Here's an interesting mental math challenge. Given two different positive real numbers, which we'll call a and b, which is greater, a^b or b^a? If you're able to calculate both exponents mentally, such as 2³ vs 3², then that's probably the simplest way to go. What happens if one or both exponential expressions are too hard to mentally calculate?

To solve this, we'll need to find a general rule. Some advanced calculation will be required to find it. However, once we do the work to find the rule, you'll see that almost no math will be needed to solve these types of problems!

SEARCH FOR THE RULE: Consider that 2³ < 3², but 3⁴ > 4³. Just looking at the placement of smaller and bigger numbers, there doesn't seem to hard and fast rule which applies. The best place to start is just by assuming what we wish to discover:



From that point, let's see if we can separate a and b somehow. The quickest way is to raise both sides to the power of 1/ab:



In English, then, when the ath root of a is greater than the bth root of b, then a^b > b^a. Don't worry, though. You won't have to do any roots in your head. Instead, just look at this Desmos graph where y=xth root of x. If you click in the box marked 1 at the upper left, Desmos highlights two points, the point with the minimum y value, which is (0, 0), and the point with the maximum value, (2.718, 1.445).

Wait a minute! The value of 2.718 sounds familiar. Could the maximum value of xth root of x really be e? Sure enough, Wolfram|Alpha verifies that the maximum is e!

Look at both sides of the graph, then. From 0 to e, the graph increases. From e on upwards, the value of the xth root of x will steadily decrease. In fact, the value will just keep getting closer and closer to 1.

This means we've found the start of our rule. When you have two numbers, both of which are equal to or greater than e, the smaller number x will always yield a greater xth root of x than the higher number. This tells us that the ath root of a is greater than the bth root of b when a is the smaller number. Working backwards to our original assumption, this means that:

When a and b are both equal to or greater than e, a^b is always greater than b^a when a is the smaller number and b is the larger number.

There's our rule!

WORKING WITH INTEGERS: What about when one or both of a and b are less than e? In the case of positive integers, this means we only have to consider the cases involving 1 and 2.

The case of 1 is easy. Assume that a equals 1 and b has a value of 2 or more. 1^b will always be 1, and b¹ will always be b, which is 2 or more. That covers every case involving 1.

What about 2? Let's work through each case individually, always assigning a to a value of 2, and b a value other than 2. If b is 1, we just covered that case. b can't be 2, as it would be the same as a. What about when b is 3? We already covered the fact that 2³ < 3², so we remember that special case. What happens when b is 4? This is a very unique case, as 2⁴ is exactly equal to 4²! In fact, this is a well-known special case, as it's the only time when a^b equals b^a when a and b are different integers. When b is 5 ore greater, as you can see in this graph, we can fall back on the rule we set above.

A CLASSIC CHALLENGE: Which is greater, π^e or e^π? There's a problem you probably never thought you'd solve in your head!

Let's go see if our primary rule applies. Are a and b both equal to or greater than e? e, of course, is exactly equal to e (2.71828 and so on). π, as any Grey Matters readers should already know, is roughly 3.1415, and therefore also greater than e.

This means we can apply our primary rule! e is smaller than π, which means that e^π must be greater than π^e. A quick verification with Wolfram|Alpha shows that this is correct.

About 4 years ago, Presh Talwalkar wrote up several approaches to this classic problem. Interestingly, this approach wasn't included. The 4th method does work with the xth root of x as we did, but it uses a deeper approach involving calculus.

Back about 5 years ago, I taught a simple method for estimating the square roots of non-perfect squares. I recently learned an improvement that makes this far more accurate, and even a little more impressive.

The new approach works well with the original approach I taught. I'll start with a review of the original method I taught, and then I'll delve into the details of the improved method.

Quick Refresher: As mentioned in the original post, you should be comfortable with mentally squaring 2-digit numbers, and being able to find the square roots of perfect squares. You'll also need to know the squares of the numbers from 1 to 31 off the top of your head, in order to handle the numbers from 1 to 1,000.

As an example number, we'll say you're given 130, and asked to find the square root. You start by partitioning the given number into the sum of the largest perfect square equal to or less than the number, plus any remaining amount. The largest perfect square equal to or less than 130 is 121, so we partition it into 121 + 9:



If you consider the sum now under the square root at this point to consist of numbers x and y (x = 121 and y = 9 in our example), then the original square root approximation is as follows:



This may seem harder than it is. Remember that x is a perfect square, so sqrt(x) is an integer, so it will be easy to work with. Applying this formula to our example, and recalling that sqrt(121) = 11, we get this approximation:



If you square that estimation, you get roughly 130.167. That's not bad, but how can we do better?

Improved Version: This past Monday, UK-based math tutor Colin Beveridge wrote a post called The Mathematical Ninja takes a square root, which is the improved version I've been mentioning.

The whole part (11 in our above example) will not change, so we'll only focus on changing the fractional part. If you consider, as in Colin's post, the fractional part to be ^a⁄_b, then the improved estimation is as follows:



So, for the new numerator (top number of the fraction) in our example, you'd multiply 9 × 22 to get 198. For the new denominator (bottom number of the fractions), you'd have to square the original bottom number, 22 squared = 484, and then add the original top number, 9, for a total of 493. Our new estimation for the square root of 13 becomes:



If you square that estimate, you get roughly 129.997, which is much closer to 130 than 130.167!

If this approach interests you, do take the time to read Colin's original post, as it has plenty of nice touches. Most importantly, he suggests using the closest perfect square above or below the given number, which will improve the accuracy of your estimate as compared to always using the perfect square less than or equal to the given number.

I apologize for the irregular posting over the past few months. I've had to deal with some personal issues (don't worry, everything is fine!). The good news is that, with this entry, everything should start returning to normal.

Having said that, let's dive into August's snippets!

• James Grime and Katie Steckles made a video about a seemingly simple game:



First, if it's on Grey Matters, you know all is not always what it seems. Long-time fans of Grey Matters may remember this when I described it under the name Wythoff's Nim. It winds up having some very interesting math behind it. James went on to make a solo video explaining the mathematics behind it in more detail:



• We can't ignore Katie Steckles' game video after all that! Katie teaches 2 games (or does she?). The first one involves numbered fishes, and the second one involves cards with stars and moons on them:



It's a little bit surprising that these are actually the same game! Back in June's snippets, there was a multiplication version of this. Like this and Scam School's game of 15, they all go back to Tic-Tac-Toe. If you want to see some other interesting variations of this same idea, read Martin Gardner's Jam, Hot, and Other Games column.

• There's usually more than one way to use your knowledge. In my tutorial about mental division, I teach a simple method for mentally dividing the numbers 1 through 6 by 7. Presenting it as an exacting feat of mental division is one thing. How else could you present it? Take a look at how Scam School presents the same feat:



If you watch the full explanation, you'll notice another difference between the way I teach it and the way Brian teaches it. He puts emphasis on the last digit, which works well for performing the feat this way. In my version, I teach how to work out the first few digits, as you'll need those first when giving the answer verbally. This is a good lesson in the benefits of changing your point of view!

On August 21, 2017, there will be a total solar eclipse visible in the United States, which hasn't happened since 2012, and won't happen again until 2024!

It's the perfect time to present an impressive knowledge of the moon, such as being able to estimate the moon phase for any date in 2017! Does learning this feat sound difficult? Surprisingly, it's much easier than you may think.

For 2017 only, the phase of the moon formula is simple: (Month key number + date) mod 30. The result is the age of the moon in days from 0 to 29. I'll explain each part of this formula below.

Month key number: January's key number is 3, February's key number is 4, and all other months' keys are their traditional numbers; March is 3, April is 4, May is 5, and so on up to December, which is 12.

Date: This is simply the number represented by the particular date in the month. For the 1st, add 1. For the 2nd, add 2. For the 3rd, add 3, and so on.

mod 30: If you get a total of 30 or more, simply subtract 30. Otherwise, just leave the number as is.

As an example, let's try the published date of this post, July 28th. July is the 7th month and the date is the 28th, so we work out 7 + 28 = 35. Since that number is greater than 30, we subtract 30 to get 35 - 30 = 5. This result tells us that the moon's age will be 5 days on July 28th, 2017.

But what does it mean to say that a moon is some number of days old? Here's a simple explanation:

0 days = New moon (the moon is as dark as it's going to get)

0 to 7.5 days = Waxing crescent (Less than half the moon is lit, and it's getting brighter each night)

7.5 days = 1st quarter moon (Half the moon is lit, and getting brighter each night)

7.5 to 15 days = Waxing gibbous (More than half the moon is lit, and getting brighter each night)

15 days = Full moon (The moon is as bright as it's going to get, and will start getting darker each night)

15 to 22.5 days = Waning gibbous (More than half the moon is lit, and it's getting darker each night)

22.5 days = 3rd quarter moon (Half the moon is lit, and it's getting darker each night)

22.5 to 29 days = Waning crescent (Less than half the moon is lit, and it's getting darker each night)

So, our 5-day-old moon would be a waxing crescent (Less than half the moon is lit, and it's getting brighter each night). Sure enough, Wolfram|Alpha confirms this estimate! Moon Giant confirms this estimate, as well.

There are a couple of finer points to note. First, this simple method happens to work only in 2017. The method won't be this simple again until 2036 and then 2055. If you want to learn to calculate the moon phase for any date in the 1900s, you can learn the full feat over in the Grey Matters Mental Gym.

Second, remember that this is an estimate. The actual error margin is plus or minus one day. So, getting an estimate of 5 days means that the moon is somewhere from 4 to 6 days old. When the overlap includes a first quarter moon, a full moon, or a 3rd quarter moon, I usually describe this as, "While it might not be technically accurate, most people would look up and describe it as a 1st quarter moon" (or full moon, or 3rd quarter moon).

This method is a simplification of John Conway's original moon phase estimation formula from Winning Ways for Your Mathematical Plays, vol. 4. Practice it and have some fun amazing your friends and family during the coming solar eclipse!

Many regular Grey Matters readers will be familiar with Michael Daniels' Mind Magician site, where he teaches numerous math and memory feats, such as calculating cube roots in your head instantly. He's recently written a new ebook on the 4-by-4 magic square, titled Perfectly Possible. I found it to be well worth the time and money invested, and wanted to share my thoughts with Grey Matters readers.

This is going to be a difficult review, as I can't give too much away, but I also want to share with you the quality of this method. I'll start with the qualities promoted by Michael Daniels himself:

• Completely impromptu. No set-up, gimmicks, or cribs.
  
• New, improved method - minimal memory and the simplest of calculations.
  
• Suitable for close-up or stage performances.
  
• Produces elegant magic squares.
  
• Can be immediately repeated for different totals.
  
• Includes a browser application that helps you to learn and practice (Internet connection not required).
  

Let's clarify a few points here. Yes, it is completely impromptu. This is a calculation method, but the calculations are minimal, quick, and will quickly become second nature during practice. Speaking of practice, the included browser application is very handy. It's similar to the magic square practice app posted at mindmagician.org, but streamlined for the new routine.

What does "elegant magic squares" mean? One problem with many magic square approaches is that the number can appear unbalanced, such as when 12 of the numbers are less than 15, and the other 4 are over 30. This can give your audience clues about the method. With the Perfectly Possible method, you don't have to worry about that. You're guaranteed a balanced magic square. Elegant also means that you're guaranteed at least 36 different ways in which some combination of 4 squares gives the magic total. Under the right circumstances, this method can yield as many as 52 different combinations!

As with any magic square, the ability to repeat the square immediately with different totals is, of course, essential. Even more impressive, though, is that if 2 people give you the same total, you can still generate a different magic square! Naturally, the same total requires the numbers used to be in the same general range, but this method will allow you to put different numbers in each of the squares with very little difficulty.

That quality is really what makes Perfectly Possible stand out. Unlike the rigid approaches behind most magic squares, the ability to take multiple approaches gives the performer more freedom while disguising the method very effectively. When a change is as constrained as the magic square, finding an approach like this that offers you remarkable degrees of freedom like this is incredible!

If you're interested in creating magic squares, I can't recommend Michael Daniels' Perfectly Possible ebook enough. It's available for $6 on its own, or $8 in combination with Mostly Perfect, its predecessor. If you're seriously consider this as a performance piece, I would also recommend the Unknown Mentalist's Why A Magic Square Should Not Be A Magic Square ebook. It teaches many very effective original presentations that disguise the principle, and will help preserve the mystery by showing you how to prevent audiences from simply searching for "magic square" on the internet during or after your performance.

Over in the Mental Gym, I have a tutorial on squaring numbers, starting with simpler techniques for multiples of 10 and 5, and working up to squaring numbers as large as 125.

Naturally, I always like to see how much farther I can go, especially when I can still keep things relatively simple. With the technique I'll be teaching in this post, if you're comfortable with squaring numbers up to 125, you're ready to move on to squaring numbers up to 199 (Well, actually 200, since that's not difficult to square).

Rightmost Two Digits

We're going to generate the answer from right to left in this technique, working with no more than 2 digits at a time. To get the rightmost 2 digits of the answer, simply square the rightmost 2 digits of the given number. For example, if you're given the number 112 to square in your head, you'd square the rightmost 2-digits of that number, 12, to get 144. You write down the rightmost 2 digits, 44 in this example, and remember the remaining digits to the left, such as the 1 in this example (underscores are used to hold places for numbers not yet written):



Middle Two Digits

For the middle 2 digits, simply double the rightmost 2 digits of the given number, then add the number you remembered, if any. Write down the 2 rightmost digits of this answer to the left of the digits previously written, and remember any digits to the left of that. Returning to our 112 example, we look at the rightmost 2 digits, 12, and double that to get 24. Adding the number we remembered from the previous step, 1, we get a total of 25. The rightmost 2 digits of this answer are 25, and there's no digits to the left of those to remember:



Leftmost Digit(s)

For the leftmost digits, take any amount you have remembered at this point, and simply add 1 to it. Write down that total to the left of all the digits you've previously written, and you're done! In our 112 example, we didn't remember anything for this stage, so we just write down 1, resulting in:



You can check for yourself that 112²=12,544.

Tips

Single-digit numbers: When working on either the rightmost 2 digits or the middle 2 digits, you may wind up working with a single-digit answer. These steps always require working with 2 digits, so for single-digit answers, just place a 0 to the left of it to make it a 2-digit number.

For example, when squaring 103, you start by squaring 3 to get 9, with nothing to remember:



Next, you'd double 3 to get 6, again with nothing to remember:



Since you don't remember anything from the previous steps, just add a 1 to the left of this answer:



This tells us that 103²=10,609.

Remembering multiple digits: Just so you have an example of working with larger numbers, let's try squaring 178. Start by squaring 78, which is 6,084. Write down the 84, and remember the 60:



Next, double 78 to get 156, and add the 60 you remembered from the previous step, giving a total of 216. Write down the 16, and remember the 2:



Finally, add 1 to the number you're remembering and write that down to the left of the previous digits. In this case, we add 1 and 2 to get 3:



The result of 178² is 31,684.

Interest for 2 time periods: This technique is especially handy for quickly calculating how many times the principal will grow at interest rates of 99% or less for 2 time periods. The only additional step is to put a decimal point between the ten-thousands and thousands place. For example, since 103²=10,609, that means that principal invested at 3% per year for 2 years will grow to 1.0609 times its original size. Although you're not likely to ever see it happen, principal invested at 78% per year for 2 years would grow to 3.1684 times its original size, because 178² is 31,684.

Practice this, have a little fun with it, and you'll have an impressive new skill!

Back in January of 2012, I wrote about the Chinese Remainder Theorem. Also, Martin Gardner taught the basics well in his book Aha!: Insight, including a trick where you can determine someone's secretly chosen number between 1 and 100 just from hearing the remainders when divided by 3, 5 and 7. Going over that post again, I've developed a few improvements to this trick that make it seem much more impressive, and maybe even easier to do.

The first problem is getting the remainders. With a standard calculator, it's not easy to do. The answer here is to simply have them divide the number by 3, 5 and 7, and have them tell you ONLY the number after the decimal point. Using the amount after the decimal point, you can work out the remainder. When dividing by 3, there are 3 possibilities for numbers after the decimal point:

• Nothing after the decimal point: Remainder = 0
• Number ends in .3333...: Remainder = 1
• Number ends in .6666...: Remainder = 2

To find the remainders when dividing by 5:

• Nothing after the decimal point: Remainder = 0
• Number ends in .2: Remainder = 1
• Number ends in .4: Remainder = 2
• Number ends in .6: Remainder = 3
• Number ends in .8: Remainder = 4

If you read my Mental Division: Decimal Accuracy tutorial, you'll get familiar with the 7s pattern. It's trickier than 3 or 5, but easily mastered. You only need to pay attention to the first 2 digits after the decimal point:

• Nothing after the decimal point: Remainder = 0
• Number ends in .14: Remainder = 1
• Number ends in .28: Remainder = 2
• Number ends in .42: Remainder = 3
• Number ends in .57: Remainder = 4
• Number ends in .71: Remainder = 5
• Number ends in .85: Remainder = 6

Using the decimals makes the trick seem more difficult from the audience's point of view, but it only requires a little practice to recognize which numbers represent which remainders.

The other improvement involves the process itself. Have them start by dividing their number by 7 and telling you the numbers after the decimal point. Using the steps above, you can quickly determine the remainder from 0 to 6. If the remainder is between 0 and 5, you can remember them by touching that many fingers of your left hand to your pant leg (or table, if present). For a remainder of 6, just touch 1 finger from your right hand to your pant leg or table.

What ever the remainder, imagine a sequence starting with this number, and adding 7 until you get to a number no larger than 34. For example, if the person told you their number ended in .42, you know the remainder is 3, and the sequence you'd think of is 3, 10, 17, 24 and 31 (we can't add anymore without exceeding 34). If the remainder was determined to be 4, instead, the sequence you'd think of would be 4, 11, 18, 25 and 32.

Next, ask the person to divide by 5, and tell you the part of the answer after the decimal point. Once you get this number, recall your earlier sequence (which can be recalled via the number of fingers resting on your pant leg or table), and subtract the new remainder from each number, until you find a multiple of 5. For example, let's say that when dividing by 7, their remainder was 4, so the sequence was 4, 11, 18, 25 and 32. Let's say that when dividing by 5, their remainder was 3. Which number from your initial sequence, when it has 3 subtracted from it, is a multiple of 5? is 4-3 a multiple of 5? No. Is 11-3 a multiple of 5? No. Is 18-3 a multiple of 5? YES! Now you have the number 18.

Whatever number you have at this point, is the smallest of 3 possible numbers. The other 2 possibilities are 35 more than that number and 70 more than that number. At this point, you can know that the chosen number is either 18, 53 (35+18) or 88 (70+18). The remainder when dividing by 3 will determine which one of these is the correct answer. For example, if they say their number, when divided by 3, ends in .3333..., you know the remainder is 1. So, run through all 3 numbers quickly and ask yourself which one is 1 greater than a multiple of 3. Is 18 - 1 a multiple of 3? No. Is 53 - 1 a multiple of 3? No. Is 88 - 1 a multiple of 3? YES! Therefore, their number must be 88.

There are certainly other approaches. In fact, a magician's magazine called Pallbearer's Review once presented this trick as a challenge, asking their readers to supply their methods. They received a wide variety of entries and many approaches. Most of them involved far more difficult methods than the above approach, which I prefer for actual performance.

A little over 4 years ago, I wrote a post inspired by Alexander Craig Aitken's approach to dividing by numbers ending in 9, called Leapfrog Division. It's a remarkably fun and simple technique, so if you haven't already checked it out for yourself, give it a read.

I wrote 3 more posts in that series. Leapfrog Division II dealt with dividing by numbers ending in 1. My last 2 posts before this year were Leapfrog Division III, which dealt with dividing by numbers ending in 8, and Leapfrog Division IV, which dealt with dividing by numbers ending in 2.

These were fine, but the latter 3 posts employed rules that became increasingly cumbersome, and were tricky to apply quickly and without error. Recently, however, I learned a far simpler approach that makes these later posts much simpler!

Let's start with credit where credit is due. The thanks should go to Saurabh G over at Hubpages. He had a post at Owlcation titled Divide Numbers Easily Using Vedic Mathematics: Fast and Easy Division Techniques. Towards the end of the post, there's a section with the heading, “How Do You Use Vedic Division When the Divisor Is More Than One Digit?”. The approach taught with numbers ending in 9 is demonstrated a little differently, but is mathematically identical to the original Leapfrog Division post.

The next section, “Multi-Digit Divisor Ending in 8 Example,” is what really struck me the most. He teaches an almost identical approach, and casually mentions that the quotient needs to be doubled before adding the remainder to the 10s place. Click on that link, read through the example, and then compare that approach to what I taught in Leapfrog Division III. I do mention a doubling idea, but the rules I taught are far more complex.

Using Saurabh G's vedic math approach, here's how his example (73 ÷ 138) would look written in the style of the Leapfrog Division posts. We start with the idea that 14 won't go into 7.3, so we start with a 0 and decimal point at the beginning of the answer, and work from there:

• 73 ÷ 14 = 5 (remainder 3)
  (Double the 5 to get 10, add the 3 in the 10s place to get 40)
• 40 ÷ 14 = 2 (remainder 12)
  (Double the 2 to get 4, add the 12 in the 10s place to get 124)
• 124 ÷ 14 = 8 (remainder 12)
  (Double the 8 to get 16, add the 12 in the 10s place to get 136)
• 136 ÷ 14 = 9 (remainder 10)
  (Double the 9 to get 18, add the 10 in the 10s place to get 118)
• 118 ÷ 14 = 8 (remainder 6)
  (Double the 8 to get 16, add the 6 in the 10s place to get 76, and continue on from here, if desired....)

If you wrote down the quotients (the number in bold above) as you went, you'd have written down 0.52898 at this point, which is correct as far as we've gone.

Just to round things out, he includes a chart showing how to adapt this approach for numbers ending in 9 all the way down to 1! Think about that: It took me 3 years and 4 posts to cover numbers ending with 4 different digits, and increasingly difficult rules. Someone else comes along, teaches 2 examples in 1 post, and leaves his readers confident they can handle 9 different digits.

I tip my hat to you, Saurabh G! Thank you for sharing this approach, and giving me and my readers a simpler and more effective mental division tool.

While 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 ¹⁷⁄₃₂. As in the technique for dividing by numbers ending in 1, we will always start by reducing the numerator by 1, giving us ¹⁶⁄₃₂. 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 ¹⁵⁄₂₂. 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 ¹⁵⁄₃.

You shouldn't be surprised that we're going to divide this out using quotients and remainders. Starting with ¹⁵⁄₂, 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 ¹⁷⁄₃₂ (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 ¹⁷⁄₃₂ equals!

Now that you understand the steps, let's work out ¹⁹⁄₂₂ as a second example. We start by reducing 19 by 1, which is ALWAYS the first step, giving us ¹⁸⁄₂₂. Half of 22 is 11, and 18 is greater than 11, so we decrease it by 1 again, leaving us with ¹⁷⁄₂₂. Rounding the denominator down and dividing by 10, our starting problem should be ¹⁷⁄₂. 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 ¹⁹⁄₂₂ ≈ 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...

• ...9? Use Leapfrog Division.
• ...8? Use Leapfrog Division III.
• ...7? Triple dividend and divisor, so divisor ends in 1, and use Leapfrog Division II.
• ...6? Double dividend and divisor, so divisor ends in 2, and use this technique.
• ...4? Double dividend and divisor, so divisor ends in 8, and use Leapfrog Division III.
• ...3? Triple dividend and divisor, so divisor ends in 9, and use Leapfrog Division.
• ...2? Use this technique.
• ...1? Use Leapfrog Division II.

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.

In 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 ¹³⁄₁₈. 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 ¹³⁄₂₀, and dropping the 0 from the denominator changes this to ¹³⁄₂.

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 ¹⁴⁄₂ 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 ¹³⁄₁₈ 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 ⁹⁄₂₈. 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 ⁹⁄₃:

• 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 ⁹⁄₂₈ 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 ¹⁷⁄₂₄, for example, just double both numbers, resulting in ³⁴⁄₄₈, 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 ³⁴⁄₄₈ on your own, right now.

3 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 6⁴⁸⁄₄₉ 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 8²⁵⁄₄₈ ≈ 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!

Ever 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.

2005

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.

2006

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.

2007

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.

2008

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.

2009

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.

2010

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.

2011

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.

2012

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.

2013

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.

2014

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.

Wrap-up

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!

Something about the challenging nature of calculating compound interest keeps drawing me back, as in my Mental Financial Wizard post and my recent Estimating Compound Interest post. Or, maybe I'm just greedy.

In either case, here's yet another way to get a good estimate of interest compounded over time. It's a little tricky to do in your head alone, so you'll probably prefer to work this one out on a sheet of paper.

It turns out that compound interest is based on the binomial theorem. This means we can use relatively simple math concepts from Pascal's Triangle (also based on the binomial theorem). The method I'm about to teach you has its roots in the approach used to work out coefficients in The Easy Peasy Binomial Expansion Trick (jump down to the paragraph which reads, "So now comes the part where the coefficients for each term are written. This is very easy to do with the way we set up our example.").

What we're going to be estimating is the total percentage of interest alone. Once this is done, you can calculate the original investment into the problem. As a first example, let's work out 5% interest per year for 10 years. To keep things simple, we'll work with 5% as if it represented 5, instead of 0.05.

To get a starting point multiply the interest rate by the time, as if you were working out simple interest. In our 5% for 10 years example, we would simply multiply 5 × 10 = 50. You need to make a table with that number expressed two ways: As a standard number, as as a fraction over 1. For this example, the first row of the table would look like this:

Number	Fraction
50	50⁄1

From here, there are 2 repeating steps, which repeat only as many times as you wish to carry them.

STEP 1: You're going to create a new fraction in the next row, based on the existing fraction. Take the existing fraction, increase the denominator (the bottom number of the fraction) by 1, and decrease the numerator by the amount of the annual interest.

In our example, starting from ⁵⁰⁄₁, we'd increase the denominator by 1, turning it into ⁵⁰⁄₂, and then decrease the denominator by 5, because we're dealing with 5% interest, to give us ⁴⁵⁄₂. The table, in this example, would now look like this:

Number	Fraction
50	50⁄1
	45⁄2

STEP 2: Take the number from the previous row, and multiply this by the new fraction, in order to get a new number for the current row. Divide the result by 100, and write this number down in the new row. This can seem challenging without a calculator, but if you think of a fraction as simply telling you to divide by the denominator and multiply by the numerator, it becomes simpler.

Continuing with our example, we'll multiply the number from the previous row (50) times our new fraction (⁴⁵⁄₂). That's 50 × 45 ÷ 2 = 25 × 45 = 1,125. 1,125 ÷ 100 = 11.25, so we add that number to the new row like this:

Number	Fraction
50	50⁄1
11.25	45⁄2

From here, we can repeat steps 1 and 2 as many times as we like, depending on what kind of accuracy is needed. Repeating step 1 one more time, we get this result (do you understand how we got to ⁴⁰⁄₃?):

Number	Fraction
50	50⁄1
11.25	45⁄2
	40⁄3

After repeating step 2, we work out 11.25 × 40 ÷ 3 = 11.25 × 4 × 10 ÷ 3 = 45 × 10 ÷ 3 = 450 ÷ 3 = 150. Don't forget, as always, to divide by 100, which gives us 1.5 for the new row:

Number	Fraction
50	50⁄1
11.25	45⁄2
1.5	40⁄3

Most of the time, I stop the calculations when the number in the bottommost row is somewhere between 0 and 10. I find this is enough accuracy for a decent estimate.

Once you've stopped generating numbers, all you need to do to estimate the interest percentage is add up everything in the the Number column! In our above example, we'd add 50 + 11.25 + 1.5 to get 62.75. In other words, 5% for 10 years would yield roughly 62.75% interest. If we run the actual numbers through Wolfram|Alpha, we see that the actual result is about 62.89% interest. That's not bad for a paper estimate!

Back in 2012, a question was posted at math stackexchange which could've benefitted from this technique. In under 3 minutes, answer the following multiple choice question without using a calculator or log tables:

Someone invested $2,000 in a fund with an interest rate of 1% a month for 24 months. Consider it to be compounded interest. What will be the accumulated value of the investment after 24 months?

A) $2,437.53
B) $2,465.86
C) $2,539.47
D) $2,546.68
E) $2,697.40

Let's use this technique to work this out:

Number	Fraction
24	24⁄1
2.76	23⁄2
0.2024	22⁄3

Hmmm...24 + 2.76 + 0.2024 = 26.9624, so that would give us about a 26.96% return, or a little less than 27%. Multiplying this by 2,000 is easy, since we can multiply by 2, then 1,000. This lets us know there must be just under $540 in interest on that $2,000. A and B are way too low, E is way too high, and D is just over $540 in interest. That eliminates every answer except C. Sure enough, Wolfram|Alpha confirms that $2,539.47 is the correct answer!

Two years ago, about this time, I reviewed Presh Talwalkar's previous Infinite Tower book.

Since then, not only has Presh not only been hard at work on his Mind Your Decisions blog, but also another book guaranteed to interest Grey Matters readers! This newest book is titled The Best Mental Math Tricks. Presh was kind enough to send me an advance copy, so I'll share my review in this post.

Probably the first thing to stand out about this book, when reading the table of contents, is that it's organized almost exactly backwards to most arithmetic, and even most mental math books. It starts out with a variety of mental math shortcuts for specific situations, then moves on to squaring shortcuts, followed by multiplication shortcuts, then division shortcuts, and it closes with another variety of shortcuts.

There's nothing bad about this approach. As a matter of fact, since the subject is mental math, this actually allows the shortcuts to be described in a rough order of simpler to more complex. It's also a nice change from the standard order of adding/subtracting to multiplication/division to roots/powers.

When you learn this book is put out by the author of a blog, you might be concerned that this is just a collection of previous mental math blog posts that you could access online for free. While there is some overlap, there's plenty of material in the book that has never been posted on the author's blog. Conversely, there are also several mental math shortcuts on his blog which don't appear in the book, so Presh's book and site wind up complementing each other quite nicely.

Even when there is crossover, the entry isn't simply copied straight from the blog to the book. For example, Presh wrote a post titled Understanding the rule of 72: a popular rule that has little practical value that was highly critical of this standard shortcut. In the book, however, the rule of 72 is taught with a less critical review, while still giving the reader an understanding of when the rule is and isn't appropriate to use.

The structure of each shortcut is also well thought-out. Each one starts with a description of the shortcut itself, followed immediately by practice problems which help you internalize it. Just before providing the answers to the practice problems, however, Presh explains the proof behind each shortcut, so you can get a better understanding of why it works. This is probably one of the most useful and important aspects of the book. It's one thing to learn a rule, but another thing to understand the reasoning behind it.

If you're already familiar with mental math shortcuts, you're still likely to find enough new shortcuts to make this book worthwhile. If you're new to mental math, this book is a definite treat for the mind!

At this writing, The Best Mental Math Tricks isn't available yet, but Presh Talwalkar assures me that it will be released in the near future. When it is released, The Best Mental Math Tricks is now available at Amazon.com. I recommend to anyone interested in improving their mental math skills!

February's snippets are here. Thanks to some old favorites, and some new favorites, we have a good selection to share with you this month.

• Just 2 days ago was Friday the 13th, so MindYourDecision.com's Presh Talwalkar decided it was a good time to teach how to divide by 13 in your head:



This is a handy technique, and you really only need to learn how to do this up to 12, which isn't too difficult.

If you'd like to learn similar tricks for dividing by 2 through 15, check out the Instant Decimalization of Common Fractions video.

• Like me, Presh seems to have plenty of fun with mental math techniques. Here's a mathematical magic trick of sorts, in which you apparently divine a crossed out digit:



Are you curious as to why this works? Presh has a detailed proof on his blog.

For those who are worried that just multiplying by 9 may seem too obvious, scroll down to the end of my Age Guessing: Looking at the Roots post. The section entitled “Sneakier ways of getting to a multiple of 9” has several useful and clever ways to disguise the method.

• IFLScience just posted 21 GIFs That Explain Mathematical Concepts. More than a few of these will be familiar to regular Grey Matters readers. Many are from LucasVB's tumblr gallery, and others are from videos I've shared over the years. Nevertheless, it's nice having all these in one place.

• Steve Sobek, who has a wide variety of videos on his YouTube channel, has recently made several mental math-related videos that are worth checking out. For example, here's his video teaching a trick for mentally subtracting large numbers:



You can find more of his mental math tricks at AmysFlashcards.com.

A recent question on the Mathematics StackExchange about mentally the compound interest formula caught my attention.

It got me thinking about good ways to work out a good mental estimate of compound interest.

Part of what makes it so tricky, is that compound interest doesn't work in a straight line, like much of the math with which we're familiar. Compound interest builds on itself exponentially (not surprising since the formula is a exponential expression). This is a good point to re-familiarize ourselves with the basics of the time value of money:



For a more detailed guide to interest rate mathematics, I suggest reading BetterExplained.com's A Visual Guide to Simple, Compound and Continuous Interest Rates.

BINOMIAL METHOD: The Mental Math wikibook suggests the following formula: To estimate (1 + x)ⁿ, calculate 1 + nx + ^n(n-1)⁄₂ x².

It is an interesting formula, especially considering that the first part, 1 + nx, is basically the simple interest formula. However, after using Wolfram|Alpha to compare the actual compound interest rate formula to this binomial estimation of compound interest method, you see that it really only gives close answers when x is 5% or less, and n is 5 time periods or less.

If your particular problem qualifies, that's not bad, but what about longer times?

RULE OF 72 AND OTHERS: Last July, I wrote another post about estimating compound interest discussing the rule of 72 for determining doubling time, as well as the rules for 114 (tripling time) and 144 (quadrupling time). Note especially that you can work out the effects of interest of long time periods with a little simple addition.

Yes, the rule of 72 has been explained many places, such BetterExplained.com's Rule of 72 post, and critically analyzed, such as MindYourDecision.com's Understanding the rule of 72 post and the related video, but as long as you understand its proper use and caveats, it's an excellent tool.

Understanding where the rule of 72 comes from, you can actually work out other rules for other multiples of your original amount, which is how the rules of 114 for tripling and 144 for quadrupling came about. If you want to estimate how long your money takes to grow to 5 times the original amount (quintupling), there's the rule of 168. Similar to the other rules, you can work out quintupling as ¹⁶⁸⁄_time ≈ interest rate (as a percentage) and ¹⁶⁸⁄_interest ≈ time.

To help increase the accuracy of needed estimates, you can also remember the 50% increases between each of the above rules. For a 50% increase, for example, there's the rule of 42. For 2.5 times, there's the rule of 96, for 3.5 times, there's the rule of 132, and for 4.5 times, there's the rule of 156.

This may seem like too many rules to remember, but there are a few things that help. First, keep the rules of 72, 114, 144, and 168 in mind as primary markers for 2×, 3×, 4×, and 5× respectively. Note that these are all multiples of 6, and that the “half-step” rules are also multiples of 6, and fall between the other numbers. So, if you forget how to work out 2.5×, you can realize that the rule is somewhere between 72 and 114, and then recall that 96 is the rule you need! Here is a handy Wolfram|Alpha chart for which rules go with which amounts.

“RULE” EXAMPLES: In the video above, Timmy needs to find out how long it will take to get 10 times his money at a 10% interest rate. Since we only have rule up to 5, how do we work this out? Well, 10 times the money is basically the same as quintupling the money, then doubling that amount of money. So, we can work out the quintupling times and doubling times at 10%, then add them together!

For quintupling, we use the rule of 168 to find that ¹⁶⁸⁄₁₀ ≈ 16.8 years. Since these are estimates, you can usually round up to the nearest integer to help with the accuracy. So, his money will quintuple in about 17 years. How long will it take to double from there? We use the rule of 72 for doubling, so ⁷²⁄₁₀ ≈ 7.2 years, and rounding up gives us 8 years. Since it will take about 17 years to quintuple, and about 8 years to double, then getting 10 times the amount should take roughly 17 + 8 = 25 years. In the video, they note that it will take 26 years, so we've got a good estimate!

In his Impress by doing compound interest in your head post, Martin Lewis describes the following interest problem: “What is the APR, ie annual interest rate, if you borrowed £80,000 and had to repay £200,000 six years later?”

Since £200,000 ÷ £80,000 = 2.5, we can use a simpler approach than Martin Lewis did in his article, as we know that the rule of 96 with the 6 year time span to work out the annual interest rate. ⁹⁶⁄₆ ≈ 16% interest, the same answer Martin worked out!

USING e: Once you start getting too far beyond 25 time periods (25 years for annual interest rates), you should start using e (roughly 2.71828...) to estimate compound interest over the long term. Last March, I wrote Calculate Powers of e In Your Head! to help with this exact task. At this point, you're probably more concerned with the scale of the answer, rather than the exact answer, so working out just the equivalent power of 10 is all you really need.

SHORT VERSION: So, instead of providing one way to estimating compound interest, here are 3 methods for different scale problems. If the rate is 5% or less, and the interest is applied 5 or fewer time, the binomial method is the way to go. If your problem is larger than that, and covers less than 25 time periods, then use the rules approach (rule of 42, 72, 96, etc.). If the interest is applied more than 25 times, use e to get an idea of the scale.

It may not be the simplest estimation approach for compound interest, but if you're stuck without a calculator, this will help you get by until you can more accurately crunch the numbers.