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!

## Review: The Best Mental Math Tricks

Published on Sunday, February 22, 2015 in books, math, mental math, products, reviews, self improvement

## Still More Quick Snippets

Published on Sunday, February 15, 2015 in magic, math, mental math, snippets, videos

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.

## Estimating Compound Interest

Published on Sunday, February 08, 2015 in mental math, money, self improvement, videos

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} *x*^{2}.

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.

## Cards and Dice

Published on Sunday, February 01, 2015 in fun, magic, math, playing cards, videos

Presh Talwalkar, from the *Mind Your Decisions* blog, recently shared a fun magic trick.

It involves playing cards and dice. Because it's mathematically based, however, you might just fool yourself while performing it. You'll really only fool yourself if you don't analyze the math behind the trick, which is exactly what we're going to do in this post.

First, let's take a look at Presh's video, from this January 2015 post:

There's quite a bit going on here, so let's break this down piece by piece.

**CARDS:** Let's ignore the dice for the time being, and focus only on the playing cards. During the dealing process, cards fall into 1 of 2 categories: Either they're dealt individually, or they're in the remainder that is placed on top of the stack. Let's look at each of these categories separately.

*DEALT CARDS:* We'll start with the first example from the video, where 10 cards were used, 7 of which were dealt. What happens to those 7 cards. The card in position 1 is dealt first, and will obviously become card 10. The card at position 2 will wind up as the 9th card, and so on. Take a look at where these 7 cards end up in the final stack:

```
Starting position Ending position
----------------- ---------------
1 10
2 9
3 8
4 7
5 6
6 5
7 4
```

See the pattern? The starting position plus the ending position always add up to 11, in this example! Why 11? What would happen if we used, say, 18 cards instead? Well, 1 would become 18, 2 would become 17, and so on. In that example, everything totals 19. The resulting total will always be 1 more than the number of cards involved.We can use this to work out a formula for the dealt cards. The total number of cards plus 1, minus the starting position of a dealt card, will give you the ending position. So, in our example with 10 cards, of which 7 are dealt, we can work out the total number of cards (10) plus 1 (equals 11) minus the original position (say, 3, so 11 minus 3 equals 8) gives us the ending location of that card (so, we can easily say that, in this example, the card that started at the 3rd position will wind up at the 8th position).

A simpler way to say this is to use S for starting position, E for ending position, and N for the total number of cards. So, our formula for dealt cards could be written as E = (N + 1) - S

That's OK for the dealt cards, but what about the undealt cards?

*UNDEALT CARDS:*If you have 10 cards with 7 cards dealt, cards 8, 9, and 10 will not be dealt. They are simply placed on top of the dealt cards as a group.

In this case, the original 8th card becomes the first card, the 9th card becomes the 2nd card, and the 10th card becomes the 3rd card. Let's chart these positions and find a pattern:

Starting position Ending position ----------------- --------------- 8 1 9 2 10 3This pattern is even simpler! The starting position, minus the number of dealt cards, gives the new ending position. Using D for the number of dealt cards, and the same variables from the first formula above, we have E = S - D for the undealt cards.

OK, we've got 2 formulas to handle our 2 cases, so let's bring the dice back in.

**DICE:**Since the dice are used to choose random numbers, we'll refer to them with the letter R, and since high, low, and medium numbers are important, we'll use R1 for the dice with the lowest number, R2 for the dice with the middle number, and R3 for the dice with the highest number.

*FOLLOW THE TOP CARD:*As explained, knowing the top card is the key to this trick, so we're only going to follow that particular card.

How many cards are used? In the first performance, the dice rolled with R1=2, R2=3, and R3=5, which means that 10 cards are used. More generally, the dice total determines the number of cards used, so N (total number of cards) = R1 + R2 + R3. In our example, this was 5 + 3 + 2 = 10 cards.

*Step 1:*Note that, when the cards are dealt initially, the top card ALWAYS becomes the bottommost card. So, the starting position for the predicted card is always at the bottom as well. In other words, S (the starting point) = R1 + R2 + R3, as well. So, the predicted card starts at 10 in the first example. In other words, S = 10.

*Step 2:*For the next deal, R2 is removed (3 in the first example), and R1 + R3 dice (D = 7) are dealt. The card as starting position 10 (S=10) is obviously not going to be dealt, so we'll apply the undealt card formula (E = S - D). E = 10 - 7 = 3. So, the predicted card winds up at position 3 in the example.

Let's look at this more generally. The card starts out at position S, which is also R1 + R2 + R3. You're dealing D cards, and D = R1 + R3. So, E = S - D can be re-written as E = R1 + R2 + R3 - (R1 + R3), which simplifies to E = R1 + R2 + R3 - R1 - R3, which further simplifies to E = R2.

In other words, after the first dealing of D cards, the predicted card will wind up at the position denoted by R2, the removed dice! That's interesting and unexpected.

*Step 3:*So, now we want to see what happens to the card at position R2 on the next deal. Because R2 must always be less than the total of R1 + R3 (removing the middle number ensures this), R2 will always be among the cards dealt in this phase. This means we need to follow the dealt card formula from above (E = (N + 1) - S).

In the 10 cards total/7 cards dealt example, we're now tracking the 3rd position, so E = 10 + 1 - 3, which simplifies to E = 8, so our 3rd card winds up in the 8th position.

The current general starting position is, as we already know, is R2. We can turn the formula, then, into E = (N + 1) - R2. Further, since the total number of cards, N, is R1 + R2 + R3, we can change the formula into E = R1 + R2 + R3 + 1 - R2. This simplifies into E = R1 + R3 +1.

Interpreting that general formula, that means the predicted card has now moved to the position denoted by the remaining dice plus 1. Sure enough, in our running 10/7 example, 7 remains, and the predicted card has moved to the 8th (7 + 1) position!

*Step 4:*This should be pretty clear. The card we're following is at position R1 + R3 + 1, and we're going to deal R1 + R3 cards off of it. In our example, the card we're tracking is at position 8, and we're going to deal 7 cards from above it. Either way, the card will be moved to the first position!

**SHORT VERSION:**If you read the above carefully, you can start to see

*WHY*this card trick works. The predicted card starts at the bottom of the pile. Next, it moves to position R2, followed by a move to position R1 + R3 + 1, and finally to position 1.

**CREDITS AND OTHER THOUGHTS:**As noted in Presh's original post, he developed this after reading the Low Down Triple Dealing routine, as found in the book

*Mathematical Card Magic*by Colm Mulcahy.

There's limitless variations to this type of routine, one of which was created by Jim Steinmeyer for use by David Copperfield's

*Fires of Passion*TV special.

If you'd like to play around with this principle without having to constantly deal cards, I've created a formula in Wolfram|Alpha that will effectively deal the cards for you.

Returning one last time to our 10 card pile (N=10) with 7 cards dealt (D=7) example, you simply set N and D, and the calculation will do the rest. The output you get from this run is {8, 9, 10, 7, 6, 5, 4, 3, 2, 1}. The numbers are the starting positions (S), and each number's placement is their ending position (E). 8 being placed first means that it started in the 8th position and moved to the first position. 9 being in the second position means that the cards which started out 9th has been moved to the 2nd position, and so on.

As a matter of fact, since the total cards (N) and the number of dealt cards (D) stay constant in most routines, you can use this one output to track the card through multiple deals. I'll show you what I mean by following through 3 deals of 7 cards from a pile of 10.

What does {8, 9, 10, 7, 6, 5, 4, 3, 2, 1} tell us about the original 10th card? One quick glance tells us it became the 3rd card (because there's a 10 in the 3rd position. Where does the 3rd card go from there? Another quick glance tells us that, since 3 is in the 8th position, the 3rd card must move to the 8th position next. Finally, what happens to the 8th card? It winds up on top, because we can see the 8th card at position 1!

That's how an entire routine with multiple deals can be explored using only one simple mathematical result!

Thanks for creating and sharing this routine, Presh! I'd love to hear about any variations my readers develop in the comments, as well!

## Even More Quick Snippets

Published on Sunday, January 18, 2015 in calendar, fun, math, mental math, products, self improvement, snippets, videos

The first snippets of 2015 are ready!

This time around, I have some clever and fun approaches to math to share. I think you'll be surprised by them, even (or especially) if you don't usually like math.

• This January marks the 28th anniversary of *Square One TV*, an educational program that taught math with the use of skits, songs, and other fun approaches. While it's not on TV anymore, YouTube user Anton Spivack has been making full episodes available. I've been gathering them together in playlists by season if you want to experience this show for yourself:

Square One TV: Season 1

Square One TV: Season 2

Square One TV: Season 3

Square One TV: Season 4

Square One TV: Season 5

Square One TV: Mathnet

• While I'm thinking about YouTube channels, check out Funza Academy's site, as well as their YouTube channel. Being interested in math shortcuts, I especially enjoy their *Math Concepts and Tricks* playlist, as it teaches some impressive math shortcuts, including rapidly multiplying any 2-digit numbers together!

• Magic Cafe user RedDevil, author of the *RedDevil Mentalism* blog, recently shared a great tip for my *Day One* routine. *Day One* is my approach to minimizing the work required for the classic Day of the Week For Any Date feat.

RedDevil took this one step further by pointing out that you don't need to remember all the year information I teach in there. Instead, you can only memorize just the leap years, and move 1, 2, or 3 days forward as you go 1, 2, or 3 years ahead respectively.

If you have *Day One*, you'll understand this. If you don't have *Day One*, it's still available for only $9.99! If you're a member of the Magic Cafe with at least 50 qualifying posts, you can read his tip in more detail in RedDevil's original thread.

Yes, the snippets are short and sweet this month, but there's still plenty to explore in these links if you take the time to learn and enjoy them!

## Day & Moon Phase For Any Date in 2015

Published on Sunday, January 04, 2015 in calendar, fun, math, memory, memory feats, mental math, self improvement, site features

Happy New Year!

With a new calendar year, you deserve a couple of new calendar feats to go with it. In this post, you'll learn how to quickly give the day of the week AND the moon phase for any date in 2015.

Even better, both of these feats are much easier than they sound!

**DAY OF THE WEEK FOR ANY DATE IN 2015:** The method to do this is quite simple, and is known as the Doomsday method, originally developed by John Horton Conway.

Start by going to last week's post, *Calendar Calculation Made Simple*, and learning the simple calendar calculation techniques taught there.

To work out dates in only 2015, all you have to remember is that 2015's "Doomsday" is Saturday. If you think about it, you can already work out any date in February using just this knowledge.

For example, Valentine's Day, Feb. 14th, must also be a Saturday, because it's exactly 2 weeks before Feb. 28th. How about Feb. 2nd (Groundhog Day)? Well, Feb. 7th is a Saturday, and Feb. 2nd is 5 days before that. What's 5 days before a Saturday? The answer is Monday! Therefore, Groundhog Day will be on Monday in 2015.

On which day will Christmas fall in 2015? We know from the technique taught in last week's videos that December 12th is a Saturday, so 2 weeks later, December 26th, is also a Saturday. Since Christmas is one day before that, it must be on a Friday!

When is July 4th this year? It's exactly 1 week before July 11th, so it must be a Saturday, as well.

St. Patrick's Day, March 17th, is 3 days after March 14th (Pi Day, mentioned in the videos from last week), so it's 3 days after a Saturday, making it a Tuesday in 2015.

January 15th is Martin Luther King, Jr.'s birthday, but what day does it fall on in 2015? January 3rd is a Saturday this year, and so is January 17th (2 weeks later). Take back 2 days, and we get January 15th being a Thursday this year!

With the knowledge from last week's videos, and a little practice, you can quickly and easily determine the day of the week for any 2015 date. You could get practice at the Day For Any Date (Mentalist Challenge) page, changing the year to 2015, and then trying to determine the date before you click the Show button.

When you're demonstrating this ability for someone, it's nice to be able to prove that you're right about the date. I use Wolfram|Alpha and/or timeanddate.com's calendars.

**MOON PHASE FOR ANY DATE IN 2015:** 2 years ago, I posted a new tutorial about determining the moon phase for any date. Similar to the year calculations, focusing on a particular year, such as 2015, greatly simplifies the required calculations. Like the doomsday algorithm above, this formula was also developed by John Conway.

In fact, working out the moon phase for any date in 2015 is even simpler than working out the date! How simple is it?

(Month key number + date + 8) mod 30

It's probably best if I explain each part:

* 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.

*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.*

**Date:***The addition of 8 takes the starting point of 2015 into account, which is why this particular formula works ONLY for 2015.*

**+ 8:***If you get a total of 30 or more, simply subtract 30. Otherwise, just leave the number as is. Betterexplained.com has an intuitive explanation of modular arithmetic.*

**mod 30:**The resulting number will be the approximate age of the moon in days, from 0 to 29. This formula only gives an approximation, so there's a margin of error of ±1 day.

As an example, let's figure the phase of the moon on July 4, 2015. July is the 7th month, and the 4th is the date, so we work out (7 + 4 + 8) mod 30 = (11 + 8) mod 30 = 19 mod 30, which is just 19.

In that example, we estimate the age of the moon to be 19 days old.

What exactly does the age of the moon in days mean in practical terms? Here's a quick guide:

- 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 th moon is lit, and it's getting brighter each night)
- 7.5 days = 1st quarter moon (Half the moon is lit, and gets 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 gets darker each night)
- 22.5 to 29 days = Waning crescent (Less than half the moon is lit, and it's getting darker each night)

If you have any experiences or thoughts you'd like to share about memorizing the dates and moon phases for the 2015 calendar, I'd love to hear about them in the comments below!

## Calendar Calculation Made Simple

Published on Sunday, December 28, 2014 in calendar, fun, math, mental math, self improvement, videos

As we wrap up the year, it's natural for thoughts to turn to the calendar.

Yes, I've talked about calendar calculation many times before, but I'm always on the lookout for better methods and better teaching. *MindYourDecision.com*'s Presh Talwalkar, whom you may remember from last week's post, is back this week with some great calendar calculation lessons!

Let's face it, calendar calculation can sometimes seem daunting. One of the best approaches to learning such a skill for the first time is to ease yourself into it. Presh's starting approach is to teach you how to work out New Year's Day for any date from 2000 to 2099:

Try practicing this skill for yourself. First, use Wolfram|Alpha to get a random year in the 2000s, workout the day of the week for New Year's Day as above, and then verify your answer with Wolfram|Alpha.

It's not difficult, and once you get the hang of this skill, you're ready to move on to the next step.

Using the New Year's Day skill as a starting point, Presh then introduces you to John Horton Conway's well-known Doomsday approach to calendar calculation:

Once again, practice is the key here. You can use Wolfram|Alpha to generate a random date in the 2000s, work it out using the above method, and then verify the correct date using Wolfram|Alpha again.

If you're interested in calendar calculation in general, I not only have several posts about it here on Grey Matters, but numerous lessons, including quizzes, about calendar calculation over in the Mental Gym. Once you get the knack, it's amazing where you can take this skill!

## 12 Days of Christmas

Published on Sunday, December 21, 2014 in fun, math, memory, mental math, puzzles, site features, TV, videos

Note: This post first appeared on Grey Matters in 2007. Since then, I've made it a sort of annual tradition to post it every December, with the occasional update. Enjoy!

Since the focus of this blog is largely math and memory feats, it probably won't be a surprise to learn that my favorite Christmas carol is The 12 Days of Christmas. After all, it's got a long list and it's full of numbers!

On the extremely unlikely chance you haven't heard this song too many times already this holiday season, here's John Denver and the Muppets singing The 12 Days of Christmas:

The memory part is usually what creates the most trouble. In the above video, Fozzie has trouble remembering what is given on the 7th day. Even a singing group as mathematically precise as the Klein Four Group has trouble remembering what goes where in their version of The 12 Days of Christmas (Their cover of the Straight No Chaser version):

Just to make sure that you've got them down, I'll give you 5 minutes to correctly name all of the 12 Days of Christmas gifts. Those of you who have been practicing this quiz since I first mentioned it back in 2007 will have an advantage.

Now that we've got the memory part down, I'll turn to the math. What is the total number of gifts are being given in the song? 1+2+3 and so on up to 12 doesn't seem easy to do mentally, but it is if you see the pattern. Note that 1+12=13. So what? So does 2+11, 3+10 and all the numbers up to 6+7. In other words, we have 6 pairs of 13, and 6 times 13 is easy. That gives us 78 gifts total.

As noted in Peter Chou's Twelve Days Christmas Tree page, the gifts can be arranged in a triangular fashion, since each day includes one more gift than the previous day. Besides being aesthetically pleasing, it turns out that a particular type of triangle, Pascal's Triangle, is a great way to study mathematical questions about the 12 days of Christmas.

First, let's get a Pascal's Triangle with 14 rows (opens in new window), so we can look at what it tells us. As we discuss these patterns, I'm going to refer to going down the right diagonal, but since the pattern is symmetrical, the left would work just as well.

Starting with the rightmost diagonal, we see it is all 1's. This represents each day's increase in the number of presents, since each day increases by 1. Moving to the second diagonal from the right, we see the simple sequence 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, and 12, which can naturally represent the number of gifts given on each day of Christmas.

The third diagonal from the right has the rather unusual sequence of 1, 3, 6, 10, 15, 21, 28, 36, 45, 55, 66, 78, 91. This is a pattern of triangular numbers.

But what can triangular numbers tell us about the 12 days of Christmas? If you look at where the 3 in this diagonal, it's southwest (down and to the left) of the 2 in the second rightmost diagonal. If, on the 2nd day of Christmas, you gave 2 turtle doves and 1 partridge in a pear tree, you would indeed have given 3 gifts, but does the pattern hold? On the 3rd day, you would have given 3+2+1 (3 French hens, 2 turtle doves and a partridge in a pear tree) or 6 gifts total, and sure enough, 6 can be found southwest of the 3! For any of the 12 days, simply find that number, and look to the southwest of that number to see how many gifts you've given by that point! Remember when figured out that the numbers 1 through 12, when added, totaled 78? Look southwest of the 12, and you'll find that same 78!

Let's get really picky and technical about the 12 days of Christmas. It clearly states that on the first day, your true love gave you a partridge in a pear tree, and on the second day your true love gave you two turtle doves and a partridge in a pear tree. You would actually have 4 gifts (counting each partridge and its respective pear tree as one gift) by the second day, the first day's partridge, the second day's partridge and two turtle doves. By the third day, you would have 10 gifts, consisting of 3 partridges, 4 turtle doves and 3 French hens.

At this rate, how many gifts would you have at the end of the 12th day? Sure enough, the pattern of 1, 4, 10 and so on, known as tetrahedral numbers, can be found in our Pascal's Triangle as the 4th diagonal from the right.

If you look at the 2nd rightmost diagonal, you'll see the number 2, and you'll see the number 4 two steps southwest (two steps down and to the left) of it, which tells us you'll have 4 gifts on the second day. Using this same method, you can easily see that you'll have 10 gifts on the 3rd day, 20 gifts on the 4th day, and so on. If you really did get gifts from your true love in this picky and technical way, you would wind up with 364 gifts on the 12th day! In other words, you would get 1 gift for every day in the year, not including Christmas itself (also not including February 29th, if we're talking about leap years)! Below is the mathematical equivalent of this calculation:

If you're having any trouble visualizing any of this so far, Judy Brown's Twelve Days of Christmas and Pascal's Triangle page will be of great help.

One other interesting pattern I'd like to bring up is the one that happens if you darken only the odd-numbered cells in Pascal's Triangle. You get a fractal pattern known as the Sierpinski Sieve. No, this won't tell you too much about the 12 days of Christmas, except maybe the occurrences of the odd days, but it can make a beautiful and original Christmas ornament! If you have kids who ask about it, you can always give them the book The Number Devil, which describes both Pascal's Triangle and Sierpinski Sieve, among other mathematical concepts, in a very kid-friendly way.

There's another 12 Days of Christmas calculation that's far more traditional: How much would the 12 gifts actually cost if you bought them? PNC has been doing their famous Christmas Price Index since 1986, and has announced their results. Rather than repeat it here, check out their site and help them find all 12 gifts, so that you can some holiday fun and then find out the total!

Since my Christmas spending is winding up, I'm going to have to forgo the expensive version, in favor of Miss Cellania's internet-style version of The 12 Days of Christmas. Happy Holidays!