3 Planets and a Rational War

Published on Friday, March 17, 2017 in , , ,

Still of Clint Eastwood from The Good, The Bad and The UglyDuels are a classic staple of westerns, but it took Sergio Leone's 1966 film “The Good, The Bad and The Ugly” to immortalize the truel (neologism for a three-way duel) in the popular mind!

Back in 1975, a high school teacher named Walter Koetke took the truel into outer space in a puzzle from the May/June 1975 issue of Creative Computing. I first ran across it in the 1980s, in The Best of Creative Computing.

This puzzle concerned 3 planets, Mutab, Neda and Sogal, who have agreed to an unusually rational war. Each planet would fire planet-destroying missiles at one another in turns, with the order of the truns being determined by random drawing. The rocket launching continues in order until only 1 planet remains. However, each planet has different capabilities. Mutab is the most advanced, and when fired at a target, have a 100% chance of destroying that target. Neda is the next most advanced, but their missiles only have an 80% chance of destroying their target. Sogal is the least advanced of the three, with missiles that only have a 50% chance of destroying their target. The puzzle is this: What are each planet's chances of surviving a war?

Even when I first ran across the problem as a teen, I remember thinking how difficult this problem must be. I realized that any calculations between the less-than=100% chances of battles between Neda and Sogal would be difficult to deal with, as they could go on for several turns, much like encountering long heads-only or tails-only runs in coin-flipping. The problem did stick in the back of my mind, though, as I thought it would be interesting to analyze when I knew more about math.

That day arrived when I first learned about Markov chains. If you're already familiar with Markov chains, you're ready for my solution. I've set up a site that takes you through the solution step-by-step here. Start by clicking the “Click to work out 2-world battles” button, and then read through the logic and tables until you get to the next button. Keep proceeding that way all the way to the end. It's written using Bootstrap, so it should be viewable using any size screen (the larger tables can be swiped left and right on smaller screens). There are lots of labeled tables, explanations of what they mean, and even several links to calculations via Wolfram|Alpha. All the math is being worked out by your browser as the program runs, so you may need to be patient with slower computers.

If you're still reading because you're unsure what Markov chains are, they're not that difficult a concept, as long as they're explained clearly. The rest of this post will deal with the basic concepts. Below is an excellent and clear (and short!) introduction to the concept.

What I realized is that I could represent the targeting of one planet by another as one state, and each of the possible outcomes as individual states. From this, I could build a transition probability matrix, as explained in the video and this visual explanation of Markov chains.

Now, the video discusses the idea of a starting probability vector as a starting point. In the Mutab/Neda/Sogal puzzle, however, we're really only interested in the long term probabilities themselves, the effects over the long term, and not so much a particular starting point, referred to as a stability distribution matrix.

As an example, imagine a 2-planet battle between Mutab (100% deadly) and Neda (80% deadly). If we list starting states from top to bottom (yes, I know this is opposite of how Markov chains are usually done) as, “Mutab fires against Neda”, “Mutab wins”, “Neda fires against Mutab”, and “Neda wins”, and label the resulting states the same way from left to right, then we can assign corresponding probabilities. The probability of going from “Mutab fires against Neda” to “Mutab fires against Neda” is 0, the probability of going from “Mutab fires against Neda” to “Mutab wins” is 1, and the probabilities of going from “Mutab fires against Neda” to “Neda fires against Mutab”, or “Neda wins” are both 0, so our first row would read (0 1 0 0). Continuing through all the probabilities between just Mutab and Neda, we get the following matrix:

Notice the 1s in the 2nd and 4th rows. Those are absorbing states. The 1 in the 2nd row basically says, the probability of going from “Mutab wins” to “Mutab wins” is 1 (100%). The 1 in the 4th row says that the probability of going from “Neda wins” to “Neda wins” is 1. They serve as an endpoint.

Where does this all wind up? Well, if we run these probabilities over 1,000 generations to get the long-term outlook (the stability distribution matrix mentioned earlier) by raising the above matrix to the 1,000th power, we get the following results:

Read with the original labels, this tells us that, over the long term, if Mutab is the first one to fire, then it's probability of survival is 1, and Neda's is 0. Conversely, if Neda is the first one to fire in that two-world battle scenario, then it's probability of survival is 80%, while Mutab's is only 20%. Yes, we figured that much out early on, but when it comes to tougher probabilities, such as the long-term probabilities between Neda and Sogal, this is a very helpful tool.

Once you've understood all that, you're ready to go through the step-by-step explanation I developed!


Grey Matters is Back!

Published on Tuesday, March 14, 2017 in , , ,

Mehran Moghtadaei's Pi Digit GraphicYes, I did take more than an extended break after my 10th blogiversary post, not posting at all in 2016, but with Pi Day coming around again, it seemed like a good time for a return! I probably won't have a regular posting schedule for sometime, as other commitments are keeping me busy, but I promise not to ignore this blog as I have been for about the past 1½ years.

While I've been thinking about returning on my own, one force that really pushed me over the edge to return to blogging is Brian Brushwood from Scam School. He recently used two of my submissions, and even encouraged people to go to Grey Matters!

The first of these two was posted back on January 18th. This was an update to Penney's Game, which was first taught on Scam School in their Sept. 15, 2010 episode (to which I also contributed!). Replacing cards with coins not might seem like such a big deal, it actually does have a significant effect. Coins can't run out of heads or tails, but cards can run out of reds and blacks. Watch the full episode and read more detail about the game here to understand it better.

Moving from cards and coins, we turn to dice. Here's an unusual dice scam from their March 1st episode. This one involves people picking how many dice of 5 they want, leaving you with at least 1 of them, and you're always able to predict the outcome. It seems like the odds are against you, but watch closely. At about the 3:30 mark, Brian asks the ladies how they think it's done. The theories included things such as particular numbers to pick, loaded dice, and changing the prediction based on previous evidence. The answer, as it so often is, is simple probability.

Even with all the troubles people have with most branches of math, probability is often the easiest to misunderstand and get wrong. As a result, many probability concept come across as counter-intuitive.

That's all for now. But I assure you, the time between this post and the next post will not be as long as the last post and this one.


Leapfrog Division IV

Published on Thursday, October 29, 2015 in , , ,

x divided by y fraction iconWhile I normally do my Leapfrog Division posts about a year apart, I though I'd wrap up this mental division series just 1 week after the previous entry.

In this post, you'll learn to mentally divide by numbers ending in 2!

STARTING POINTS: This is the most advanced technique of all of the Leapfrog Division posts, so you should be familiar with and practice the the previous techniques. Not only does this employ the basic ideas taught in the original Leapfrog Division post, but also the subtraction from 9 idea used in Leapfrog Division II, AND both the doubling and halfway-comparison concepts from Leapfrog Division III. If you comfortable with all of these concepts, then you're ready to move on to this version.

This version also introduces a new idea to the Leapfrog Division series: The stopping rule. In the previous versions, you could stop either when you realized the numbers were going to repeat, or when you didn't need any further precision. While the same is true here, in this version, you'll also need to stop when you get a quotient of 5, and a remainder of 0. You'll understand this better, including the exceptions, as you work your way through the technique.

THE TECHNIQUE: For our first example, we'll use 1732. As in the technique for dividing by numbers ending in 1, we will always start by reducing the numerator by 1, giving us 1632. Similar to the method for dividing by numbers ending in 8, we're also going to compare numbers to the half of the denominator. Anytime the numerator is greater than or equal to half of the original denominator, we'll reduce it by 1.

In the case of 32, half of that is 16, so we ask if the current numerator is greater than or equal to 16. In this example, it's currently 16 (exactly equal to half!), so we reduce by 1 more, giving us a division problem of 1522. As with all the Leapfrog Division techniques, we're now going to round the denominator to the nearest multiple of 10, and then divide it by 10. So, the problem becomes 153.

You shouldn't be surprised that we're going to divide this out using quotients and remainders. Starting with 152, we get:

  • 15 ÷ 3 = 5 (remainder 0)
Naturally, you write the quotient down right away, with a 0 and a decimal in front, as in 0.5. Ordinarily, the stopping rule might tell us to stop here. After all, we have a quotient of 5, and a remainder of 0. However, since 1732 (our original problem) isn't exactly 0.5, there's definitely more numbers to calculate, so we'll continue. We should still keep the stopping rule in mind for later, however.

From here, you're going to use the doubling idea as taught in Leapfrog Division III, in which you double numbers, but only keep the ones (units) digit. 5 doubles to 0, because 5 × 2 = 10, and we only keep the ones digit, which is 0. Next, as in Leapfrog Division II, you're going to subtract the quotient from 9. In this example, 9 - 0 = 9, so the new quotient is now 9. Leapfrogging the remainder, 0, to the front of the quotient, we know have 09, or simply 9.

Before dividing, we need to ask whether 9 is greater than or equal to 16. It isn't, so we don't decrease the number at this stage. After that question, only then do we do the division again:
  • 9 ÷ 3 = 3 (remainder 0)
We write the 3 down, giving us 0.53 so far. 3 doubled becomes 6, and 9 - 6 = 3. Leapfrogging the remainder of 0 in front, we have 03, or just 3. Is 3 greater than or equal to 16? No, so there's no decrease this time, either. The next division problem yields:
  • 3 ÷ 3 = 1 (remainder 0)
So far, our mental work has given us an answer of 0.531. Repeat once more. Double the 1 to get 2, subtract 9 - 2 to get 7, and put the remainder 0 in front of the 7, giving us a new divisor of 07. Is 7 greater than or equal to 16? No, so there's no decrease. Moving on to the next division:
  • 7 ÷ 3 = 2 (remainder 1)
Now we have 0.5312. 2 doubled becomes 4, and 9 - 4 = 5. Leapfrogging the 1 in front gives us 15. Is 15 greater than or equal to 16? No, so there's no decrease. Obviously, we can move on to the next division:
  • 15 ÷ 3 = 5 (remainder 0)
At this point, with 0.53125 as our current answer, you'll note we have a quotient of 5 and a remainder of 0. This means that the stopping rule kicks in. So, our final result is 0.53125, which is exactly what 1732 equals!

Now that you understand the steps, let's work out 1922 as a second example. We start by reducing 19 by 1, which is ALWAYS the first step, giving us 1822. Half of 22 is 11, and 18 is greater than 11, so we decrease it by 1 again, leaving us with 1722. Rounding the denominator down and dividing by 10, our starting problem should be 172. We start there, and work through the problem this way:
  • 17 ÷ 2 = 8 (remainder 1)
    (8 doubles to 6, 9-6=3, 1 makes it 13, which is MORE than 11, so 13 - 1 = 12.)
  • 12 ÷ 2 = 6 (remainder 0)
    (6 doubles to 2, 9-2=7, 0 makes it 07, which is less than 11.)
  • 7 ÷ 2 = 3 (remainder 1)
    (3 doubles to 6, 9-6=3, 1 makes it 13, which is MORE than 11, so 13 - 1 = 12.)
Since we're already back to dividing by 12, you can see that this is going to repeat. Writing down just the quotients, we get the correct answer of 1922 ≈ 0.863!

TIPS: Yes, this has more steps than any of the other approaches taught in the Leapfrog Division series, and it's not difficult to confuse the steps of the various versions. The solution, as always, is practice, practice, practice!

You may have noticed that I referred to this as the last post in the Leapfrog Division series. Why is that? Because using the 4 different techniques I've taught, you can actually handle dividing by most numbers with just a little adjustment. How do you handle numbers ending in...
So, every number except those ending in 5 or 0 are covered. That's good incentive to practice these techniques, as you can amaze many people with the ability to handle almost every division problem thrown at you. I hope you've found this series enjoyable and useful.


Leapfrog Division III

Published on Friday, October 23, 2015 in , , ,

x divided by y fraction iconIn 2013, I posted about Leapfrog Division, which was A.C. Aitken's approach for mental division by numbers ending in 9. In 2014, I built on this method with Leapfrog Division II, an approach for mentally dividing by numbers ending in 1.

It's 2015, so it's time for another update to the Leapfrog Division technique. This time, you'll learn the method for mental division by numbers ending in 8!

STARTING POINTS: You'll want to be very familiar with the process of dividing numbers by 9, as taught back in the Leapfrog Division post. There are a few extra steps in this version, as compared to the original version, so being well versed in the original is imperative. You may also find it helpful to have practiced the technique for dividing by numbers ending in 1, as taught in Leapfrog Division II, but that's not as essential to this approach.

When teaching this technique, I'm going to be referring to doubling a given number, but they're doubled in a special way. As used in this technique, you double the number, but only keep the ones (units) digit. Ordinarily, you would double 5 to get 10, but here you only need to remember the 0. In a similar manner, 6 doubled will give you 2 (12, with the tens digit dropped), 7 doubled will give you 4, 8 doubled will give you 6, 9 doubled will give you 8, and 0 doubled will give you 0.

THE TECHNIQUE: As our very first working example, we'll work out the decimal equivalent of 1318. Just as before, you're going to start by rounding up the denominator (the bottom number) to the nearest multiple of 10, and then drop the 0. In our example, that means that our fraction gets changed to 1320, and dropping the 0 from the denominator changes this to 132.

We're going to ask a question here which will be asked over and over again, and this question will help give us the correct total. Is our current numerator greater than or equal to half of the original (before rounding) denominator? If so, we MUST add 1 to it. For example, half of our original example denominator, 18, comes to 9. So, we're going to be asking at the start, and several points afterward, whether our current numerator is equal to or greater than 9. To start, we realize that 13 is equal to or greater than 9, so we add 1, giving us 142 as our actual first problem to solve.

We're going to work this out in a similar manner as before, solving this division problem with a quotient and a remainder. As you go, you're going to write down quotients as you go, and keep remainders in your head. Our first division yields:

  • 14 ÷ 2 = 7 (remainder 0)
At this point, you can write down 7 as the first number after the decimal point, and begin developing and solving the next problem.

How do we take the next step? First, the quotient (the 7 in our example answer above) must be doubled. Don't forget that we drop the tens digit when doubling! So, we double 7 to get 14, and drop the tens digit, leaving us with 4. The remainder (0, in the problem above) then “leapfrogs” to the front of the 4, giving us 04 as our new numerator (which is just equal to 4, of course).

We ask ourselves one more time, is our current numerator (4) greater than or equal to 9? In this case, 4 isn't greater than 9, so we don't add 1. After that, we divide by 2 again to get:
  • 4 ÷ 2 = 2 (remainder 0)
So, we can write down 2 as the next digit in the decimal answer, and move on to the next digit.

2 (the quotient) gets doubled again, to make 4, and the remainder of 0 leapfrogs in front to give us 04, or 4, once again. Is this new 4 greater than or equal to 9? No, so we leave it alone. Dividing by 2 one more time yields:
  • 4 ÷ 2 = 2 (remainder 0)
You write down the quotient 2 again. At this point, you can probably already see that this is going to repeat endlessly. If you check against a calculator, you find that 1318 is indeed 0.722..., with the 2 repeating endlessly.

Just to help lock in the technique, let's try and work out the decimal equivalent of 928. We have to remember to keep asking ourselves about half of the original denominator, which is 14 this time. Is 9 equal to or greater than 14? No, so we won't add 1 at this point. The denominator gets rounded up to 30, and we drop the 0 to leave us with a starting calculation of 93:
  • 9 ÷ 3 = 3 (remainder 0)
    (3 doubles to 6, 0 in front makes 06, which is less than 14.)
  • 6 ÷ 3 = 2 (remainder 0)
    (2 doubles to 4, 0 in front makes 04, which is less than 14.)
  • 4 ÷ 3 = 1 (remainder 1)
    (1 doubles to 2, 1 in front makes 12, which is less than 14.)
  • 12 ÷ 3 = 4 (remainder 0)
    (4 doubles to 8, 0 in front makes 08, which is less than 14.)
  • 8 ÷ 3 = 2 (remainder 2)
    (2 doubles to 4, 2 in front makes 24, which is MORE than 14, so 24 + 1 = 25.)
  • 25 ÷ 3 = 8 (remainder 1)
    (8 doubles to 6, 1 in front makes 16, which is MORE than 14, so 16 + 1 = 17.)
  • 17 ÷ 3 = 5 (remainder 2)
    (5 doubles to 0, 2 in front makes 20, which is MORE than 14, so 20 + 1 = 21.)
  • 21 ÷ 3 = 7 (remainder 0)
    (7 doubles to 4, 0 in front makes 04, which is less than 14.)
  • 4 ÷ 3 = 1 (remainder 1)
    (1 doubles to 2, 1 in front makes 12, which is less than 14.)
Double checking with our calculator once again, we find that 928 is equal to 0.32142857..., with the 142857 repeating. Actually, if you know the 142857 pattern from knowing your 7ths, and you realize that 28 is a multiple of 7, you should realize that you'll eventually run into the 142857 pattern from there.

TIPS: As always, the biggest tip is practice, practice, practice! Once you can divide by numbers ending in 8, you should also realize that you can divide by numbers ending in 4. If you want to divide by a number ending in 4, just double both numbers in the problem. If you need to work out 1724, for example, just double both numbers, resulting in 3448, and work the problem out from there, as described in the technique section. As a matter of fact, you'll get a great deal of practice if you work out 3448 on your own, right now.


Simple math? Not so simple

Published on Sunday, May 10, 2015 in , ,

mattbuck's Bayes' Theorem neon sign photoJust over a month ago, TheWeek.com posted an article titled The simple math problem that blows apart the NSA's surveillance justifications. It concerned the probability of detecting terrorists, when you have a near-perfect terrorist-detecting machine.

It turns out that the simple math isn't so simple.

Let's start with the question itself:

Suppose one out of every million people is a terrorist (if anything, an overestimate), and you've got a machine that can determine whether someone is a terrorist with 99.9 percent accuracy. You've used the machine on your buddy Jeff Smith, and it gives a positive result. What are the odds Jeff is a terrorist?
A better way to state the question is, “Given that the machine has identified Jeff as a terrorist, what is the probability Jeff is actually a terrorist?” Questions like this are known as conditional probabilities, and it turns out that Bayes' Theorem helps answer questions like this very effectively. If you're not already familiar with Bayes' theorem, read that post and watch the videos to better understand it before proceeding.

Unfortunately, the linked article above doesn't employ such computations, so we have to go about it ourselves. Let's assume the 99.9% (0.999) accuracy of the machine applies to detecting not only terrorists, but to identifying innocent people, as well. In turn, that means that the machine has a 0.1% (0.001) chance of identifying an innocent person as a terrorist, or identifying a terrorist as an innocent person. So, we have four different probabilities:

Chance that an actual innocent is identified as a terrorist: 0.001 (False +)
Chance that an actual innocent is NOT identified as a terrorist: 0.999 (True -)
Chance that an actual terrorist is identified as a terrorist: 0.999 (True +)
Chance that an actual terrorist is NOT identified as a terrorist: 0.001 (False -)

Let's put these numbers in the following table:

  Is a terrorist Is innocent
Identified as terrorist 0.999 (True +) 0.001 (False +)
Identified as innocent 0.001 (False -) 0.999 (True -)

Now that we've got the probabilities in order, let's see what happens when 1 terrorist and 999,999 innocent people are thrown into the mix. We'll multiply both entries in the “Is a terrorist” column by 1, to represent the 1 terrorist, and both entries in the “Is innocent” column by 999,999, to represent the 999,999 innocent people:

  Is a terrorist Is innocent
Identified as terrorist 0.999 (1 × 0.999) 999.999 (999,999 × 0.001)
Identified as innocent 0.001 (1 × 0.001) 998,999.001 (999,999 × 0.999)

We can double-check that the table has been correctly constructed, because all the numbers add up to 1 million. This covers all the data, so now we're ready to tackle the original question.

Remember that the question itself is “Given that the machine has identified Jeff as a terrorist, what is the probability Jeff is actually a terrorist?” In other words, we aren't concerned with the possibility of being identified as an innocent, as identification as a terrorist is already a given. All we have to do here is trim the “Identified as innocent” row out of the table completely:

  Is a terrorist Is innocent
Identified as terrorist 0.999 999.999

At this point, don't forget the basic probability formula: Probability = (targeted outcome) ÷ (total possibilities). What are the total possibilities here? 0.999 + 999.999 = 1000.998. What is the targeted outcome? It's that Jeff is a terrorist, which is 0.999. So, the probability is 0.999 ÷ 1000.998 ≈ 0.000998, or about a 0.0998% chance.

In more practical terms, once the 99.9% accurate machine has identified Jeff has a terrorist, there's still only a 1 in 1,002 chance that he's actually a terrorist! Granted, this isn't radically different from the 1 in 1,000 chance posted in the original article. However, in math, the path you take is just as important as the results.


Cheryl's Birthday Round-Up

Published on Monday, April 20, 2015 in , , , , ,

normanack's birthday cake photoThanks to a Singapore math exam, the internet is being driven crazy by the biggest problem in birthdays since the birthday paradox!

Here's the problem: Albert and Bernard want to know Cheryl's birthday, but Cheryl isn't willing to tell them directly. Instead, she gives them a list of 10 possible dates: May 15, May 16, May 19, June 17, June 18, July 14, July 16, August 14, August 15, and August 17. She then whispers only the month to Albert and the date to Bernard. The following discussion then takes place between Albert and Bernard:

Albert: "I don't know when Cheryl's birthday is, but I know that Bernard does not know, too."

Bernard: "At first, I didn't know when Cheryl's birthday was, but I know now."

Albert: "Then I also know when Cheryl's birthday is."

When is Cheryl's birthday? We'll look at how to find the answer in this post!

The simplest and most direct explanation of this puzzle I've found is in Presh Talwalkar's post, When Is Cheryl’s Birthday? Answer To Viral Math Puzzle. The included video makes the answer seem so straightforward:

Another helpful approach is Mark Josef's interactive Cheryl's Birthday page, on which you can click each of the dates to see why that the logic determines that date to be right or wrong. Both Cahoots Malone and The Washington Post have also featured simple and straightforward video explanations of this puzzle.

For a more detailed look at the solution, check out Numberphile's thorough explanation, as well the extra footage:

Ever the intrepid explorer, however, James Grime takes an even closer look at Cheryl's Birthday, and finds that the intended answer may not necessarily be the right answer:

Has this puzzle driven you crazy? Did you manage to solve it? If so, how? I'd love to hear your answers in the comments below!


Estimating Roots

Published on Sunday, March 29, 2015 in , , ,

purzen's (openclipart.org) thought bubble surrounding David Vignoni's square root icon over Josh Green's (subtlepatterns.com) Old Mathematics pattern3 years ago, I posted a tutorial about estimating square roots of non-perfect squares, including tips and tricks.

Since then, I've wondered if there was a general formula for estimating other roots, such as cube roots, fourth roots, and so on. Reddit user InveighsiveAd informed me that there's a simple general formula very similar to the method I've taught for square roots! Once you pick up the basic idea of this method, you'll be able to astound friends, family, and teachers.

The approach for estimating roots originates from an approach developed by Leonhard Euler, and involves taking derivatives, so I won't delve into the math behind why this works here. I'll focus more on the resulting formulas, which can be used to

The method I taught for estimating square roots basically boiled down to this formula, where a was a perfect square equal to or less than x, and b was equal to x - a:

With Euler's method, we'll be estimating roots using the same basic approach of breaking up a number into a number which is a perfect power (square, cube, 4th power, etc.) and the difference between that power and the targeted number. The following formula may look scary at first, but it's simpler than it looks:

y is simply the root we wish to know. For square roots, y would equal 2, for cube roots, y would equal 3, and for 4th roots, y would equal 4. As a matter of fact, I'm not going to concern this article with anything past 4th roots, as this quickly becomes complex. Here are the formulas for square, cube and 4th roots individually:

These look worse than they really are. Remember that a is always chosen to be a perfect power, so you're working with an easily determined number. If you were going through this process for cube root, and using 729 for a, the cube root of 729 would be 9. So, any where you see the cube root of a, you can mentally replace it with 9, in this example.

Obviously, knowing perfect squares up through 31 will be of help, as in the original method. Knowing the perfect cubes from 1 to 10, as many Grey Matters readers already do, will allow you to estimate cubes of number up to 1,000. Memorizing or being able to quickly calculate perfect 4th powers will allow you to estimate 4th powers up to 10,000!

For those confused by the ± symbol in the equations, it simply means that we're going to choose a to be the closest perfect power, and adjust b accordingly. For example, if we want the cube root of 340, then we'd use 343 (73), and work it out as the cube root of 340 as the cube root of (343 - 3).

Let's estimate the cube root of 340 as a full example. As explained above, we've already broken this up into the cube root of (343 - 3). Your mental process might go something like this:

How close is 64849 to the cube root of 340? The two numbers are very close, as this Wolfram|Alpha comparison shows!

Colin Beveridge, of Flying Colours Maths has helpfully pointed out that the error in the method will increase as you get approach the geometric mean of two closest consecutive perfect powers. For example, when using this method to find the cube root of 612, which is close to 611 (the approximate geometric mean of 512 and 729), you'll be farther off.

Let's find out exactly how far off we would be. The cube root of 612 could be worked out as (729 - 117), but (512 + 100) is closer, so we'll use the latter. Working this out, we'd get:

Wolfram|Alpha shows that 82548 ≈ 8.52, while the actual cube root of 612 ≈ 8.49. It's off by about 3 hundredths, but that's still a good estimate!

As an added bonus, if you wind up with a fraction whose denominator ends in 1, 3, 5, or 7, you can use the techniques taught in Leapfrog Division or Leapfrog Division II to present your estimate with decimal accuracy! Yes, it's just the same number presented differently, but working out decimal places in your head always comes across as impressive. Personally, I reserve the decimal precision for when I know the root is close to a perfect power.

Try this approach out for yourself. If you have any questions, feel free to ask them in the comments!


Grey Matters' 10th Blogiversary!

Published on Saturday, March 14, 2015 in , , , , , , , , , , , , ,

Mehran Moghtadaei's Pi Digit GraphicEver since I started this blog, I've been waiting for this day. I started Grey Matters on 3/14/05, specifically with the goal of having its 10th blogiversary on the ultimate Pi Day: 3/14/15!

Yes, it's also Einstein's birthday, but since it's a special blogiversary for me, this post will be all about my favorite posts from over the past 10 years. Quick side note: This also happens to be my 1,000th published post on the Grey Matters blog!

Keep in mind that the web is always changing, so if you go back and find a link that no longer works, you might be able to find it by either searching for a new place, or at least copying the link and finding whether it's archived over at The Wayback Machine.


My most read posts in 2005 were 25 Years of Rubik's Cube (at #2), and Free Software for Memory Training (at #1). It was here I started to get an idea of what people would want from a blog about memory feats.


In the first full January to December year of Grey Matters, reviews seemed to be the big thing. My reviews of Mathematical Wizardry, Secrets of Mental Math, and Mind Performance Hacks all grabbed the top spots.


This year, I began connecting my posts with the interest of the reader, and it worked well. My series of “Visualizing” posts, Visualizing Pi, Visualizing Math, and Visualizing Scale were the biggest collectively-read posts of the year.

Fun and free mental improvement posts also proved popular in 2007. Unusual Lists to Memorize, my introduction to The Prisoner's Dilemma, and my look at Calculators: Past, Present, and Future (consider Wolfram|Alpha was still 2 years away) were well received! 10 Online Memory Tools...For Free! back-to-back with my Memorizing Poetry post also caught plenty of attention.


I gave an extra nod to Pi this year, on the day when Grey Matters turned Pi years old on May 5th. The most popular feature of the year was my regularly update list of How Many Xs Can You Name in Y Minutes? quizzes, which I had to stop updating.

Lists did seem to be the big thing that year, with free flashcard programs, memorizing the elements, and tools for memorizing playing card decks grabbed much of the attention in 2008.


Techniques took precedence over lists this year, although my series on memorizing the amendments of the US Constitution (Part I, Part II, Part III) was still popular. My web app for memorizing poetry, Verbatim, first appeared (it's since been updated). Among other techniques that caught many eyes were memorizing basic blackjack strategy, the Gilbreath Principle, and Mental Division with Decimal Precision.


This year opened with the sad news of the passing of Kim Peek, the original inspiration for the movie Rain Main. On a more positive note, my posts about the game Nim, which developed into a longer running series than even I expected, started its run.

As a matter of fact, magic tricks, such as Bob Hummer's 3-Object Divination, and puzzles, such as the 15 Puzzle and Instant Insanity, were the hot posts this year.

Besides Kim Peek, 2010 also saw the passing of Martin Gardner and Benoît Mandelbrot, both giants in mathematics.


The current design you see didn't make its first appearance until 2011. Not only was the blog itself redesigned, the current structure, with Mental Gym, the Presentation section, the Videos section, and the Grey Matters Store, was added. This seemed to be a smart move, as Grey Matters begin to attract more people than ever before.

The new additions to each section that year drew plenty of attention, but the blog has its own moments, as well. My list of 7 Online Puzzle Sites, my update to the Verbatim web app, and the Wolfram|Alpha Trick and Wolfram|Alpha Factorial Trick proved most popular in 2011.

My own personal favorite series of posts in 2011, however, was the Iteration, Feedback, and Change series of posts: Artificial Life, Real Life, Prisoner's Dilemma, Fractals, and Chaos Theory. These posts really gave me the chance to think about an analyze some of the disparate concepts I'd learned over the years when dealing with various math concepts.


In 2012, I developed somewhat of a fascination with Wolfram|Alpha, as its features and strength really began to develop. I kicked the year off with a devilish 15-style calendar puzzle, which requires knowing both how to solve the 15 puzzle and how to work out the day of the week for any date in your head! Yeah, I'm mean like that. I did, however, release Day One, my own original approach to simplifying the day of the week for any date feat.

Estimating Square Roots, along with the associated tips and tricks was the big feat that year. The bizarre combination of controversy over a claim in a Scam School episode about a 2-card bet and my approach to hiding short messages in an equation and Robert Neale's genius were also widely read.


After we lost Neil Armstrong in 2012, I was inspired to add the new Moon Phase For Any Date tutorial to the Mental Gym. A completely different type of nostalgia, though, drove me to post about how to program mazes. Admittedly, this was a weird way to kick off 2013.

Posts about the Last Digit Trick, John Conway's Rational Tangles, and Mel Stover were the first half of 2013's biggest hits on Grey Matters.

I also took the unusual approach of teaching Grey Matters readers certain math shortcuts without initially revealing WHY I was teaching these shortcuts. First, I taught a weird way of multiplying by 63, then a weird way of multiplying by 72, finally revealing the mystery skill in the 3rd part of the series.


Memory posts were still around, but mental math posts began taking over in 2014. A card trick classically known as Mutus Nomen Dedit Cocis proved to have several fans. The math posts on exponents, the nature of the Mandelbrot set, and the Soma cube were the stars of 2014. Together, the posts Calculate Powers of e In Your Head! and Calculate Powers of π In Your Head! also grabbed plenty of attention.


With 999 posts before this one, this barely even scratches the surface of what's available at this blog, so if you'd made it this far, I encourage you to explore on your own. If you find some of your own favorites, I'd love to hear what you enjoyed at this blog over the years in the comments below!