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!
The first snippets of 2015 are ready!
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.
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.
+ 8: The addition of 8 takes the starting point of 2015 into account, which is why this particular formula works ONLY for 2015.
mod 30: 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.
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!
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!
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!
Imagine having someone think of a number from 1 to 100, having them cube the number using a calculator, telling you only the result, and you're able to calculate the cube root of their result (the original number they put in the calculator)!
Learning to work out cube roots of perfect cube is an impressive feat, but it's far less difficult than it appears.
We'll get right to the method, taught in the video below. You can read the MindYourDecisions.coom post New Video – Calculate Cube Roots In Your Head for further details.
Over in the Mental Gym, I have a more detailed tutorial on working out cube roots of perfect cubes, including a cube root quiz.
Back in March of 2013, Scam School also taught the cube root feat in their own unique way. If you like this feat and want to take it a step further, check out Numberphile's fifth root feat tutorial. Surprisingly, this is even easier than the cube root feat!
I apologize for not posting the past 2 Sundays, but my internet connection was down.
Grey Matters is back this week, however, with some Sunday afternoon football...postgame conference, anyway. How does this relate to math or memory? Read on!
In the November 30 Texans vs. Titans game, Texan player Ryan Fitzpatrick threw 358 yards for 6 touchdowns, setting a record for the franchise.
You'd think that would be the big talking point of the postgame press conference, but Fitzpatrick's son Brady winds up stealing the show with his mental math skills (starting about 1 minute into the video):
I'm not sure exactly how long Brady has been performing this feat, but I've found an excellent candidate. It seems that just 10 days before that conference, the Mind Your Decisions blog posted about how to perform this exact feat. You can learn it below, including how to handle numbers in the 80s:
With a little practice, you can multiply numbers like these as quickly as Brady Fitzpatrick. The next step, performing this on TV, is a little trickier, however. However, you can still perform this for your friends and family!
Sorry about missing a post last week. It turned out to be a busier weekend for me than I originally planned.
I'm back this week, and I've brought plenty of snippets with me to make up for the missing post!
• Back in September, in DataGenetics' Grid Puzzle post, an intriguing puzzle was posed: Imagine there is a grid of squares. You draw a line connecting one intersection to another intersection on the grid. The question is: How many squares does this line cross through?
The link works through the answer, including an interactive widget which may help you work it out. It turns out that the answer involves only simple arithmetic and being able to work out the greatest common divisor (GCD) of 2 numbers.
If you can work out the GCD of 2 numbers in your head, you can turn this into a mental math presentation. How do you go about that? Math-magic.com has the answer, in both webpage form and PDF form! Page 94 of Bryant Heath's Number Sense Tricks also has some great tips on finding the GCD of 2 numbers in your head.
• Also back in September, I wrote about Knight's Tours on a calendar, and included sample calendars for both September and October 2014. I figured it's time for the November 2014 version, included below. This month, 6 is most date-to-move matches I could find:
• OfPad.com seems to have a similar goal to Grey Matters, which is to improve your genius. There's plenty to explore as a result, but I'd like to draw your attention to their Mental Math Tricks posts in particular. There's lots of great tricks, some of which you may not have seen before.
• To wrap up this month's mental math snippets, I'd like to focus on one particular technique for multiplying any two 2-digit numbers together. I've seen this technique before, but the way dominatemath teaches this multiplication technique, it just made so much more sense than the other instructions I've seen for this approach.
That's all for this month's snippets. Have fun exploring them!
I've wanted to write about factoring numbers in your head for a while, but never really had a good approach to analyze and discuss.
Recently, I've come across some strategies that mesh well with what I've discussed before on this site. Learning to factor a number in your head can be tricky, but it can be done.
BASICS: Starting from a given number between 1 and 10,000, you're only going to test for divisibility from the number 2, up to the square root of the given number. If you're familiar with estimating square roots, you only need the whole number part.
For example, if you're given the number 447, you only need to estimate the square root as 21, to realize you only need to be concerned with numbers from 2 to 21.
To narrow things down even further, you're only going to test for divisibility by prime numbers from 2 up to the limit you determine (21 in the above example). Between 2 and 100, there are only 25 prime numbers (2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, and 97), so testing only for these minimizes the time it will take.
Certainly, divisibility tests for 2 (is the rightmost digit even?), 3 (do the digits add up to a multiple of 3?), and 5 (is the rightmost digit a 5 or a 0?) are well known, but how do you go about testing, and remembering the tests, for the higher primes?
TECHNIQUES: NUMBERS ENDING IN 9 - Back in April, I discussed how to work out divisibility tests for numbers ending in 9. For divisibility by 19, you would divide by 20, add up the quotient and remainder from that divisibility problem, and see if they added up to a multiple of 19. For 29, you'd run through the same process, but divide by 30, instead. For 59, you'd divide by 60, and so on.
When testing for various primes, however, working out all these quotients and remainders can quickly get confusing. Thankfully, it turns out that there's an alternative method, and it's even a little quicker!
Let's start with the test for 19. Instead of dividing by 20 and running through the process described above, split the number into the rightmost number and the rest of it, multiply that rightmost digit by 2, and add that total to the rest of the number. As an example, let's test 285 for divisibility by 19. You'd split 285 up into 28 and 5. Doubling the rightmost number, 5 × 2 = 10, so we'd add that 10 to the rest of the number, 28, to get 38. 38 is a multiple of 19, so 285 must be a multiple of 19! Wolfram|Alpha confirms that 285 is evenly divisible by 19.
For bigger numbers, you may not be sure about the result you got, so you simply run the test on your new total. For example, when testing 2,527 for divisibility by 19, you'd split it up into 252 and 7. Doubling the 7 makes 14, so you'd add 252 + 14 to get 266. Not sure whether 266 is evenly divisible by 19? We run the same test on 266, by splitting it up into 26 and 6. 6 doubled is 12, and 26 + 12 = 38, which we know is a multiple of 19! Therefore, 2,527 is a multiple of 19!
For 29 instead, you would divide up the number in the same way, but multiply the last digit by 3 instead, and then add . Why? Because 29 is right next to 30, and 30 divided by 10 equals 3 (similar to the way we used 2 as a test for 19, because 19 is next to 20). This same pattern holds for the other prime numbers ending in 9. For 59, you'd multiply the rightmost digit by 6 (59 + 1 = 60, 60 ÷ 10 = 6) and add it to the rest of the number. When testing divisibility by 79, you'd multiply the last digit by 8 before adding, and when testing divisibility by 89, you'd multiply the last digit by 9 before adding. Get the idea?
That's great for those numbers, but what about primes which don't end in 9?
SCALING NUMBERS TO END IN 9 - As also discussed in The Great Divide, you can also use this test on numbers which can be scaled up to end in 9.
This approach is particularly hand for primes ending in 3. For example, 13 × 3 = 39. Since 39 is right next to 40, you should quickly realize that you can test for divisibility by 13 by splitting the number as before, multiplying the rightmost number by 4 (40 ÷ 10 = 4), and adding that total to the rest of the number.
Is 507 evenly divisible by 13? 507 split becomes 50 and 7, multiplying 7 by 4 gives us 28, and 50 + 28 = 78. 78 is a multiple of 13, so 507 is evenly divisible by 13!
In the same way, 23 can be tested by multiplying the last digit by 7 and adding it to the rest of the number, because 23 can be scaled up to 69, and 69 is next to 70. Need to test for divisibly by 43, you split the number, multiply by 13 (do you see why?), and add as before.
Just using this approach covers 10 of the 25 prime numbers from 2 to 97. The well-known tests for 2, 3, and 5 add another 3, so you should comfortably be able to test for divisibility by more than half the prime numbers below 100!
Naturally, that brings up the question of how to handle the other half.
NUMBERS ENDING IN 1 - In my follow-up to The Great Divide, I cover a similar technique for numbers ending in 1. The big difference between the ending-in-9 technique and the ending-in-1 technique is that adding is replaced by subtracting.
Let's start by testing for divisibility by 11. The test will begin with the same splitting as before, and since 11 is next to 10, we'll multiply the last digit by 1. However, this time we'll subtract that number from the other numbers.
Is 341 evenly divisible by 11? We split 341 into 34 and 1, multiply 1 by 1, giving us 1, and subtract that from the rest of the number, 34 - 1 = 33, and since we know that 33 is evenly divisible by 11, then so is 341!
I'm sure you have the idea by now. For 31's divisibility test, you'd multiply the rightmost digit by 3 and subtract, for 41's test, you'd multiply by 4 and subtract, and so on.
SCALING NUMBERS TO END IN 1 - Just as before, this technique also applies to scaling numbers up to end in 1. For testing prime numbers ending in 7, including 7 itself, this is a big help.
The test for 7 begins by realizing that 7 can be scaled up to 21. This tells us that the rightmost number must be multiplied by 2 (remember why?) before subtracting from the rest from the rest of the number.
Is 665 evenly divisibly by 7? Let's use our knowledge of the 21 test to find out. The number is split into 66 and 5, and we double the 5 to get 10. 66 - 10 = 56, and we know 56 is a multiple of 7, so therefore 665 is evenly divisble by 7!
WIND-UP: At this point, we've covered the classic divisibility tests for 2, 3, and 5, as well as how to recall and perform easy divisibility tests for every prime number below 100 which ends in 1, 3, 7, and 9 - in other words, every test you need!
You'll need to recall that 9s and 3s require multiplying then adding, and 1s and 7s require multiplying followed by subtracting. When subtracting, note that you can always subtract the larger number from the smaller, which can prevent having to deal with negative numbers.
Since you're performing multiple tests, this won't be the quickest of mental math feats. However, being able to recall and perform tests for such a wide variety of prime numbers is still an impressive, and even useful, feat!