I recently ran across a number of videos I figured would be interesting to regular Grey Matters readers, so I thought I would share them.

We'll start things off with a little math magic, courtesy of Tom London and his appearance on *America's Got Talent* earlier this week:

Yes, I could explain the method, but I don't want to ruin the fun and the mystery. Just enjoy the magic of the prediction for what it is, since that's how it's meant to be enjoyed.

If you want mathematical explanations, however, I highly recommend checking out PBS Digital Studios' *Infinite Series*. These are videos on assorted advanced mathematical topics, yet they're taught in a very accessible way. Back in March, I discussed a puzzle which required the understanding of Markov chains to solve. Compare that to their video *Can a Chess Piece Explain Markov Chains?*, which also happens to employ my favorite chess piece, the knight:

If you enjoy Grey Matters, you may also the work of 4-time USA memory champion Nelson Dellis, who focuses on both mental and physical fitness. He has a series of memory technique videos, as well as interviews with masters of mental skills. Both of these are available on his YouTube channel, as well. As a taste of his skilll, watch his video, *Memorizing 28 names in less than 60 seconds!*:

Curious how he's able to do that? He explains in the next video in the series, *HOW TO // "Memorizing 28 names in less than 60 seconds!"*.

At this point, I'll wrap things up so you can get started on a potentially mind-expanding journey.

## Magic, Math and Memory Videos!

Published on Sunday, June 25, 2017 in Knight's Tour, magic, math, memory, memory feats, self improvement, videos

## Yet Still More Quick Snippets

Published on Sunday, June 11, 2017 in magic squares, math, snippets, videos

It's now been a few months since Grey Matters was back, so now it's time to bring Quick Snippets back!

This time around, we have plenty of mathy goodness, so it's best to just jump right in!

• Besides the Clay Institute's famous selection of Millennium Problems, which will make you a millionaire if you prove or disprove any one of them, there's a lesser known set, known as John Conway's $1,000 problems. Not long ago, the 5th conjecture, which claims that working through a certain procedure (described in the link an video below) will always end in a prime, was disproven by physicist James Davis. The Numberphile video below details the problem and James Davis' counterexample:

For more about the million-dollar Millennium Problems, watch the BBC's *Horizon* documentary, "A Mathematical Mystery Tour of Unsolved Mathematical Problems."

• Speaking of fun discoveries in recreational mathematics, check out Allan William Johnson's "Magic Square of Squares", discovered back in 1990, and just recently posted over at Futility Closet.

• James Grimes introduces us to a different sort of "Square of Squares", in his latest Numberphile video, "Squared Squares". The challenge here is to make a perfect square shape from a set of smaller square shapes:

• Presh Talwalkar, of Mind Your Decisions, posted an interesting puzzle recently. It's titled, "The Race To 32,768. Game Theory Puzzle". Read the article up to the point where you're challenged to work it out yourself, or watch the video below up to the 2:12 mark, and try and figure it out for yourself. If you get stuck, try going over my *Scam School Teaches the Game of 15* post for inspiration.

• Late last month, mathematical video maker 3Blue1Brown posted a must-watch video on the visualization of all possible Pythagorean triples. Even if you remember everything from you math classes about Pythagorean triples, this video is both eye-popping and an eye opener:

• We'll wrap this set of Quick Snippets up with help from Mathologer. His videos are always interesting, but his latest one is one of those unusual approaches to math that makes you appreciate its beauty. This video is titled, "Gauss's magic shoelace area formula and its calculus companion", and it teaches an simple but unusual method for working out the area of any polygon that doesn't intersect itself. The host even goes on to show how this approach can be adapted in calculus to work out the area contained by curves!

## Tour the U.S.!

Published on Sunday, May 28, 2017 in fun, Knight's Tour, puzzles, self improvement

*Starting in Delaware, you must tour the 48 contiguous United States, visiting each state exactly once. Where will you finish?*

That's a puzzle from Futility Closet. I'll link directly to it later, so as not to get too far ahead. It's really a puzzle within a puzzle, however.

As anyone who has tried the Knight's Tour knows, moving around with the limitation of landing in each space only once can be quite a challenge.

The original Futility Closet puzzle can be solved either by thinking about that puzzle on its own with a little analysis. However, I think it might be a little more fun to try and solve the puzzle by trying various ways to get around the 48 contiguous United States, visiting each state exactly once.

To that end, I've written this as a puzzle you can play below. It's an interactive map, so you can scroll (by clicking and dragging outside of the U.S.) and zoom (using the + and - in the upper left corner of the map) as needed, which helps when you're trying access smaller states.

You start by clicking on any state. That state will turn blue, representing your current location. Some states will also appear in green, and these are states which border your current state, and to which you haven't already traveled.

To move to a new state, simply click on any one of the green states. Clicking on any other states won't have any effect. Once you click a green state, that state will become blue (again, denoting your current location), and a new set of bordering states will become green. Also, the state which you just left becomes red, which denotes that you can never move to that state again.

Quick side note: The way I've programmed it, you can't move directly from Arizona to Colorado or directly from New Mexico to Utah, as I didn't consider meeting at a single point to be a bordering state.

If you want to work on the original Futility Closet puzzle, zoom in on Delaware, and click on it to start. Otherwise, start on any state. Try and see if you can get to all 48 states just once each time. You'll get an alert if you've either solved the puzzle, or become trapped. In either case, the map will go blank and you can try again.

Even if you don't get all 48 states, try and beat your highest number of states each time. As you play, perhaps you'll even get a realization that will help you solve the original Futility Closet puzzle.

The more you try this, the better you'll get, and you'll probably be able to solve the 48-state U.S. tour puzzle and the Futility Closet puzzle before too long. At this point, you can challenge your friends, and show your skills at solving the tour. What kind of experience did you have solving the puzzle? I'd love to hear about it in the comments below!

## The Collective Coin Coincidence

Published on Sunday, May 21, 2017 in fun, magic, math, money, psychology, Scam School, videos

This week, Diamond Jim Tyler demonstrates a new take on an old trick. Regular Grey Matters readers won't be surprised to learn that I like it because it's based on math, and it's very counterintuitive. We'll start with the new video, and then take a closer look at the trick.

This week's Scam School episode is called *The Collective Coin Coincidence*, and features Diamond Jim Tyler giving not only a good performance, but also a good lesson in improving a routine properly:

Brian mentions that this was an update from a previous Scam School episode. What he doesn't mention is that you have to travel all the way back to 2009 to find it! The original version was called *The Coin Trick That Fooled Einstein*, and Brian performed it for U.S. Ski Team Olympic gold medalist Jonny Moseley. It's worth taking a look to see how the new version compares with the original.

Brian and Jim kind of rush through the math shortly after the 4:00 mark, but let's take a close look at the math step-by-step:

Start - The other person has an unknown amount of coins. As with any unknown in algebra, we'll assign a variable to it. To represent coins, change or cents, we'll use: c

1 - When you're saying you have as many coins (or cents) as they do, you're saying you have: c

2 - When you're saying you have 3 more coins than they do, the algebraic way to say that is: c + 3

3 - When you're saying you have enough left over to make their number of coins (c) equal 36, that amount is represented by 36 - c, so the total becomes: c + 3 + 36 - c

Take a close look at that final formula. The first c and the last c cancel out, leaving us with 3 + 36 which is 39. If you go through these same steps with the amount of coins (in cents, as it will make everything easier) as opposed to the number of coins, it works out the same way. This is what Diamond Jim Tyler means when he explains that all he's saying is that he has $4.25 (funnily enough, he says that just after the 4:25 mark).

As long as we're considering improvements, I have another unusual use for this routine. If you go back to my *Scam School Meets Grey Matters...Still Yet Again!* post, I feature the Purloined Objects/*How to Catch a Thief!* episode of Scam School, which I contributed to the show. It's not a bad routine as taught, but my post includes a tip which originated with magician Stewart James. This tip uses the Coin Coincidence/Trick That Fooled Einstein principle to take the Purloined Objects into the miracle class! I won't tip it here, so as not to ruin your joy of discovery.

## Chinese Remainder Theorem II

Published on Sunday, May 14, 2017 in fun, Martin Gardner, math, mental math

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

- 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

*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

*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.

## Yin-Yang Challenge

Published on Sunday, April 30, 2017 in Martin Gardner, math, Pi, puzzles

For this post, I'd like to turn to a variation of a classic Henry Dudeney puzzle, from his book *Amusements in Mathematics*. It can also be found in Martin Gardner's October 1960 *Scientific American* column, or his book, *New Mathematical Diversions*, as the 5th puzzle ,“Bisecting Yin and Yang”, in chapter 12, “Nine Problems”.

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=πr^{2}, 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}⁄_{2}πr^{2}. 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.

First, angles are usually measured in relation to the positive x axis, so we're actually talking about a 125° (35° + 90°), or ^{25π}⁄_{36} radians (Confused? Read *Intuitive Guide to Angles, Degrees and Radians*). In geometry, we'd say we were trying to calculate rise over run (rise ÷ run). In trigonometry, we're trying to calculate the opposite site of the angle by the adjacent side (Confused? Read *How To Learn Trigonometry Intuitively*), and that means we need to use the tangent!

So, the equation is y = tan(^{25π}⁄_{36})x, or y = tan(125°)x, if you prefer. The actual slope is an irrational number which is roughly equal to -1.428148. There you have it, the equation to a line which divides the design into equal parts of light and dark, as shown below.