## Saturday, June 30, 2012

### Memory Tricks

Ran across an interesting TED Talk on cultivating memory, embedded below:

The premise of the video is that people with "super human memories" aren't necessarily super human --  but most of their abilities are trained -- that you may can use a relatively simple technique to remember many different things. Joshua Foer calls it building a "memory palace" which is a series of images which you can use to associate things.  He opens the speech with a very vivid account of several images, including nudist bikers outside his front door, cookie monster greeting him on a talking horse in his living room, the four wizard of oz characters in his kitchen and Britany Spears in his hallway. This palace reminded him of all the talking points he was going to make in his speech, which he revealed when he decoded the dream at the end.

His talk inspired me to think of ways that I can make my classes more memorable -- and so I hope by including the video in this blog, I will remember to consciously teach with these techniques in mind.

I have used other techniques for memories -- mostly chunking.  That involves breaking a sequence of numbers, words, or objects into groups. I remember the decimal notation of 1/7th that way: 14 (which is twice 7) 28 (which is twice 7) and then 57, which I "just remember" which gives me .142857.  I initially remembered it as 56 is twice 28... but it was a little off. Regardless, it has stuck in my head, and to this day when I say the number, I say it in chunks: onefour, twoeight, fiveseven....  My mom memorized her aunts phone number in a very similar, mockable way -- but she never forgot it.

Another example of chunking I've used is for Philippians 4:8: Whatever is true, whatever is noble, etc...
I grouped the first three together and remembered that i sing TeNoR (true, noble, right).  I grouped the second three together because I PLAy music (pure, lovely, admirable) and the last two (excellent, praiseworthy) reminded me of an acronym from waterskiing, an Exceptional Performance or EP. So I learned the verse by Tenor, Play, EP. Incidentally, I wrote a song for that verse and the surrounding verses which clumps them together like that.

But, this memory trick is about picture association, so here's my attempt at building a picture to help remember the seventh's:

Since there are six digits that repeat, I'm going to imagine six passengers in my minivan. Driving the van is a pirate, who has one leg, and an eye-patch over one eye and has a one-winged parrot who has dropped a small single drop of white poop with black speckles onto his shoulder.  He forces a soccer ball over to a 4-point white tailed deer who is standing with all four legs on the seat. He head butts the ball behind him to a terrible two year old orange bicycle named Bo who bounces the ball off both handle bars and behind him to an red octopus wearing a white border who promptly ate the soccer ball and farts which causes the five year old Abraham Lincoln who is stuck in his five-point safety harness to laugh uncontrollably, kicking the seat of his seven sisters, collectively named Pleiades, who spill their rainbow colored slurpee on the pirate and his bird who says "do it again -- squawk!"....

I could probably use some practice.... What have you memorized before?  Comment below.

## Wednesday, June 27, 2012

### Perspective in astronomy pictures

One movie that had a profound effect on me was E.T. Besides opening my eyes to a much larger world besides the earth we live in, I remember being amazed at Spielberg's beautiful filmwork and storytelling. Especially striking however, an image that often pops in my head is the picture of Eliot and the alien riding their bike transiting the moon. I must admit being tricked by the picture for many years and longing to see a moon that was that big! I always loved seeing the moon on the horizon because of the optical illusion it provides, but the illusion was never THAT big!  I want the moon to fill up more than half the sky like that!

Of course, I learned later the moon itself was zoomed in on, the biker superimposed on top, but still, what a great picture!

The thought for today is to calculate how far away are Elliot and the alien, given what we know about the true size of the moon? And we'll look at another "true" photograph from yesterday to calculate how far away the fisherman were.

One important fact that we need to know is the angular width of the moon. If you've been following the Summer Math Series so far, you'll remember I told you the moon and sun both occupy approximately 1/2 degree in the sky. This coincidence is a gift of God allowing for eclipses and measuring the solar system, but more about that later.

I "photoshopped" the image of E.T. to see how wide the bike and boy are. It looks like 4.25 "bikers" fit across the moon, so that means the angular size of a biker is:
$\frac {\tfrac{1}{2}^{\circ}}{moon } \frac{1 moon}{4.25 bikers} \approx 0.1^{\circ} / biker$

Using some trigonometry we can calculate the distance away the biker is. First, a triangle picture.  I am assuming that the biker in the picture is pretty much perpendicular to me, and so I draw a right triangle as shown:
The angle, height, and distance are all related by the tangent function:
$tan(\Theta )= \frac {opposite}{adjacent} = \frac{h}{distance}$
That means that if we know two out of those three things, we can calculate the third. Earlier we measured the angle, and if I assume the height of the biker to be approx 5 feet, we can calculate the distance by rearranging the equation:
$distance = \frac {h}{tan(\Theta )} = \frac {5 ft}{tan(.1^{\circ})} \approx 3000 ft$
Notice, if you type this into your calculator, you might get a more precise number -- I actually got 2864.786067 feet. That amount of precision is misleading, because I made two pretty uncertain measurements in the height of the biker (how do i know he isn't 5' 2" or 4' because he's sitting down?) and in the angular width of the biker (maybe its more like 4.5 bikers per moon, or only 4?)  When this situation arises, one rule of thumb is to round to as many numbers as you had when you made your assumptions. Since 5' had only one digit, I'm going to round to the nearest 1st digit.

So, if this picture was real -- the bikers would have had to be about a half mile away, giving you some idea of the telephoto lens required to make the shot.

Using similar analysis, we can return to the real picture that we looked at yesterday and calculate how far away the fisherman are:
The fisherman are approximately 2 suns tall, or 1 degree.  Assuming that they are adults of approximately 6 foot in height, I calculate:
$distance = \frac {h}{tan(\Theta )} = \frac {6 ft}{tan(1^{\circ})} \approx 343 ft$
To me that looks about normal for a foot ball field away, so this picture was probably taken with very little zoom.

## Tuesday, June 26, 2012

### How much daylight is left?

In yesterday's post, we learned about the angle, the foundation measurement tool for observing things in the sky. Today you'll use those tools, and your hand, to get a quick estimate of the amount of daylight left.

First, we should figure out how quickly the sun moves across the sky. Because the earth rotates once every 24 hours, the sun appears to move across the sky once every 24 hours. To be more specific:
$\frac {360^{\circ}}{24 hours} = 15^{\circ}/hour$

Since the width of your hand spans approximately 10-15 degrees (mine is relatively "fat" and covers 15), you can use it to approximate how far the sun will move in an hour. So, what I've done on many occasions is counted how many hands up from the horizon the sun is, and approximated how long till sunset. Since you have 4 fingers, they make decent 15 minute approximations.

A few notes -- first, the sun does not travel straight down, but at an angle towards its final resting place. In the Northern Hemisphere (specifically North of the Tropic of Cancer line) where I'm guessing any of you readers are from, the sun will move further north as it sets, so you might need to tilt your hand somewhat to accommodate.

Secondly, it doesn't become instantly dark once the sun sets, but there is plenty of twilight to help you. I typically figure on an additional hour of twilight before it gets too buggy or dark to want to be outside.

If we know the width of the sun, we can calculate how long sunset will take, from the moment the sun first touches the horizon till it dips behind the horizon. The "width" of the sun, and the moon for that matter, is about 1/2 degree. So once the edge of the sun touches the horizon, you'll have:$\frac {\tfrac{1}{2}^{\circ}}{ } \frac{1 hour}{15^{\circ}} \frac{60 min}{hour}= 2 min$ to enjoy the sunset before its gone.

You can also use this fact if your clock on your camera ever goes bad to tell when a picture was taken. In photoshop (who am I kidding... i just used MsPaint, I can't afford photoshop) I projected where the sun was going to travel, and more importantly how many "suns" were left in the sky. Since five suns were left, this picture was taken approximately 10 minutes before "sunset" which you can look up for any particular place and day.

## Monday, June 25, 2012

### Astronomical Angle Measurement

As many of you know, I get excited about astronomy. So this post is the first in a series devoted to understanding the math behind the measurements of our solar system.

Next clear night, go take a look at the stars, and try to identify some of the constellations. If you want some help, sky maps for June and July are available that I like. For the sake of this post, try especially to identify the big dipper -- which will be visible in the Northwest, about halfway between the horizon and the "zenith" which is the point straight above you.

If you've never done it before, hold out your hand straight in front of you, about as far as you can reach, and see how much of the big dipper you can cover up. Don't worry about looking silly or foolish -- that just means you're doing something cool. I do it all the time and no one ever mocks me about it.

Every time you do this, you should notice that the dipper looks about the same size. What you have done is effectively measured the size of the constellation, in a repeatable way. Now you can try to find Cassiopeia (the W in the North) or the teapot (part of Sagittarius, low in the southeast) and compare the size of those constellations with something more familiar.

Maybe you'll be fortunate enough to see the moon on this night, it will be about half full. Hold up your pinky and I bet you can make it disappear? Try it when it looks "HUGE" near the horizon and you'll realize how it's just an illusion. Try the same thing with the sun today or the next day and see if you can compare how big they look.

Astronomers have tools a little more sophisticated than their hands by which to measure things, but you can estimate the same sorts of measurements they make.  What you are measuring specifically is an angle. Angles can be measured from lots of places, but usually they are measured up from the horizon, in which case they are called "altitudes."  Below I am demonstrating two altitudes approximately 20 degrees and 10 degrees respectively.
 Approximately 20 degrees
 Approximately 10 degrees
The outstretched hand represents approximately 20 degrees for most people -- and the closed hand width represents about 10 degrees. You can measure how accurate your hands measure angles by trying to measure how high it is from the horizon to the zenith -- straight above you. It is exactly 90 degrees, and so you should be able to measure 4-5 hand-spreads up, or about 9 hand widths.

Some other measurements I often use are the width of my thumb or pinky:
 Approximately 2 degrees
 Approximately 1/2 degree
So? How big is the big dipper? How many degrees is it from the big dipper to the North Star?

## Saturday, June 23, 2012

### How many games of chess are possible?

I like playing chess, and so I couldn't escape the opportunity to talk at least once about some mathematics related to chess.

In my lifetime, I've played probably 500 or more games of chess, and I often wonder -- how many more games do I have to play to have played them all?  Even though I've played hundreds of games, I've never seen the same game twice, so there must be over 500.

To begin, let's try to figure out how many different opening moves are possible. First, a quick explanation of the pieces that are possible:
 From left to right: King, Queen, Bishop, kNight, Rook, pawn
Each piece in chess has different moves.

• King: The king is the most valuable piece -- the point of the game is to keep your king alive, and to attack the opponents king. The king can move one piece in any direction -- horizontally, vertically, or diagonally.
• Queen: The queen is the most powerful piece on the board. As long as there are no pieces in her way (only the kNight is allowed to jump over pieces) she can move as many spaces as she wants horizontally, vertically, or diagonally.
• Bishop: The bishop is similar to the queen, but is restricted to just diagonal motion, and as such, is a weaker piece
• kNight: (by the way, I've purposely miscapitalized the word because when annotating moves, the letter N is used to indicate kNight moves, because the K is used for kings). The night has the most interesting move -- it always moves in an L shape, two spaces vertically and one space horizontally. If desired you can also move two spaces horizontally and one space vertically. If there are pieces in the way, the kNight can jump over them.
• Rook: The rook, or castle, moves similarly to the queen, but is restricted to only horizontal or vertical motion.
• Pawn: The pawn moves only forward, usually only one space at a time. On a pawn's first move, they can move two spaces, but after that it's only forward and one space at a time. When they want to attack the opponent, they move diagonally forward one space.
There are other rules, but this brief description will do for today.  So, here's the opening position of every game:
White always moves first, and has a limited number of moves -- let's count them:
• The king (on e1) is currently trapped by its own pieces, and so cannot move yet
• The queen (on d1) is also trapped, and unable to move. Typically players will use some of their first moves to open up lanes for the queen to move.
• The bishops (on c1 and f1) are also blocked in and unable to move (fun game eh?)
• The knights (on b1 and g1) DO have options, because they are allowed to jump over those pawns that are blocking them in. They move forward to the 3rd row (rank is the typical word used) and over one space to the left or the right. So the knight on b1 can move to a3, or c3, or the knight on g1 can move to f3, or h3. Thus, 4 possible moves (Na3, Nc3, Nf3, Nh3)
• The rooks (on a1 and h1) cannot move yet, as they are blocked in.
• Each of the pawns (on a2 through h2) can move forward either one space or two spaces forward, that's 2*8 = 16 possible moves.
As a whole, white has 20 possible options.

After white moves, black is faced with the exact same set of options. To calculate how many total games are possible after just the first two moves, you need to multiply each of whites moves with each of blacks, to get 20x20 = 400 possible positions. Though I've played 400 games, I know for a fact I have not played every one of my 20 possible first moves (I stick to the immensely popular d- or e- pawn moves), so I am a long ways away from playing every possible game. And we're a long ways away from being done with the game.

Every opening move will open up new opportunities for the other pieces, and so on white's second move, he will have more twenty options available. The most freeing move is moving the e-pawn forward to e3 or e4. That move gives the bishop on f1 5 possible moves, the king 1 move, and the queen 4 possible spaces. Depending on where black moved, white would have possibly the original 20 - 1 (because the pawn no longer has the option to move forward two spaces) plus the additional 10, or 29 possible moves. (By the way, the worst move is moving an a or h pawn forward one space -- that only frees up one additional rook move, but takes away a knight and pawn move and would reduce your options to only 19).

Analysis from here becomes much complicated because you need to account for the possibility of interactions among pieces, and the fact that different moves allow different amounts of moves. I won't go into an analysis of how many moves are possible for each of the 400 different positions so far -- I'll assume there are 24 possible moves now for each player -- a rough average of the worst (19) and best (29) possible outcomes.

That leaves (20*20)*(24*24) or 230,400 possible positions after just two moves! Most of these positions will leave players with even more options available on their next moves. (By the way, 8 of those possible positions result in checkmate (game over) with black winning -- how many of you knew it was possible to lose in just two moves?) Lets assume, conservatively, that each player has 28 moves available now.  That leaves (20*20)*(24*24)*(28*28) or 180,633,600 possible positions after just three moves. To put that in perspective, if you paired everyone in the United States up (311 million/2 players per game) and assigned everyone else a different sequence of moves, you still wouldn't cover all the possible positions after just three moves.

During the majority of the game, players typically have around 20-30 possible moves to consider, although most of those can be filtered out as "blunders" like moving your piece into a space where it can be taken without consequence. Typically, I find I am considering 5-10 "real" moves per turn, and I try to look at about 3 of my opponents responses for each of those, so I'm juggling maybe 15-30 possible positions in my head.  If from this point on, we assume there are 5 moves worth considering per player, and games typically last at least 20 moves, I estimate (on the low side) that there are at least
180,633,600 *(5*5)^20 = 1.64*10^36 different games. That's around 1 billion, billion, billion, billion different games (of only 20 moves, many games go much longer!)

As of this writing, the true answer of how many different games are possible is yet to be determined.  The often quoted number, called the Shannon number, is 10^120, or 100 billion billion google games. But even that is probably an underestimate, as it cuts games off at 40 moves.
So, I guess I just need to keep playing.

By the way, many of the positions after three moves are more popular than others. The most popular positions typically have names associated with them, like the Ruy Lopez, the Sicilian, Nimzo-Indian defense, etc.  These names typically refer to one of the 180,633,600 possible positions after three or so moves, in an attempt to bring some order and description to the moves. See my chess page on my class website for links to videos others have made regarding these openings if you're interested. My favorite opening is probably the London.

## Friday, June 22, 2012

### 153

In John 21, John recalls a time when they were fishing, and Jesus asked them to throw their nets on the other side of the boat, and when they obeyed they were unable to haul the nets in. This was the second time this had happened, one of the times Jesus appeared to the disciples after his resurrection.

I always loved the little detail John throws in verse 11: there were 153 fish.

That particular number always seemed a little odd to me. (yes, pun intended... you can stop laughing and start reading again anytime...)

Besides being odd, 153 has a lot of other interesting properties, and the one I want to describe is a remarkable fact that 1^3 + 5^3 + 3^3 = 1 + 125 + 27 = 153.  How remarkable is that?! When you cube the digits of a number, and add them up, you get the same number back! Now, does that always happen? That is, when you do this little operation--I'm going to call it a "cubic sum"-- do you always get your number back?

That's easy enough to check -- lets find the "cubic sum" of 972.
9^3 = 729
7^3 = 343
+ 2^3 = 8
1080

Nope. In fact, 153 is one of very few numbers that equals the "cubic sum" of its digits. Wikipedia calls that a "Narcissistic Number" but I call it cool. The only other numbers with that property (for cubes at least) are 0, 1, 307, 371, and 407.  Lets check 371 a second:
3^3 = 27
7^3 = 343
+ 1^3 = 1
371

Now if that was all that I knew on the topic, that would be cool enough, and your day would be better for having read it, but there's one more fact about 153 and these "cubic sums" that needs to be told.  Lets go back to our first effort of 972.  When we found its cubic sum, it gave us 1080, and we were bummed that we didn't get what we were hoping for. Lets find 1080's cubic sum:
1^3 = 1
0^3 = 0
8^3 = 512
+  0^3 = 0
513
So what?  Well, let's keep going... find the cubic sum of 513 and you get ... drum roll please ... 153!  I have a feeling that's why these numbers are called narcissistic -- because this process often leads back to themselves, even when starting with other numbers!

Now, not every number chosen at random leads to 153, here's a chain I found that leads to 370, having started with the number 259.

259 --> 862 --> 736 --> 586 --> 853 --> 664 --> 496 --> 1009 --> 730 --> 370 --> 370 ...

524 --> 197 --> 1073 --> 371.

In fact, most of the numbers I initially tried went to 371 -- perhaps its even more narcissistic than 153?    Actually, I was just unlucky in the numbers I tried -- because every multiple of 3 will cycle around to 153. I had forgotten that fact... Try 51 = 17*3  and you'll find:
51 --> 126 --> 225 --> 141 --> 66 --> 432
--> 99 --> 1458 --> 702 --> 351 --> 153

I know its even possible to create a loop of numbers that just cycle among themselves, like a clique one such clique of numbers is 133 --> 55 --> 250 --> 133 ...

## Thursday, June 21, 2012

### Work-Cost of Activities

Inspired by a quote on Dave Ramsey, and a comment by my wife, I thought I would add a little bit to the post on the cost of gas to get somewhere and the money it costs to accomplish various tasks.

So you'd need to make $10 working in GR to make the trip break even. That's why I didn't like summers working at FCS where I'd be scheduled for 2-3 hours and make$20. Half of my paycheck would be going toward gas!
So, here's a handful of calculations, based on an hourly rate of $10. Do I really make$10 per hour? I suppose that depends on how many hours you divide my salary into, but I'm not going to reveal that all, and $10 is easy to divide, so I'll stick with it. How much work do I need to do to take my wife out to the movies? Assuming the new$10 price per ticket, $9 for popcorn,$15 for babysitting, and the gas out there and back (9.60) we're looking at basically 55 dollars: $Movies: \frac{\50}{ } \frac{hour}{\10} = 5 hours$ To fill up the gas tank? $Fill Up Tank: \frac{\60}{ } \frac{hour}{\10} = 6 hour$ To come into work? $To Come To Work: \frac{\9.60}{ } \frac{hour}{\10} \frac{60min}{hour}= 58 min$ To buy that iPad you've always wanted? $iPad: \frac{\499}{ } \frac{hour}{\10} = 49.9 hours$ How much work for that Starbucks Mocha Frappucino: $Mocha: \frac{\4.25}{ } \frac{hour}{\10} \frac{60 min}{1 hour} = 25.5 min$ A different way to think about it is to figure out how long it takes to earn a dollar, and than start referring to dollar bills in those terms instead. For me that is: $\frac{\1}{ } \frac{hour}{\10} \frac{60 min}{hour} = 6 min$ So now I think: wow! Gas costs almost 24 minutes of work a gallon! ## Wednesday, June 20, 2012 ### How many digits of pi does your calculator know? Most calculators will display only a certain number of digits. The calculator I normally use (a TI-84) only displays 10 digits, so when I ask it what pi is, it spits out the familiar 3 and nine decimal places: 3.141592654 Calculators store more digits in their memory than they can display however, and through 'shifty' 'trickery' that only Mathmagicians know, you can squeeze out a few more digits. I always feel like I'm torturing the calculator when I do this, like taking years away from Wesley's life in Princess bride, but here goes anyway. Simply subtract the 3, and recalculate: I get .141592653 The first thing I noticed is that I got most of the same digits, except the 4 has become a 3, and now a 6 shows up. That 6 (or maybe its a 5... we'll see) is the 10th digit of pi, and because it was 5 or above, the ninth digit 3 had to be rounded up to a 4 in the first example. I think the calculator has more he isn't telling us though -- lets suck more out it! If I multiply the current number by 10 and then subtract the part before the decimal place you can get another digit: 1.415926536 - 1 becomes .4159265359 and I see that the 10th digit of pi is actually a 5, not six, and the 11th digit is a 9 (or is it an 8?). This process can be repeated over and over again to squeeze more digits out, but only to a point. For the TI-84, that is only 13 digits long, yielding a stored value of pi which is: 3.1415926535898 or 3.1415926535897 As shown above, we can't know for sure if the 13th digit has been rounded or not because we don't know what the 14th digit is, so it's either a 7 or an 8. Wolfram Alpha to the rescue reveals that it is indeed a 7, and lists many more digits for you, if you'd like. How many digits are there? As of October 17, 2011, over 10 trillion were known, which is enough to give everyone on earth a digit, and still have plenty left over. Computers are constantly working at finding more digits, using even shiftier trickery than this. What's really creepy, is that your phone number shows up in pi! And so is mine, and my neighbors. It must be stalking us! ## Tuesday, June 19, 2012 ### Converting a decimal to a fraction If you haven't done so yet, read yesterday's post on decimal representations of fractions, as some of the concepts discussed today will reference those ideas. When you calculate a number on a calculator, sometimes you are left with a decimal number, such as 3.125. Sometimes you round that part off, but other times it is useful to rewrite that decimal part as a fraction instead. Here are the methods I know for how to do that: 1. Duh--It's one that you memorized. 2. Use the calculator 3. Write it over a factor of ten 4. Write it over a bunch of nines 5. Solve an equation 1. After years of working with numbers, I have a huge number of fraction/decimal number memorized, not because I made flash cards of them and studied them like the times tables, but because familiarity comes with frequency. How many songs can you recite the words to? I immediately recognized the .125 as one of the 8ths, and if you know it, then you're done! I always list the simplest approaches first, because sometimes my students forget that they know things, and try to use the more difficult approaches on questions they shouldn't. 2. Many calculators have a fraction button, or have conversion tools built into them. I love the TI-30's implementation of fractions -- but TI 82, 83, and 84 decided to hide this functionality in the Math menu. You can still find it easily -- type the decimal number, then press math, and the first option converts it to a fraction. This is usually the second thing I teach my algebra students when they get their new TI84's each year. 3. If the decimal version is a non-repeating fraction, like .125, but one you don't have memorized, you can easily write it as a fraction. Since the decimal notation has 3 digits in it, write those three digits in the numerator, and write it over 10^3, or a 1 with three zeros. That will produce a fraction which may or may not be reducible. Any power of ten is only divisible by 2's or 5's, and when you reduce, simply try to divide by 2 or 5 as often as possible. $0.125 = \frac{125}{1000}\frac{\div 5}{\div 5} = \frac{25}{200}\frac{\div 5}{\div 5} = \frac{5}{40}\frac{\div 5}{\div 5}=\frac{1}{8}$ 4. If the decimal version is of the repeating variety, then as we learned in yesterday's post, it can be written over a bunch of nines. Like in #3, figure out how many digits repeat in the pattern, and write it over the same number of repeated nines. As an example, consider the decimal $\inline 0.027027027027... = 0.\overline{027}$ which you might recognize from yesterday. It has three digits repeating, and so I'll put 027 in the numerator and 999 in the denominator, and reduce it if possible. Here the reducing might be more difficult as it will involve all sorts of possible factors, except 2's and 5's. $0.027027027027... = 0.\overline{027} = \frac{27}{999}\frac{\div 3}{\div 3} = \frac{9}{333}\frac{\div 3}{\div 3} = \frac{3}{111}\frac{\div 3}{\div 3}=\frac{1}{37}$ 5. The hardest type of fraction to rationalize (i.e. write as a fraction) is one that is delayed repeating fraction, such as 0.52727272727... The process I've learned for this type of fraction is to start by writing it as an equation equal to x, multiply by powers of ten till you have only repeated fractions, subtract, and divide. I will simply show the process below, and if you want more explanation, examples, or a proof for why it works, I'll consider expanding it in a future post. First write it as an equation: x = 0.52727272727... Next multiply both sides by 10 enough times to remove one cycle of repeating digits. In this particular case, the repeating part is 2 digits long, so I'm going to multiply by 100 to shift 2 decimal places. That yields the equation: 100x = 52.72727272727... The next step is to subtract these two equations and divide. Notice how the shifted equation and the original have the repeated parts aligned perfectly, so they subtract themselves away. \begin{align*} 100x &= 52.7272727272... \\ - x &= 00.5272727272... \\ 99x &= 52.200000000... \\ 990x &= 522 \\ x &= \frac{522}{990} \left (\frac{\div 18}{\div 18} \right ) =\frac{29}{55} \end{align*} Not the most intuitive method in the world, but it's a nice one to have in your bag of tricks. ## Monday, June 18, 2012 ### Cool Decimal Representations of Fractions I believe that God created the world in such a way that we can discover it, by creating things with pattern and order. Over the years, I've learned a lot of cool patterns -- and quite a few of them in the realm of fractions. I wanted to share just a few of them with you. The first are the 'sevenths', shown below: $\begin{matrix} \tfrac{1}{7} = 0.{\color{Blue} 142857}14285714... \\ \tfrac{2}{7} = 0.2857{\color{Blue} 142857}1428... \\ \tfrac{3}{7} = 0.42857{\color{Blue} 142857}142... \\ \tfrac{4}{7} = 0.57{\color{Blue} 142857}142857... \\ \tfrac{5}{7} = 0.7{\color{Blue} 142857}1428571... \\ \tfrac{6}{7} = 0.857{\color{Blue} 142857}14285... \\ \end{matrix}$ I remember first being amazed when showed me that 1/7 was what's called a repeating decimal -- that is, the digits 142857 repeat over and over again in that order forever, which is often written as $\inline 0.\overline{142857}$. I remember trying hard to remember that sequence of six digits. As amazing as that was, I was blown away when I looked at the other seventh's fractions, and notice that same string of digits show up in each of their representations! The only difference between them is where in the sequence they start! Here's a few other sequences that are easy to remember: $\inline \begin{matrix} \tfrac{1}{9}= 0.111111111... \\ \tfrac{2}{9}= 0.222222222... \\ \tfrac{3}{9}= 0.333333333... \\ \tfrac{4}{9}= 0.444444444... \\ \tfrac{5}{9}= 0.555555555.. \\ \tfrac{6}{9}= 0.666666666... \\ \tfrac{7}{9}= 0.777777777... \\ \tfrac{8}{9}= 0.888888888... \\ \end{matrix}$ Each of the 'ninths' is simply the numerator repeated forever. In fact, this is just the simplest case of a much broader pattern -- that of any denominator of all 9's. Any fraction with all 9's in the denominator will have a representation that simply repeats the numerator over and over again. (A small catch: you may have to add a few zeroes to pad the numerator so it has as many digits as the number of nines in the denominator). Here's a few examples: $\frac{{\color{Blue} 237}}{999} = 0.{\color{Blue} 237}237237237237... \\$ $\frac{{\color{Blue} 8514}}{99999} = 0.{\color{Blue} 08514}08514085$ If I haven't bored you enough, I'll show a few others that I've learned $\begin{matrix} \tfrac{1}{11} = 0.09090909090909... \\ \tfrac{2}{11} = 0.18181818181818... \\ \tfrac{3}{11} = 0.27272727272727... \\ \tfrac{4}{11} = 0.36363636363636... \\ \tfrac{5}{11} = 0.45454545454545... \\ \tfrac{6}{11} = 0.54545454545454... \\ \tfrac{7}{11} = 0.63636363636363... \\ \tfrac{8}{11} = 0.72727272727272... \\ \tfrac{9}{11} = 0.81818181818181... \\ \tfrac{10}{11} = 0.90909090909090... \\ \end{matrix}$ The pattern here may not be quite as obvious, but 'elevenths' have a nice trick to them too. Take the numerator, and multiply it by nine, and that's the number that gets repeated. Finally, a pair of fractions that might have been twins separated at birth: $\begin{matrix} \frac{1}{{\color{Red} 27}} = 0.{\color{Blue} 037}037037037037037... \\ \\ \frac{1}{{\color{Blue} 37}} = 0.{\color{Red} 027}027027027027027... \\ \end{matrix}$ It's ok if you just said "Awesome" or something similar. As I said above - I think God designed these patterns to help us learn and understand math better, and this lost pair is not the only pair that exists, just the first one I was ever shown. I won't list anymore, but I will point out that 27*37 = 999, and that is perhaps enough of a hint to get you on a path to discovering your own. If you find some others, feel free to post them in the comments below! ## Saturday, June 16, 2012 ### ... is 3 Yesterday's puzzle was "What is the last digit of 3^2153?" Today I'll show a technique for answering the question. The simplest approach is to simply get out the calculator and type in 3^2153 and see what the last digit is. Unfortunately, my calculator isn't powerful enough to answer that question, giving me the screenshot off to the side. The reason for this is that my calculator, (which is the same TI 83 or TI 84 that most of my students have access to) is only capable of working with numbers that have less than 100 digits. Worse, it will only display approximately 10 of those digits (though there are tricks you can use to display a few more!). Just how many digits does this number have anyway? When asking that question, the tool to use is the log function. Typing log(3^2153) would hypothetically give you the answer (you'll have to round up to the nearest whole number) but even that is impossibly big for the calculator. I can use a property of logarithms to rewrite it as 2153* log (3) which is calculable: $log_{10}(3^{2153})=2153 \cdot log_{10}(3) \approx 1027.24$. Practically, this means that the number 3*2153 has 1028 digits in it, and I'm interested in just the last one. Interestingly, this subquestion of how many digits a huge number has been able to be answered for far longer than we have had calculators, by using log tables, but that's an entirely different story. Back to the question at hand. If we aren't going to just cheat and look at all the digits of a supercalculator, how can we answer the question? Someone always suggests just multiplying it out by hand. As ridiculous as that would be (simply multiplying the last step would probably take 5 minutes, and that's after getting there in the first place!) the approach actually reveals something very important that can apply, so lets try it. The numbers are already getting pretty big, but the final digit of each is all that's important to this discussion, and if you pay attention, you'll notice that there is a definite pattern to those digits. The last digits are 3, 9, 77, 1, 3, 9, 7, 1, and so on. This pattern will continue up long past the ability of our calculators (or pencils and papers) to handle, and all that is necessary to answer the question is where in this sequence of numbers 3^2153 falls. Notice, this cycle of numbers repeats every four digits. Specifically, every multiple of four (3^4, 3^8, and 3^12) ends in a 1. The closest power of four to 2153 is 2152, and so: And so we have it -- 3^2153 ends in a 3. I encourage you to try to answer the rest of these questions, which can all be answered using similar techniques: Last digit of 2^2153 Last digit of 4^2153 Last digit of 5^2153 Last digit of 6^2153 Last digit of 7^2153 Last digit of 8^2153 Last digit of 9^2153 ## Friday, June 15, 2012 ### The Last Digit of 3^2153... Quick puzzler for Friday: What's the last digit (the ones digit that is) of 3^2153? This question, or some variant of it, has been a common extra credit question in my classes every year. I'll post the answer tomorrow. I tried to make sure it was a question you couldn't just google. Unfortunately, if you really want to cheat, you can still Wolfram Alpha it. Mathematically speaking, Wolfram Alpha kicks Google's butt, and is a resource you ought to check out frequently. Below are some pretty amazing things Wolfram Alpha can do for you: How many meters is 52 inches? ## Thursday, June 14, 2012 ### What's the real price of fresh baked bread? My wife loves making bread with the little bread maker we own. The other day she pondered making bread so we wouldn't have to buy loaves anymore, and so naturally, I thought I needed to use this as a math post, and see if it's economically sound. For our two pound recipe, the ingredients include: Water: 1.25 cups + 2 TBL Oil: 2 TBL Sugar: .25 Cup Salt: 2 tsp Dry Milk: 2 TBL White Flour: 3.5 cups Wheat Flour: 0.5 cups Active Dry Yeast: 2.25 tsp Except water, the items can be bought at the store "in bulk" and so I researched the prices online: Water: ?? Assumed to be free - perhaps a later post will calculate that? Oil3.19 / 48 oz bottle
Sugar: $2.54 / 5 lb bag Salt:$0.49 / 26 oz
Dry Milk $6.49 / 25.6 oz box White Flour$1.49 / 5Lb bag
Wheat Flour $3.49 / 5Lb bag Active Dry Yeast: 4.99 / 4oz jar Converting each of the following into prices per recipe will prove to be a bit challenging, as many items were sold by weight but used by volume. There is no single direct conversion between these two types of measurements, because the same volume of flour will weigh less than the same volume of water, because it has a lighter density. I was tempted to use fluid ounces as a common unit for all items, but because of their different densities, this would incorrect. In fact, flour is so much lighter than water that it would be nearly nearly 2 times more costly to naively assume that (8 / 5.5 to be more precise). So I will demonstrate the calculation for flour: $WhiteFlour: \frac{3.5cup}{recipe} \frac{5.5 oz}{cup} \frac {1 lb}{16 oz} \frac {\1.49} {5 lb} = \.36$ Using similar logic for the rest of the items, I got the following prices*: $Oil: \frac{2 Tbls}{recipe} \frac{.25 cup}{4 Tbls} \frac { 8oz}{cup} \frac {\3.19} {48 oz} = \0.06$ $Sugar: \frac{\tfrac{1}{4}cup}{recipe} \frac{7 oz}{cup} \frac {1 lb}{16 oz} \frac {\2.54} {5 lb} = \.06$ $Salt: \frac{2 tsp}{recipe} \frac{.25 cup}{12 tsp} \frac { 8oz}{cup} \frac {\.49} {26 oz} = \0.01$ $DryMilk: \frac{2 Tbls}{recipe} \frac{.25 cup}{4 Tbls} \frac { 8oz}{cup} \frac {\6.49} {25.6 oz} = \0.25$ $WheatFlour: \frac{\tfrac{1}{2}cup}{recipe} \frac{5.5 oz}{cup} \frac {1 lb}{16 oz} \frac {\3.49} {5 lb} = \.12$ $Yeast: \frac{2.25 tsp}{recipe} \frac{.25 cup}{12 tsp} \frac { 8oz}{cup} \frac {\4.99} {4 oz} = \0.47$ This brings the grand total for a recipe of bread to:$1.33.  Unfortunately this does not yield the same usefulness in terms of number of servings, nor in shelf life, as a typical $1.50 loaf of bread, nor does it include the cost of water (probably negligible?) nor the price of electricity to run the bread maker (probably not negligible). But it sure tastes good, and makes the house smell a lot prettier than a bag of bread. And now you know how to make the same sort of calculations for your meals. Anyone want to calculate the true price of lasagna now? Before you go out and buy your own bread maker, remember that ours was a gift so I didn't include that price in the mix, but if you buy your own, you'll have to estimate how many times you will bake bread and figure in that cost too. At$100, if you bake 400 loaves you'll add another \$0.25 per loaf to the cost.