is here (opens in new window). The latter functions better on mobile devices. Clicking on the options button in the upper right corner, then selecting Projector mode is also recommended.
PRIMES AGAIN
We left off in the last post talking about prime numbers, integers (whole numbers) that are only even divisible by 1 and themselves. This seemed like a good place to continue the discussion.
If you take a look at the column (3,y), or the row (x,3) for that matter, you'll notice that only every 3rd dot isn't visible. This shouldn't be surprising by now, since we know from the previous post that only the dots that are visible are the ones that are coprime (their only common divisor is 1).
Similarly, with (x,2) and (2,y) only every 2nd one isn't visible, with (x,5) and (5,y) only every 5th one isn't visible, and so on. It's a simple pattern, but one that can help you easily pick out the prime numbers in even very large integer lattices.
In either of these, start clicking on the Xs in the icons of the formulas until you get down to the next set of text instructions. When you do this, diagonal purple lines will appear. First, notice that diagonals not highlighted by purple lines have missing dots, while the highlighted diagonals have unbroken lines of dots (at least when they're not crossing an axis or going off the other edge of the grid).
What's so special about these purple lines? In each (x, y) pair, the sum of x and y add up to prime numbers! For example, in the coordinates (1,6), (2,5), (3,4), (4,3), (5,2), and (6,1), all the pairs add up to 7, and they're all visible! These are the dots highlight with the formula y=7-x, where x is greater than 0.
It turns out that, regardless of the size of the integer lattice you use, coordinate pairs whose sum is a prime number will always have an unbroken line of visible dots (again, as long as neither coordinate is 0). But what exactly does this mean?
From the previous post, we know that the dots on the lattice are only visible when their coordinate pair is coprime. So, we can determine from this that any two non-zero positive integers whose sum is a prime number must be coprime to each other! Take another look at all the pairs above that total 7, and you'll see that none of those pairs can be reduced!
This surprising discovery, done with a grid and simple arithmetic, is part of the reason I find the integer lattice so fascinating.
ARE THERE LINES THAT WILL NEVER CROSS ANY DOT?
Imagine that the integer lattice we've been using is extended infinitely on flat plane. Is it possible to draw a line that never hits any grid point?
Going back to the previous post once again, we know that a line that goes from (0,0) through (x,y) must have a slope of y⁄x. Conversely, a line with slope y⁄x can always be drawn from (0,0) to (x,y).
Since (x,y) coordinates on the integer lattice are, by definition, always integers, we're basically looking for numbers that aren't expressible as fractions. Fortunately, there is a large category of such numbers, known as irrational numbers. The video below will explain about them in a little more detail:
On the integer lattices to which I linked at the beginning of DIAGONAL PRIMES section above, scroll down just past the instructions which read, “Click on the Xs in the icons below to display orange lines whose slopes are irrational numbers (numbers that can't be represented by a fraction b/a).”
Just below those instructions, I've set up lines with slopes of Pi and the square root of 2, both prominent irrational numbers. Click on the Xs on the icons to make those visible. You'll probably want to click on the icons of the purple lines to turn them off again.
These lines, as we've said, will never hit any integer point on the grid. Although, they can come extremely close. If you're looking at the 20 by 20 grid, you may notice that the orange line with a slope of the square root of 2 goes right through the dot at (12,17). If you zoom in far enough, using either your mouse's scroll wheel or the + and - buttons in the upper left of the grid, you'll see that it actualyy misses the dot by a very tiny amount.
In Wikipedia's entry on the square root of 2, it mentions that some known approximations for it include 3⁄2, 17⁄12, 577⁄408, and 665,857⁄470,832. So, a line drawn from (0,0) through (408,577) will appear to be very close to a line with a slope equal to the square root of 2, but they will eventually diverge.
Here's where everything comes together in a surprising way. Let's figure out what the chances are of a randomly selected coordinate pair being visible. This sounds challenging, but if we break it down, we can make it manageable.
We'll use m and n to represent two randomly chosen integers. What is the probability that m is evenly divisible by 2? That's easy, it's 1⁄2, or .5 (a 50% chance). The same is true for n.
The probability of both m and n being evenly divisible by 2, then, is 1⁄2 × 1⁄2, or 1⁄4. The probability of at least 1 of the 2 randomly chosen numbers NOT being even divisible by 2, then, is 1 - 1⁄4, which is 3⁄4.
If we wanted to work out the odds of two randomly chosen integers both being divisible by 3, that's 1⁄3 × 1⁄3 = 1⁄9. Conversely the probability of at least 1 of the 2 digits not being divisible by 3 is 1 - 1⁄9, or 8⁄9.
To check the odds of two randomly chosen numbers not having 2 or 3 as a common factor, then, is 3⁄4 × 8⁄9 = 2⁄3.
You might think that we could just continue working on this way, working out the probabilities for numbers divisble by 4, 5, 6, and so on. We actually don't need to use composite numbers (numbers even divisible by numbers other than themselves and 1) such as 4 and 6. This is because of the fundamental theorem of arithmetic, which states that every integer greater than 1 is either prime itself or is the unique product of prime numbers. To answer our question, what we need to do is work through the above process using only prime numbers.
So, the process for working this out from prime numbers from 2 on up would appear as:
The proper mathematical shorthand for this would be to say that p represents the set of all prime numbers, and then use the product operator like this:
Using Wolfram|Alpha's Prime[n] function, which returns the nth prime number, and its Product[] function, which works as the product operator above, we can work out the probability of two chosen numbers not having the first n prime factors in common.
Amazingly, as you apply this formula with more and more numbers, it approaches 6/π2 (roughly 60.79%)! That's quite surprising, since the formula we used really only involved prime numbers.
The 30-minute video The Story of Pi has a nice segment (shown below) about the integer lattice and this appearance of Pi (only plays from 18:27 to 20:17):
I apologize for the irregular posting over the past few months. I've had to deal with some personal issues (don't worry, everything is fine!). The good news is that, with this entry, everything should start returning to normal.
For this post, I'd like to turn to a variation of a classic 
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