It's been over a year since I posted a tutorial over at the Mental Gym, so I figured it was about time for a new one!

This one is a new spin on working out the calendar for a given month and year. Yes, I have an existing Day of the Week For Any Date tutorial, and even a commercially-available version, but this new one is remarkably simple!

The new tutorial is dubbed the Quick Calendar Month Creation. It's a combination of a little-known, yet surprisingly simple calendar calculation method published by W. W. Durbin and E. Rogent in 1927, Robert Goddard's *First Sunday Doomsday Algorithm*, and my own approach of creating a full-month calendar to was calculations.

For those familiar with Doomsday algorithm, this isn't yet another variation of John Conway's fine work. Instead, I started with Durbin and Rogent's unusual and simple approach to dealing with the year, and adjusted the math so it meshed with Goddard's powerful work. The exact details and credits are given at the bottom of each section of the tutorial.

The result is a calendar creation routine that's quick and simple to learn, yet powerful enough to let you create calendar for any month and year back to 45 B.C., when the Julian calendar was first used!

To help you practice it more effectively, I've also developed a quiz page. The initial quizzes are simple, in order to help you master the calculation and recall required by each step, and then there's a more complete quiz, which simply has you create a calendar month for random dates. Some of you may recognize this as a modified version of my calendar quiz from Day One.

If you've never tried to do a calendar calculation before, try this out, and you might surprise yourself. If you've tried to do calendar calculations before and given up because of the difficulty of other methods, try the Quick Calendar Month Creation tutorial, and see the ease and power of this approach. Either way, I'd love to hear what you think about it in the comments!

## New Mental Gym Tutorial!

Published on Sunday, May 25, 2014 in calendar, fun, math, memory, memory feats, mental math, self improvement, site features

## Free Math Magic Books!

Published on Sunday, May 18, 2014 in books, calendar, downloads, fun, math, mental math, nim

Last week, I gave you some free magazines to peruse.

This week, I have some books I'd like to share with you, and they're filled with a great selection of classic mathematical feats and magic!

The first book I'll share with you is a 1952 book titled *Mental Prodigies*, by Fred Barlow (embedded below). The first section is concerned purely with arithmetical prodigies, or what we might call lightning calculators or human calculators today. The next section discusses related types of prodigies, such as chess or music. Next, is a section on memory, including chapters on famous memorizers, as well as mnemonics and memory techniques for actors.

The chapter I think will be the most interesting to Grey Matters readers is the section on Mental Magic, starting on page 183. It includes many of the standard feats covered here on Grey Matters, such as day for any date, squaring, cubing, and finding roots.

However, there are some more unusual ones, such as calculating the number of farthings in a given number of guineas, or how many barleycorns there are in a given number of yards. Granted, these might not play too well today, but the same technique, as the author explains, can also be used to give the number of minutes in a given number of weeks!

After the embedded book below, scroll down further for another math magic book.

If you'd prefer more than just a few small chapters about mathematical magic, check out the 1950 classic, *Math Miracles*, by Wallace Lee.

Among the more unusual tricks here are the Ne Plus Ultra Lightning Multiplier and its variations. The Miscellanies chapter is filled with numerous quick and unusual tricks, including the 100 version of our old friend Nim!

I don't want to rob you of the joy of discovery however. Take a chance to page through these e-books, and you may find some unexpected treasures!

## Free Math Magazines!

Published on Sunday, May 11, 2014 in books, calendar, downloads, fun, Martin Gardner, math, nim, Scam School, site features

Many of you are spending today with your mother in honor of Mother's Day, so I won't strain your brain too much today.

In fact, I'll just leave a few free magazines on the table for your perusal when you have some time later.

I'll start with the brand new *Recreational Mathematics Magazine*. This magazine is available as a whole PDF, or as PDFs of individual articles. The first article that caught my attention here was “The Secrets of Notakto: Winning at X-only Tic-Tac-Toe”. It caught my attention because I'd written about Notakto strategy 2 years ago, including how to win playing on 1 or 2 boards, and then how to win when playing on 3 or more boards.

Don't let me rob you of the joy of discovery, however. The other articles, including the one about Lewis Carroll's mathematical side, the one about vanishing area puzzles, and others are all waiting to be discovered.

The next math magazine I'd like to draw your attention to is *Eureka*, published by the Archimedeans, the Mathematical Society of the University of Cambridge, since 1939. New issues are being made available online for free by mathigon.org. This is no minor mathematical publication, either. It was the Archimedeans' *Eureka* magazine that, back in October 1973, had the honor of being the first to publish John Conway's Doomsday Algorithm for calculating the day of the week for any date.

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!

The final magazine I'll set out for your perusal isn't a mathematical magazine, but rather a magic magazine called *Vanish*, which is free to download, or read online as a page-flipping e-magazine. The reason I'm including it here with math magazines is because of Diamond Jim Tyler's article on “The Game of 31”. This is variation of our old friend Nim. For a Nim variation, 31 has a surprising amount of its own variations, including a dice version, a finger dart version, and a version which you can still scam someone after teaching them the secret!

That's all for now, so I'll wish you happy reading!

## The Great Divide II

Published on Sunday, May 04, 2014 in fun, math, mental math, self improvement

My previous post, The Great Divide, focused on divisibility testing for numbers ending in 9, such as 19, or for numbers which could be scaled up to a number ending in 9, such as 13, which can be scaled up to 39.

The tests took advantage of the fact that they have neighboring numbers ending in 0, which are easy to divide by, and can easily generate a quotient and remainder.

Assuming you're familiar and comfortable with the previous approach, today's post will focus on divisibility testing for numbers ending in 1.

**DIVISIBILITY BY 11:** In the previous post, we saw that we could test for divisibility by 11 through scaling it up to 99, and taking advantage of the ease of dividing by 100.

For example, to check whether 1,353 is divisible by 11, we'd divide 1,353 by 100, giving us a quotient of 13 and a remainder of 53. Since 13 + 53 = 66, and 66 is obviously a multiple of 11, we can see that 1,353 is evenly divisible by 11.

However, since 11 is right next to 10, isn't there a way to take advantage of that relationship without scaling 11? Yes, there is!

Just as before, we divide by the neighbor to get a remainder and a quotient, but this time we *subtract* (instead of *adding*) the smaller of the two numbers from the larger of the two numbers.

Let's try this new version of the test on 1,353 again. Dividing 1,353 by 10, we get 135 remainder 3. This time, we need to subtract the smaller number from the larger, so we work out 135 - 3 = 132.

If you know enough multiples of 11, you should recognize 132 as a multiple of 11. If not, just as with the tests for numbers ending in 9, you can repeat the test on the new number. 132 ÷ 10 = 13 remainder 2, and 13 - 2 = 11, so 1,353 (and 132, for that matter) is confirmed as a multiple of 11.

**HOW CLOSE ARE WE TO A MULTIPLE?:** Also, just as before, numbers that don't pass these tests for divisibility can still show how close they are to a multiple, even through repeated testing.

As an example, let's test 4,085 for divisibility by 11. 4,085 ÷ 10 = 408, remainder 5, and 408 - 5 = 403. Testing 403 for divisibility by 11, we work out 403 ÷ 10 = 40, remainder 3, and 40 - 3 = 37. We know 37 isn't evenly divisible by 11, so 4,085 isn't evenly divisible by 11.

Since 37 is only 4 more than 33, which is a multiple of 11, we can also say that 4,085 must be 4 more than a multiple of 11. 4,085 - 4 = 4,081, and we can verify that 4,081 is a multiple of 11 by our own testing, or with the help of Wolfram|Alpha.

**WHY DOES THIS WORK?:** This works for similar reasons as the previous method works. In this case, we're testing a number *x* for divisibility by *n* by using *n* - 1 to get a quotient, *q*, and remainder, *r*, so this version looks like this:

^{x}⁄_{(n - 1)} = *q* - *r*

Solving this equation for x, we get:

*x* = (*n* - 1)(*q* - *r*)

Expanding this works out to:

*x* = *nq* - *nr* - *q* + *r* = *nq* - *nr* - (*q* - *r*)

From here, the logic is basically the same as in my previous post. Since each of the variables are defined as integers, *nq* - *nr* must be evenly divisible by *n*. Therefore, as long as subtracting *q* - *r* is evenly divisible by *n*, then *x* must be evenly divisible by *n*.

**TESTING OTHER NUMBERS BY SCALING:** Just as with the approach in my previous post, numbers can be scaled to work with this approach. For example, in the previous post we scaled 7 up to 49, so we could test for divisibility using 50.

Thanks to the new version of the test for numbers ending in a 1, we can scale 7 up to 21, and use 20 for testing instead.

Is 2,667 evenly divisible by 7? Let's use the scaling up to 21, so we can divide by 20 in our tests. 2,667 ÷ 20 = 133, remainder 7, and 133 - 7 = 126. Personally, I know enough multiples of 126 to recognize that it's a multiple of 7. What if I didn't?

I'd just test 126 again. 126 ÷ 20 = 6, remainder 6, and 6 - 6 = 0. Yes, 0 counts as a multiple of 7, so 2,667 is evenly divisible by 7.

**WIDE VARIETY OF OPTIONS:** With this new version of divisibility testing, you can now test for divisibility by any number ending in 9 or 1. Further, you can multiply any number ending in 3 by 3 to get a number ending in 9, or any number ending in 7 by 3 to get a number ending in 1. So, you've got a wide variety of tests for any number ending in 1, 3, 7, or 9 that you can recall how to do with a little practice and effort!

When testing for primes, this gives you an arsenal of tests that you can keep straight and use quickly and easily. Testing for divisibility by 11, 19, 29, and so on? No problem, as they're ready made for the test! Need to test for divisibility by 23? Scale it up to 69, and use 70 in your tests! Divisibility by 17? Scale it up to 51, and use 50 in your tests!

This is a little unusual when it's a new concept, but continued practice will give you a tool well worth having in your mathematical arsenal.