Leapfrog Division II

Published on Sunday, April 20, 2014 in , , , ,

Melchoir's, AzaToth's, Mets501's, and Sopoforic's 0.999 perspective imageA little over a year ago, I wrote up my Leapfrog Division post. It's about time for a modification!

The original technique showed you how to work out the decimal equivalent for fractions whose denominators end in 9. In today's post, you'll learn a similar technique for denominators ending in 1.

Before learning this technique, you do want to be very comfortable working with the original Leapfrog Division technique, as it has all the basics of this modified version.

In that post, we worked out 1319, so I thought it might keep things simple if we used 1321 for our first example this time around. Just like before, we're going to use 20 as a base to divide, and minimize it to the number 2 (simply by dropping the 0), but there are a few changes.

Before any dividing, the next step this time is to take the numerator and subtract 1. So, we've taken 1321, converted it to 1320, dropped the zero to get 132. Subtracting 1 from the numerator at this point gives us 122 as our starting point.

Do this calculation in your head so that you get a quotient and a remainder, just as in the original technique: 12 ÷ 2 = 6 (remainder 0)

Just as in the original, you're going to have the remainder “leap in front of” the quotient, but here's where the new extra step comes in. Before the “leaping” is done, you're going to subtract the quotient from 9, then put the remainder in front of that new result.

With our 6 (remainder 0) example, we'd work out 9 - 6 (the quotient) = 3, and then put the remainder of 0 in front of that to get 03. You keep repeating the steps in this manner as far as you wish. Starting from the 122 step:

  • 12 ÷ 2 = 6 (remainder 0)
  • 03 ÷ 2 = 1 (remainder 1)
  • 18 ÷ 2 = 9 (remainder 0)
  • 00 ÷ 2 = 0 (remainder 0)
  • 09 ÷ 2 = 4 (remainder 1)
  • 15 ÷ 2 = 7 (remainder 1)
  • 12 ÷ 2 = 6 (remainder 0)
  • 03 ÷ 2 = 1 (remainder 1)
Do you see how the new number to divide by 2 at each point. In each case, the tens digit is the remainder from the previous problem, and the ones/units digit is 9 minus the quotient from the previous problem. This is a little tricky until you get familiar and comfortable with the technique.

At this point, we can see that the pattern is already starting to repeat, which happens often. Checking with Wolfram|Alpha, we confirm that 1321 ≈ 0.61904761...

You would deal with improper fractions just as before, reducing them to mixed fractions (the links about improper fractions in the original post are still helpful here) before using the leapfrog technique. 7131, for example, should be converted to 2931 as the first step. So we have 2 something, and then we use this version of the leapfrog division technique to work out the decimal equivalent.

In this example, 931 becomes 930, which then becomes 93, and subtracting 1 from the numerator gives us 83. From here, we work out:
  • 08 ÷ 3 = 2 (remainder 2)
  • 27 ÷ 3 = 9 (remainder 0)
  • 00 ÷ 3 = 0 (remainder 0)
  • 09 ÷ 3 = 3 (remainder 0)
  • 06 ÷ 3 = 2 (remainder 0)
  • 07 ÷ 3 = 2 (remainder 1)
  • 17 ÷ 3 = 5 (remainder 2)
  • 24 ÷ 3 = 8 (remainder 1)
No, I don't see a pattern repeating yet, but that's more than enough digits to be impressive. Sure enough, Wolfram|Alpha confirms that 7131 ≈ 2.29032258...

If you take the time to become comfortable with the original version, and then this version, you have 2 very powerful tools for converting decimals to fractions. As a matter of fact, they may be more powerful than you think!

BONUS: Long-time Grey Matters readers may remember my posts on estimating square roots of non-perfect squares and a few tips and tricks. This square root estimation results in an answer in fraction form. Because you're always adding effectively adding two consecutive integers to get the denominator, you'll always have a fraction with an odd denominator.

So, the denominator in that feat will always end in either 1, 3, 5, 7, or 9. Thanks to both versions of the leapfrog division technique, you can now convert any denominators ending in 1 or 9. What about the others numbers?

It turns out that for denominators ending in 3, you can simply multiply the numerator and denominator by 3 to get an equivalent fraction with a denominator ending in 9. For example, 513 = 1539, so you can use the original version of the leapfrog division technique to work that out.

Similarly, when you have a denominator ending in 7, you can simply multiply the numerator and denominator by 3, which will give you a denominator ending in 1. 1217 = 3651, which should be easy for you with this second version of the technique.

For fractions with denominators ending in 5, this technique isn't often applicable. You may get lucky and be able to scale a fraction down to a number ending in 1 or 9, such as 2055 = 411, or 35145 = 729, but you can't always count on that. Using any whole number to scale a fraction up whose denominator ends in 5, of course, can only result in a denominator ending in 0 or 5, of course. If the denominator is just 5, though, you should have little problem working out the decimal equivalent.

Since you now realize you can handle odd denominators ending in 1, 3, 7, or 9, for the square root estimation feat, you stand a good 80% chance of being able to give a decimal equivalent, and taking the feat to an impressive new level!

I suggest practicing this, having fun with it, and impressing a few friends with your newfound skill. Enjoy!


How to Find Counterfeit Coins

Published on Sunday, April 13, 2014 in , , , , , ,

Toby Hudson's brass scale imageScam 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.

We'll start by jumping right in to the counterfeit coin puzzles as presented on this week's Scam School (alternative YouTube link):

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.








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).

There's only one way to write the binary equivalent of the decimal number 217. To find out what it is, you can either do a binary conversion with the help of a tool such as Wolfram|Alpha, or, if you've been reading Grey Matters long enough, do the conversion in your head.

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.


1062.2 grams - 1,023 grams = 39.2 grams

39.2 grams ÷ 0.1 gramscounterfeit coin = 392 coins

392 in binary = 0110001000

• Therefore, if the total weight is 1,062.2 grams, the bags containing counterfeit coins are bags 8, 7, and 3.


Calculate Powers of π In Your Head!

Published on Sunday, April 06, 2014 in , , , ,

Mehran Moghtadaei's Pi Digit GraphicAs a follow-up to last week's tutorial on calculating powers of e in your head, I'm going to teach you how to do the same for our old friend Pi!

As an added bonus, calculating powers of Pi can be slightly easier than powers of e, so even if you passed up last week's tutorial, you should still give this one a look.

BASICS: In a manner similar to the previous tutorial, we'll request a number x, and solve for y in the equation πx = 10y.

This method, just like the previous one, is also based on turning this problem into a logarithm. It becomes log10(πx) = y, which simplifies to x × log10(π) = y. This works out to about x × 0.49714987... = y.

This time around, we can take advantage of the fact that this number is close to 0.5!

TECHNIQUE: We'll use π16 as our example requested power.

The first step this time involves setting up a subtraction problem with 2 numbers, both of which start with the given number (16, in our example).

BUILDING THE SUBTRACTION PROBLEM: To begin, take the given number an multiply it by 500. This can be made simpler, if you prefer, by simply tacking “,000” on the end of your number, and then dividing by 2. This is the first number we need.

Applying this step to our example number 16, we add a comma and 3 zeroes to it (16,000), and then divide by 2 to get 8,000. We have our first number.

Note: If you're given an odd number, you will always end up with a “,500” at the end. For example, if the given number was 15, this step would result in 7,500. Knowing this is a handy way to make sure you didn't misplace the comma.

To get the second number for the subtraction problem, simply multiply the given number by 3. This should be easy enough to do without any special tips.

Doing this, our second number is 16 × 3 = 48.

SUBTRACTION:Having set up the 2 numbers for the subtraction problem the next step, not surprisingly, is to perform the subtraction by subtracting the smaller number from the larger number.

We've worked out the numbers 8,000 and 48 in our 16 example, so the subtraction problem is 8,000 - 48.

If you're like most people, though, you remember writing down subtraction problems with lots of zeroes in school, and having to borrow over multiple places. That being the case, you're probably wondering how to deal with all this in your head! The following video teaches you how to deal with problems like these without ANY borrowing:

To work out 8,000 - 48 using the above technique, it's probably better if you think of the problem as 8,000 - 048. The first step, as in the video, is to round the leftmost digit up, from 0 to 1 in this case, and seeing that 80 - 1 = 79. We already know the answer begins with 79!

How far up would you have to go from 48 cents to get to a whole dollar? Getting the answer of 52 cents shouldn't be a problem here. That's the other half of the answer.

Your running total, at this point, is 7,952. After a little practice, subtracting from zeroes in your head will seem not only less scary, but nearly effortless.

ADJUSTING FOR APPROXIMATION: We're going to add a little now to improve the accuracy of our answer. How do we do that?

Take the number you just subtracted, and throw away the ones digit. Divide the remaining digits by 2. If that ends in a .5, just throw the .5 away, as well. This is the number you add to your running total.

We just subtracted 48 to get 7,952. We take 48 and throw away the ones digit, leaving 4. Dividing that by 2, we get 2. Finally, we add 2 to 7,952 to get 7,954 as our new running total.

DIVIDING BY 1,000: To divide by 1,000, I could tell you to move the decimal point three places to the left, but there's an even simpler technique this time. All you have to do is replace the comma in the total with a decimal point!

With this approach, 7,954 instantly becomes 7.954 with very little effort.

At this point, you're done! As you can verify on Wolfram|Alpha, π16 ≈ 107.954.

THE FULL PROCESS ALL AT ONCE: To run through this at once, and to better acquaint you with the full range of situations you'll run across, let's try to work out π33 = 10y.

  • Multiply 33 × 1,000 to get 33,000, and divide by 2, getting 16,500.
  • Multiply 33 × 3 to get 99.
  • Subtract 16,500 - 99 = 16,401.
  • Throw away the 9 (the ones place of 99), leaving 9, and divide that by 2 (4.5), throwing away the .5 to leave 4.
  • Add 16,401 + 4 = 16,405.
  • Replace the comma with a decimal point, resulting in 16.405.
Once again, check for yourself on Wolfram|Alpha to see that π33 ≈ 1016.405

TIPS: Most of the tips I gave for e apply for π, as well. I'll repost the relevant ones below for convenience, modified for our Pi examples.

• By looking at the whole number part of the answer (the significand) and adding 1 to that, you can state the number of digits the full answer would have. In our 7.954 example, we take the whole number part, the 7, and add 1 to get 8, so we can state that the answer is a 8-digit number. Having worked out π33 to be about 1016.405, we can state that the answer is a 17-digit (16 + 1) number!

• You can handle exponents with a decimal in them by working them out as if they were a whole number, and then adjusting for an appropriate number of additional decimal points when you're moving the decimal point. For example, π1.6 is the same as π16, but with the decimal moved one place more to the left. Since π16 ≈ 107.954, it's easy to see that π1.6 ≈ 100.7954.

• If you want to take this a step further, and be able to say that π16 is roughly equal to 9 × 107, check out Nerd Paradise's Calculating Base 10 Logarithms in Your Head, the video Calculating logarithms in your head, and the PDF How to Quickly Calculate Logarithms to Three Decimal Places in Your Head.


Calculate Powers of e In Your Head!

Published on Sunday, March 30, 2014 in , ,

LoStrangolatore's 10,000 digits of e posterBefore getting to the main topic of today's post, I have a couple of announcements to make!

First, I've normally tried to post on Thursdays and Sundays on this blog, but regular commitments are taking priority, so I'm going to be posting on Sundays only for the time being.

Second, I recently created a new subreddit all about mental math techniques. If you're on reddit.com, please subscribe and contribute your own thoughts and links! Since it's often quicker to post links there, this may hopefully compensate for the difference in the blog posting schedule.

The rest of today's post will be about calculating powers of e in your head. Wait...what exactly is “e”?!? That's a great place to start.

BASICS: When mathematicians refer to e, they're referring to a specific number, approximately 2.71828 (it goes on forever, just like Pi), that is used to calculate continuous growth.

How can one number be used to calculate continuous growth? Why is it that particular number? The best place to understand these and other questions about e is to read BetterExplained.com's post An Intuitive Guide To Exponential Functions & e. It really is the best way to wrap your head around e as a number.

What exactly will you be working out in your head? Given an exponent x, you'll be able to take a problem of the form ex, and turn it into a power of 10 (in other words, you'll figure out y in 10y).

The technique I'll teach is based on the fact that we can take the general problem ex = 10y and isolate y by turning the problem into a logarithm: log10(ex) = y. Thanks to standard logarithm rules, we can turn this into the following form: x × log10(e) = y. A little help from Wolfram|Alpha tells us that this can further be simplified to the following form: x × 0.4342944... = y

OK, I hear you say, we've turned a seemingly complex exponential problem into a multiplication problem, but it doesn't seem easy to multiply by a number like 0.4342944 in your head! What we're actually going to be doing is approximating this number by multiplying by 0.4343. Yes, we'll be adjusting for the approximation, as well. Further, we're going to break that up into several steps, by turning it into the form (43 × 101)10,000.

Yes, I have a few tricks up my sleeve to help you make these calculations at each stage, and I'll teach them in the next section.

TECHNIQUE: Let's start with a simple example of working out e16 as we go through each of the following steps.

MULTIPLYING BY 43: Take the exponent and multiply it by 4. In our example, 16 is the exponent, and you should be able to quickly work out that 16 × 4 = 64.

Next, take the exponent and multiply it by 3. Continuing with our example, 16 × 3 gives us a total of 48.

To make things easier, we need to split this 2nd answer up into the 1s digit and everything else. In our example, instead of thinking of the answer as simply 48, think of it as “+ 4 with an 8 slapped on the end.” You'll always think of the multiple of 3 in this way. If multiplying by 3 resulted in, say, 312, you'd think of that as “+ 31 with a 2 slapped on the end.” This type of thinking will make it easier for the next step.

To complete this step, add the two answers together. In our example, we'd add 64 “+ 4 with an 8 slapped on the end.” The answer then, is 64 + 4, which is 68, with an 8 slapped on the end, which is 688. Yes, 43 × 16 is 688.

Dealing with say, 23? 23 × 4 = 92 and 23 × 3 = “+ 6 with a 9 slapped on the end.” 92 + 6 with a 9 slapped on the end is equal to 98 with a 9 slapped on the end, or 989. Practice multiplying by 43 in this way before moving on. You may be surprised how quickly you get used to it.

MULTIPLYING BY 101: When dealing with 2 digit numbers, multiplying by 101 is ridiculously easy, as you simply repeat the 2 digit number itself. 43 × 101 = 4343 and 86 × 101 = 8686.

For this feat, however, you'll have to be able to deal with multiplying 3- and 4-digit numbers by 101. Don't worry, though, it's still simple.

To multiply a 3- or 4-digit number by 101, split the original number up, with the 10s and 1s digit off to the right, the remaining digits off to the left, and imagine two placeholders in-between them.

So, continuing with our example, 688 would be split up into 6__88. That's 6, followed by 2 placeholders, followed by 88, the original 10s and 1s digit.

To fill the placeholders, simply add the numbers on either side of the placeholders. 6 + 88 = 94, so we put 94 in those empty spaces, resulting in the number 69488. Yes, you've just mentally worked out the answer to 16 * 4343 in your head!

You do have to watch for the times when the digits on either side of the place holder add up to more than 100. In that case, you have to add the 1 to the digits on the left to compensate.

For example, if you multiplied 44 by 43 earlier and worked it out to be 1892, you'd then split that into 18__92, and then add 18 + 92 to get 110. The 10 part goes in the middle, and the 1 from the hundreds place is added to the 18 to get 19. The result is 191092. This is simple, once you get used to it.

DIVIDING BY 10,000: This is the simplest part of all! Just take the number from the previous step, and move the decimal 4 places to the left.

In our continuing example, we worked out the number 69488. Moving the decimal 4 places to the left turns the number into 6.9488!

What about numbers like, say, 191092? that becomes 19.1092. Simple isn't it?

ADJUSTING FOR APPROXIMATION: Remember that the decimal approximation of the base 10 log of e is actually 0.4342944..., and we've multiplied by 0.4343. That's close, but those are still 2 different numbers.

To make your result more accurate, simply drop the last (rightmost) digit of your answer. Don't round up or down, just completely ignore the last digit!

For our example result of 6.9488, we'd drop the rightmost digit to get 6.948. With that step, you're done. In your head, you worked out that e16 ≈ 106.948. How close is that to the actual solution? According to Wolfram|Alpha, the actual answer is 106.9487, so your mental estimate is accurate as far as it goes!

Naturally, if the rightmost digit of your calculation is 0, this is done for you almost automatically.

THE FULL PROCESS ALL AT ONCE: To make sure you've got this, let's try and work out a bigger problem. This time, we'll calculate e27 = 10y.

  • 27 × 4 = 108 and 27 × 3 = +8 with a 1 slapped on the end.
  • 108 + 8 with a 1 slapped on the end is equal to 116 with a 1 slapped on the end, or 1161.
  • 1161 becomes 11__61, and 11 + 61 = 72, so we get 117261.
  • 117261 with the decimal moved 4 places to the left is 11.7261.
  • Drop the rightmost digit to get the final estimate of 11.726.
Wolfram|Alpha initially shows that e27 = 1011.726. If you click the More digits button, you'll see that 11.7259510113878 is a more accurate answer, so 11.726 is still an impressive mental estimate!

REAL-LIFE EXAMPLE: In this Business Insider story, there's an anecdote involving the absurdity of 4.5% growth for 3,000 years. You can figure that out with e3,000 × 0.045.

To get our exponent, that's 3,000 × 0.045 = 300 × 0.45 = 30 × 4.5 = 3 × 45 = 135. So, the answer we need to estimate is e135.
  • 135 × 4 = 540 and 135 × 3 = +40 with a 5 slapped on the end.
  • 540 + 40 with a 5 slapped on the end is equal to 580 with a 5 slapped on the end, or 5805.
  • 5805 becomes 58__05, and 58 + 05 = 63, so we get 586305.
  • 586305 with the decimal moved 4 places to the left is 58.6305.
  • Drop the rightmost digit to get the final estimate of 58.630 (or 58.63).
Sure enough, that's quite close to the answer of e3,000 × 0.045 = 1058.6298!

TIPS: e is all about continuous growth, which is basically all about scale. I suggest reading How to Develop a Sense of Scale, Using Logarithms in the Real World, and my own Visualizing Scale post. Viewing and recalling the classic Powers of Ten video can be useful for concrete comparisons, as well.

• By looking at the whole number part of the answer (the significand) and adding 1 to that, you can state the number of digits the full answer would have. In our 6.9488 example, we take the whole number part, the 6, and add 1 to get 7, so we can state that the answer is a 7-digit number. Having worked out e135 to be about 1058.63, we can state that the answer is a 59-digit (58 + 1) number!

• Practice estimating larger and larger powers of e, until you're comfortable dealing with numbers up into the low 200s. Since all you really need is to get comfortable with multiplying by 4 and 3, you can find tips for multiplying 2-digit numbers by 1-digit numbers here, and multiplying 3-digit numbers by 1-digit numbers here.

• To prevent yourself from having to deal with multiplying 5-digit numbers from 101, simply avoid exponents greater than 232. Being able to estimate up to e232 in your head is still impressive!

• You can handle exponents with a decimal in them by working them out as if they were a whole number, and then adjusting for an appropriate number of additional decimal points when you're moving the decimal point. For example, e1.6 is the same as e16, but with the decimal moved one place more to the left. Since e16 ≈ 106.948, it's easy to see that e1.6 ≈ 100.6948.

• If you want to take this a step further, and be able to say that e16 is roughly equal to 8.8 × 106, check out Nerd Paradise's Calculating Base 10 Logarithms in Your Head, the video Calculating logarithms in your head, and the PDF How to Quickly Calculate Logarithms to Three Decimal Places in Your Head.


Yet Again Still More Snippets

Published on Sunday, March 23, 2014 in , , , , , , ,

Luc Viatour's plasma lamp pictureMarch's snippets are ready!

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.

• Just recently, @preshtalwalkar of the Mind Your Decisions blog posted an examination of the classic four knights puzzle. Read the post up to the answer, and then try playing it yourself in my 2011 post on the same puzzle. It's a challenging puzzle, until the simple principle behind it becomes clear. Once you understand the principle behind the four knights puzzle, see if you can use it to work out the method for the Penny Star Puzzle.

• 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!

• About a year ago, @Lifehacker had a post about measuring your feet and hands to measure distances accurately without needing a ruler, which was based on this quota.com reply. To take this a step farther, I recently learned you can even judge far-off distances and even angles using just your fist and thumb! This is one of those tricks that can be handy and even impressive at the right moment.

That's all for this month's snippets, but it's plenty to explore and discover, so have fun with these links!


Enter The Matrix

Published on Sunday, March 16, 2014 in , ,

Jamie Zawinski's screenshot of the famous GLMatrix screensaverNo, you don't have to decide between the blue pill and the red pill right now, but it will be hard to escape jokes about the movie The Matrix when talking about matrices.

Sometimes, mental math isn't about being able to do math in your head quickly, but more about wrapping your head around a topic. Matrices can be especially tricky, so this post is dedicated to them.

About a year ago, TED-Ed posted a great introductory video on matrices, with a full lesson available here. I especially like the visual approach to matrix multiplication:

I do have a few criticisms of that video. First, I think they should've gone a little further, and shown a complete multiplication of a 3 by 2 and a 2 by 3 matrix, just for completeness sake. Second, did anyone else get confused by the mention of “linear algebra”? Doesn't it seem strange that a rectangular arrangement of numbers should be so important in something that begins with the word linear?

Once again, BetterExplained.com's Kalid Azar comes to the rescue with An Intuitive Guide to Linear Algebra. He takes things quite a bit slower than the video, but he does this so you can grasp all the concepts.

Kalid starts, thankfully, with an explanation of just what is so linear about matrices and linear algebra. He then shows you how to get your head around the unusual notation and multiplication. His analogy to mini-spreadsheets is a very helpful!

Probably the biggest help is the real-world example, instead of the classic vectors that are often used. He even briefly covers scary terms such as “eigenvector”.

Do take the time to go through these resources, as the matrix is a fascinating tool that shouldn't be ignored.


Grey Matters' 9th Blogiversary!

Published on Friday, March 14, 2014 in , , ,

Mehran Moghtadaei's Pi Digit Graphic135 years ago today, Albert Einstein was born. 9 years ago today, Grey Matters was born. Of course, today will always be Pi Day!

Welcome to a geeky, yet still special, day of the year that is close to my heart!

Vi Hart kicked things off with an anti-Pi rant. This might seem like a strange action to take on Pi Day, but she does it in her own inimitable style, and really does make some good points about Pi in the process:

Numberphile followed closely behind with some Pi-sinpired music, and then enlisted James Grime to talk about river lengths, and their amazing connection to Pi:

Here's the 1996 paper that inspired this claim, but there is a bit of controversy over this point. It's still an interesting concept to ponder, however.

The good people over at Plus Magazine have some great thoughts, posts, and even artwork all about Pi!

Speaking of Pi artwork, mathematical animator 1ucasvb created a new sine and cosine animation especially for Pi Day that is simple, yet informative. Lucasvb has more animations on that site (this one being a particular favorite), and even more over in his Wikimedia Commons gallery. Even if you don't understand all the math behind them, enjoy and explore the animations.

I'm off to celebrate Pi Day for now, but feel free to enjoy and post your Pi Day wishes in the comments!


Dots Crazy!

Published on Sunday, March 09, 2014 in , , ,

Scam School logoIn their recent 311th episode, Scam School featured an interesting scam that warrants a closer look.

After all, there's nothing fairer than a game board consisting solely of dots, right?

Check out Scam School's 311th episode, and see how fair it comes across versus the sneaky approach behind the game:

That's amazing and sneaky, but how would you even go about working out the math behind it?

Never fear, because Numberphile is about to come to the rescue! In a recently posted video, they take a look at a similar type of game called Brussel Sprouts. The math behind it is less complicated than you may expect:

As many regular Grey Matters readers realize, I love it when something fun like this can lead to a better understanding of the math behind it.

According to the above video, we can also expect a more detailed video about the Euler characteristic mentioned in this video. You can also find out more about Brussels Sprouts and other dot games in Martin Gardner's books Mathematical Carnival and The Colossal Book of Mathematics.

Have fun and play around with these games. You never know what else you may learn!