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## Even More Quick Snippets

Published on Sunday, November 24, 2013 in , , , , , , , November's snippets are ready, and they're chock full of some amazing and fun ways to get involved with math!

• If you enjoyed the recent Arthur Benjamin posts from MAA, It's All About The Benjamin and Fibonacci Meets Dr. Benjamin, you should know they have plenty more!

This includes Dr. Benjamin on how to square 2- and 3-digit numbers in your head (below) and an interesting little side story about how he discovered this principle on his own, even before learning algebra!

Among the other interesting goodies on their YouTube channel is James Tanton's Curriculum Inspirations, which feature challenging math puzzles. James Tanton helps you get started on these puzzles, and then encourages you to solve them on your own.

• Vi Hart, a longtime Grey Matteras favorite, has released a new video titled How I Feel About Logarithms. It's an intuitive and holistic look at logarithms that is probably quite different from any approach to the topic that you've seen before. Compare it to, say, the more linear (but still intuitive) approach used in BetterExplained.com's Using Logs In The Real World post, or Steve Kelly's Logarithms, explained.

• A while back, there was a company which produced a DVD series titled Total Breeze Mathematics. Their videos on memory and math shortcuts, which they made available for free on YouTube, were visual and very clear. Their company website and videos quickly disappeared, but fortunately there were those who managed to save the videos and have since reposted them. I've gathered them in their original order, and put them in a YouTube playlist, so you can enjoy them once again.

There are some parts missing, though. There was a 3-minute section on multiplying vertically and crosswise in one video. This can be replaced by viewing the Math Tutor videos on multiplying any 2 digit numbers, multiplying 3 digits by 1 digit, multiplying 3 digits by 2 digits, and multiplying 3 digits by 3 digits.

There's also a section on memorizing numbers that isn't included on the playlist. This free video by Dr. Benjamin on memorizing numbers covers the same technique.

• From 2007-2010, there used to be a magazine called iSquared, which focused on the interesting ways in which mathematics applied to the real world. Originally, it required a subscription, but the entire run of 12 issues has now been made available online for free! This is not only a fun read, but can also help you answer questions such as when you're going to use a particular math technique in the real world! 1

## Digging Up 3-Digit Cube Roots (Tips & Tricks)

Published on Thursday, November 21, 2013 in , , , , The method for extracting 3-digit cube roots of perfect cubes taught in my previous post is the classic approach for this task, but it's not the ONLY approach.

In this post, we'll look at some other ways to tackle the cube root challenge.

QUICK TIP: If you get a perfect cube ending in 3 zeroes, such as 238,328,000, after you put down the zero for the rightmost digit, you can ignore the zeroes in the cube, and work out the rest in the same way as you do for 2-digit cubes.

Working through our 238,328,000 example number, you can instantly put 0 down as the rightmost digit, and then think of the number as being 238,328 (effectively dividing it by 1,000). As in the standard method, you can see that 238 is greater than 63 and less than 73, so the leftmost digit will be 6, giving us 6_0 so far. Now, the rightmost digit of our modified number, the final 8, indicates that 2 is the cube root, which gives us 620 as the answer. Sure enough, 6203 is 238,328,000!

ALTERNATE MOD 11 APPROACH: In the previous post, the approach for working out mod 11 involved alternately adding and subtracting number from right to left, and could result in numbers of 11 or more, requiring further adjustments. Here's another method you may prefer, as it's more straightforward.

To help you understand the process, we're going to go back to the technique of long division. Imagine you're asked to work out 94,818,816 ÷ 11 by long division, and give the remainder, as well. Writing it out, it would look something like this:

Consider that you'll have the perfect cube, 94,818,816 in our example above, written down already. You know mod 11 is important from the beginning, so there's no real need to write that down. Finally, we're ONLY interested in the remainder, so there's no need to keep track of a huge number such as 8619892 at all.

All you really need to do in this approach, then, is do the subtractions mentally in the same way, starting with the leftmost 2 digits (if the leftmost 2 digits are 10, then start with the leftmost 3 digits), and appending the next single digit to the right after each subtraction. In the example above, you'd look at the numbers on the paper as you go through them mentally, like this: “94 - 88 = 6...68 - 66 = 2...21 - 11 = 10...108 - 99 = 9...98 - 88 = 10...101 - 99 = 2...26 - 22 = 4” That 4 means that the perfect cube you were given, when calculated mod 11, is 4.

From that point, you would work through the procedure as taught in the previous post. The nice thing about this approach is that it will work for ANY modulus number for the same reason that long division always works. Once you get the hang of doing the subtractions and adding the next digit to the right, it's not hard at all.

MOD 9 APPROACH: There's also a mod 9 approach for finding the middle digit. Many people, including Arthur Benjamin, whose videos were recently discussed here and here, prefer this approach.

This version works on the idea of digital roots, also known as digit sums. Digital roots are a very simple idea, but if you need a refresher course, this video on digit sums and casting out 9s gives a good brief introduction to the idea, and this second video gives a little more detail.

Once you've got the leftmost and rightmost digits of the cube root, you're ready to use this technique. The first step is simply to add up all the digits of the perfect cube. If the total has 2 or more digits, you add up the individual digits of that number, and keep repeating this process until you get a 1-digit number, as described in the videos linked above.

For example, let's say you're given the perfect cube of 143,055,667. At this stage, you should already know that the cube root looks something like 5_3. You'd add 1 + 4 + 3 + 0 + 5 + 5 + 6 + 6 + 7 = 37. 37 has 2 digits, so you'd add 3 + 7 = 10. 10 is a 2-digit number, so you'd add 1 + 0 = 1, so the digital root of the perfect cube is 1.

How do we use this knowledge? Here, Wolfram Alpha shows us what happens when the number from 0 to 8 are cubed and then reduced mod 9 (which is the same as taking the digital root). Any number, when cubed, will always have a digital root of either 0, 1, or 8.

If a perfect cube has a digital root of 0 (or 9, which is the same thing), then its cube root must have a digital root of either 0 (or 9), 3, or 6. If a perfect cube has a digital root of 1, then its cube root must have a digital root of either 1, 4, or 7. And if a perfect cube has a digital root of 8, then its cube root must have a digital root of either 2, 5, or 8.

This table is definitely simpler to memorize than the mod 11 version, but it does come with a price. How do you know which cube root? Fortunately, a little rough judgement is all that is needed.

In our 143,055,667 example from above, we know that the cube root looks something like 5_3, and has a digital root of 1. This means that the cube root must have a digit root of 1, 4, or 7. Let's try the various possibilities and see what happens. 5 + 3 = 8, so we could get a digital root of 1 by adding 2 in the middle (since 5 + 2 + 3 = 10, and 10 has a digital root of 1). We could also get a digital root of 4 by putting a 5 in the middle (5 + 5 + 3 = 13, and 13 has a digital root of 4), or a digital root of 7 by putting an 8 in the middle (5 + 8 + 3 = 16, and 16 mod 9 is 7).

Now we have 3 possible answers: 523, 553, and 583. How do we choose? Go back and look at the leftmost group of numbers. Is the answer closer to 500 cubed (125 million), 600 cubed (216 million), or somewhere in the middle? If it were in the middle, the number would begin with a number around 170 million, and if it were towards the high end, the number would be close to 216 million, which rules out these possibilities. 143 million is much closer to 125 million, so we must be looking for the lowest of the 3 number, which is 523.

Double checking with Wolfram Alpha, sure enough, we find that 5233 = 143,055,667!

Dr. Benjamin has his own paper on finding cube roots here (PDF), where he discusses the mod 9 approach in more detail, if you're interested.

Consider these tricks and tips, put together your own preferred version of this feat, and then go out and amaze your friends and family with your newfound skills! 0

## Digging Up 3-Digit Cube Roots

Published on Sunday, November 17, 2013 in , , , , Over in the Mental Gym, you can learn how to mentally work out the cube root of any perfect cube up to 1,000,000!

It's time to take that feat to the next level, so in today's post, you'll learn how to mentally work out the cube roots of perfect cubes between 1 million and 1 billion!

BASICS: To learn this feat, you should already be comfortable with the techniques taught in the Mental Gym for determining cube roots of perfect cubes up to 1 million. There's also a Scam School video on working out 2-digit cube roots of perfect cubes which you may find helpful.

To start, you'll need a volunteer to choose the number, and they need to be using a calculator that can display up to 9 digits. Fortunately, most smartphone-based calculators can handle this.

Have the volunteer enter any whole 3-digit number into the calculator, and then ask them to multiply it by itself and by itself again. In other words, if they chose the number 123, they'd multiply 123 × 123 × 123. Once they've done that, they should read off the number from their calculator, and you then write it down. You write it down because you'll need to study it as you work through it.

FINDING THE HUNDREDS AND ONES DIGITS: Just as with the original 2-digit version, you start by mentally breaking up the digits into groups of 3s. It's often helpful to write the number given to you with commas, which visually break up the number into groups of 3 automatically.

As an example, let's say you're given the number 15,069,223. Written that way, you can already see 15 as the leftmost group, 069 as the middle group, and 223 as the rightmost group. The leftmost group can consist of either 1, 2, or 3 digits, while the other groups will always consist of 3 digits.

You get the hundreds digit in the same way as in the 2-digit version (mentioned above in BASICS). You look at the leftmost set, in our example number of 15,069,223, that would be 15. We see that 15 is greater than 23 (8) and less than 33 (27), so the leftmost digit of the cube root would be 2 in our example.

To get the ones digit, you look at the rightmost digit, and work out which number, when cubed, would result in that final digit. Again, this is discussed in the tutorials linked under BASICS. Going back to our 15,069,223 example, we see that the last digit is 3. The only number which gives a number ending in 3 when cubed is 7, so the ones digit of the answer must be 7.

At this point, we know 2 of the 3 digits in the answer. In our example, we've already determined that the cube root must be 2 hundred-something and 7, or 2_7.

FINDING THE TENS DIGIT: To find the middle digit, we need to consider the entire number in a very particular way. What we need to know is what the remainder would be when the entire number is divided by 11. This is easy with small numbers, such as 68. We know 66 is a multiple of 11, so it's easy to work out that 68 has a remainder of 2 when divided by 11, but how do you deal with numbers of 7-, 8-, or 9-digits?

Fortunately, there are simple shortcuts you can use, and yes, they're simple enough to do in your head.

The classic trick is to start with the rightmost digit of the given number (the ones digit, in other words), subtract the number to its left, add the number to the left of the next one, and so on, alternately adding and subtracting as you go.

Our example of 15,069,223 becomes 3 - 2 + 2 - 9 + 6 - 0 + 5 - 1, or 4. In your mind, you might think of it more as, #147;3 minus 2 is 1, plus 2 is 3, minus 9 is -6, plus 6 is 0, minus 0 is 0, plus 5 is 5, minus 1 is 4.” Note that, if we double check the problem of 15.069,223 mod 11 with Wolfram|Alpha, we also get 4. Working this way, you'll always get an answer from 0 to 10, but how do we use this number?

Note that the answer we get at this point is representative of the large number in question, not the 3-digit answer. Basically, we need to ask ourselves what type of number, when cubed and divided by 11, would leave the specific remainder (4, in our example). Here's how we do that.

Each number from 0 to 10, when cubed, leaves a unique remainder when divided by 11, as shown in this table on Wolfram|Alpha. This pattern repeats starting over again at each multiple of 11. So, any number 2 greater than a multiple of 11 when cubed and divided by 11, will always leave a remainder of 8. Any number 3 greater than a multiple of 11, when cubed and divided by 11, will always leave a remainder of 5, and so on.

Since we're starting from the cubed, number, we look for the remainder we've determined, and find the number above it on the table. Our example's remainder was 4, which can be found underneath the 5 in that table. This means that the cube root must be 5 greater than a multiple of 11.

Before I continue, let me discuss memorizing the table. It may seem hard, but it relly isn't. First, notice that the cubes which, when divided by 11, leave a remainder of 0, 1, or 10, will have a cube root that, when divided by 11, will leave a remainder of 0, 1, or 10 respectively. Those are the easiest ones to recall.

If you look at 23 mod 11, note that the result will be 8. Now look at 83 mod 11, which is 6. 63 mod 11 is 7, and 73 mod 11 is 2. Those 4 numbers make a small cycle of {2, 8, 6, 7}. When working from the large perfect cube, you find the remainder after dividing by 11, look on that list, and then look for the number immediately before it for the number you need to determined the root. 7 would give you 6, 6 would give you 8, 8 would give you 2, and 2 would give you 7 (note how this cycles around).

There's one more cycle to help you memorize the table. 33 mod 11 is 5, 53 mod 11 is 4, 43 mod 11 is 9, and 93 mod 11 is 3. This other cycle, then, is {3, 5, 4, 9}. Similar to the previous cycle, you find the remainder of the large cube when divided by 11, look on the list to the number immediately before it, and you'll get the number that will help you determined the root.

Since we determined that 15,069,223 mod 11 is 4 with a quick calculation, we can recall that 4 is part of the {3, 5, 4, 9} cycle, and that since the 5 is right before the 4, the root must be 5 greater than a multiple of 11. We did this earlier by looking at the table, but you want to be able to recall the table from memory when demonstrating this feat.

At this point, we know 2 important things about the cube root. Our example cube root looks like 2_7, and is 5 greater than a multiple of 11. That's enough to narrow down the answer to one specific number!

Remember the technique of alternately adding an subtracting digits to get the remainder after dividing by 11? We're going to use it again here. So, the example problem becomes 7 - x + 2 = 5. That simplifies to 9 - x = 5, and the answer is obviously 4, because 9 - 4 = 5. That means that 4 is the digit in the middle! Our example cube root, then, must be 247. Double-checking our answer with Wolfram|Alpha, we find that 247 is 15,069,223!

More generally, you can just add the two numbers you have and subtract the remainder number you worked out. We take 2_7 and add 2 + 7 to get 9, and then subtract the remainder, 5, to get 9 - 5 = 4. Since we're talking about working through this in your head, that's the easiest way to go about it.

QUICKER EXAMPLE: This seems long only because it's new to you, and the example was interspersed with the description of the technique. Let's try a more streamlined example, and say we're given the number 241,804,367. What's the cube root?

Looking the leftmost group, we see 241 is between 63 and 73, so the answer is going to be in the 600s. The rightmost digit is a 7, and only 33 ends in 7, so our answer is along the lines of 6_3.

What is the remainder when that number is divided by 11? It's 7 - 6 = 1, + 3 = 4, - 0 = 4, + 8 = 12, - 1 = 11, + 4 = 15, - 2 = 13. 13 is actually larger than 11, so we subtract 11 from 13 to get 2.

So, the large perfect cube, when divided by 11, leaves a remainder of 2. 2 is part of the easily-recalled {2, 8, 6, 7} cycle, and right before the 2 is 7, so the cube root must be 7 more than a multiple of 11.

One last simple calculation, 6 + 3 (the two outside numbers) = 9, and 9 minus 7 (the remainder after dividing the cube root by 11) = 2, so the cube root must be 623! Sure enough, Wolfram|Alpha confirms that 6233 is 241,804,367.

PRACTICE: You can practice your newfound skill by having Wolfram|Alpha generate random perfect cubes between 1 million and 1 billion here.

While practicing, you'll also learn to watch for pitfalls. For example, I once got the number 367,061,696. I worked out that the leftmost group was between 73 and 83, so the cube root must be in the 700s. The rightmost digit is 6, which means that the rightmost digit of the cube root is 6. So far, I have 7_6.

Working out the remainder when dividing by 11, I got 6 - 9 = -3, + 6 = 3, - 1 = 2, + 6 = 8, - 0 = 8, + 7 = 15, - 6 = 9, + 3 = 12, and since 12 is greater than 11, I worked out 12 - 11 = 1. Recall that a remainder of 1 when the cube is divided by 11 means that the cube root has a remainder of 1 divided by 11, as well.

So far, so good, but we're about to encounter a small problem. We add together the two outer digits, 7 + 6 = 13, and then subtract 1 (the remainder of our cube root) for a final result of . . . 12?!? Obviously, 12 can't be the middle digit, as we're only looking for single digit. When you get a number equal to 11 or more, always subtract 11, and keep doing so until you get a number from 0 to 10. 12 - 11 = 1, so 1 is actually the middle digit we're looking. Therefore, the cube root of 716, and Wolfram|Alpha verifies this is correct.

Have fun, practice, and go out and amaze your friends and family with your newfound skill! I'd love to hear about any memorable reactions in the comments! 0

## Fibonacci Meets Dr. Benjamin

Published on Thursday, November 14, 2013 in , , , , , , , If you watched even part of Dr. Arthur Benjamin's lecture in my previous post, you get an idea of the kind of joy and enthusiasm he has for mathematics.

In this post, we feature another new video from Dr. Benjamin, focusing only on Fibonacci numbers. This video is only about 6-7 minutes long, so you can enjoy it even if you couldn't find time to watch the complete video in my previous post.

This video is titled Arthur Benjmain: The Magic of Fibonacci Numbers, and is available on YouTube, as well as TED.com:

The presentation you see behind him in this video was created on Prezi.com, and Dr. Benjamin has made it freely available to view there. If you're viewing this on an Apple mobile device, the presentation can be viewed more effectively via Prezi's free apps.

Here on Grey Matters, we have a healthy respect for the magic and fun Fibonacci numbers make possible. Back in 2010, I posted about the classic Fibonacci addition trick, and other fun with phi. A little less than a year later, Scam School featured the same trick. Playing around with it on my own, I even found a little-used way to expand the routine to a list of any length!

Fibonacci numbers have even found their way into Nim, a favorite game here on Grey Matters. There's Fibonacci Nim, which is the standard take-away game with new rules that let you win with your knowledge of Fibonacci numbers. The Corner The Lady version of Wythoff's Nim is probably the most deceptive use of Fibonacci numbers.

In a post a little over a year ago about the principle behind the classic Age Cards magic trick, there's even a James Grime video showing the Age Cards set up with Fibonacci numbers instead of the standard binary approach!

Take some time to explore and play with these magical numbers, and your time will be well rewarded! 0

## It's All About The Benjamin

Published on Sunday, November 10, 2013 in , , , , , , , , , , , Even if you're not into mathematical magic and mental math, you're probably familiar with Dr. Arthur Benjamin from one or more of his TED talks.

Another video of his mathemagical feats has surfaced on the web, but this one includes the methods of each routine!

This video lecture is titled The Magic and Math of Mental Calculation, and was held at the 2013 Martin Gardner Celebration of Mind in Washington DC, courtesy of the Mathematical Association of America and Math For America-DC.

The Magic and Math of Mental Calculation is done in full lecture style, and runs about 78 minutes. It is introduced by MAA's Ivars Peterson and Thinkfun (Amazon.com link) CEO Bill Ritchie:

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.

Coming back to the discussion of the missing digit feat, it's hard to make this much clearer than it is on the video. There is the amusing question of whether zero is an even number, which Numberphile tackled in one of their videos.

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.

Why is he doubling that first calculation? Effectively, he's breaking the problem down into (37,000 + 947)(37,000 + 947). As with any problem of the form (a + b)(a + b), Wolfram Alpha shows that the result must be a2 + 2ab + b2.

The mysterious words he's uttering are actually ways of remembering numbers. Arthur Benjamin has another free lecture available online that details how to memorize numbers like this.

As with many live lectures, this one winds up with several mentions, including that of Harvey Mudd College, where Dr. Benjamin teaches.

Several of Dr. Benjamin's books and DVDs are promoted in the lecture. Since Grey Matters is an Amazon.com affiliate, you can help support this blog by buying Dr. Benjamin's books through our affiliate link, his Secrets of Mental Math DVD (from which the above free number memorization lecture is taken), his Joy of Mathematics DVD, and/or any of the Amazon.com links listed above. 0

## Numerous Wonders

Published on Monday, November 04, 2013 in , , , , , , , , , It's time for some magic!

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!

The method behind this simple trick is described in Futility Closet's On Time post.

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 basic tapping presentation has a long history in magic. In Martin Gardner's book, Mathematics, Magic and Mystery, there's an entire section on tapping tricks. Thank to Google Books, you can read the entire section online for free, running from page 101 to page 107.

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! 