Etruscan Numerals

Etruscan Numerals


The Etruscan numerals were used by the ancient Etruscans. The system was adapted from the Greek Attic numerals and formed the inspiration for the later Roman numerals.There is very little surviving evidence of these numerals. Examples are known of the symbols for larger numbers, but it is unknown which symbol represents which number. The general argueement among the Etruscologists is :

Etruscan	Decimal
θu	1
zal	2
ci	3
śa	4
maχ	5
huθ	6
semφ	7
*cezp	8
nurφ	9
śar	10
*θuśar	11
*zalśar	12
*ciśar	13
huθzar	14
*maχśar	15
*śaśar	16
ciem zaθrum	17
eslem zaθrum	18
θunem zaθrum	19
zaθrum	20
cealχ	30
*huθalχ	40
muvalχ	50
śealχ	60
semφalχ	70
cezpalχ	80
*nurφalχ	90

Difficult Squaring

Square 2 Digit Number: UP-DOWN Method

Square a 2 Digit Number, for this example 37:

• Look for the nearest 10 boundary
• In this case up 3 from 37 to 40.
• Since you went UP 3 to 40 go DOWN 3 from 37 to 34.
• Now mentally multiply 34x40
• The way I do it is 34x10=340;
• Double it mentally to 680
• Double it again mentally to 1360
• This 1360 is the FIRST interim answer.
• 37 is "3" away from the 10 boundary 40.
• Square this "3" distance from 10 boundary.
• 3x3=9 which is the SECOND interim answer.
• Add the two interim answers to get the final answer.
• Answer: 1360 + 9 = 1369

With practice this can easily be done in your head.

Fun With Maths - Binary Card Trick

Binary Card Trick

You put a deck of cards in your pocket, and invite anyone in the audience to call out a number between 1 and 15. Then you reach into your pocket, you take out a set of cards whose sum is the number that was called! How can you perform this magic trick? The Math Behind the Fact: This mathematical magic trick can be found in the reference and is based on the properties of binary numbers. Every number between 1 and 15 has a unique representation as a sum of some collection of the numbers 1, 2, 4, and 8. (To see which collection, just take the given number and successively subtract the largest number of 1, 2, 4, and 8 that is less than the given number. That number is part of your collection. The subtraction yields a new number; now repeat the process with this number, over and over, until you get 0.) The collection of numbers you obtain reveals the binary decomposition of the given number into sums of powers of two (in contrast to the usual representation of a number into sums of powers of ten). Now before the trick starts, pick an Ace, 2, 4, and 8 and put them on top of the deck, and then put the deck in your pocket. Then when a number between 1 and 15 is called out, take the binary decomposition of the number, and use that to determine which of the first four cards you will pull out. No one needs to know that you never need to use the other cards!

Fun With Maths - Perfect Shuffles

Perfect Shuffles

Figure 1 We know from the Fun Fact Seven Shuffles that 7 random riffle shuffles are enough to make almost every configuration equally likely in a deck of 52 cards. But what happens if you always use perfect shuffles, in which you cut the cards exactly in half and perfectly interlace the cards? Of course, this kind of shuffle has no randomness. What happens if you do perfect shuffles over and over again? There are 2 kinds of perfect shuffles: The out-shuffle is one in which the top card stays on top. The in-shuffle is one in which the top card moves to the second position of the deck. Figure 1 shows an out-shuffle. Surprise: 8 perfect out-shuffles will restore the deck to its original order! And, in fact, there is a nice magic trick that uses out and in shuffles to move the top card to any position you desire! Say you want the top card (position 0) to go to position N. Write N in base 2, and read the 0's and 1's from left to right. Perform an out-shuffle for a 0 and and in-shuffle for a 1. Voila! You will now have the top card at position N. (See the reference.) Example. Since 6 is 110 in binary notation, then the sequence IN-IN-OUT will move the top card (position 0) to position 6 (the seventh card). Presentation Suggestions: Have students go home and determine how many in-shuffles it takes to restore the deck to its original order. (Answer: 52.) You can also have students investigate decks of smaller sizes. As a project, you might even tell them part of the binary card trick and see if they can figure out the rest: whether 0 or 1 stands for an in/out shuffle, and whether to read the digits from left to right or vice versa. The Math Behind the Fact: This fact may come as somewhat of a surprise, because there are 52! possible deck configurations, and since there is no randomness, after 52! out-shuffles, we must hit some configuration at least twice (and then cycle from there). But 8 is so much smaller than (52!). See Making History By Card Shuffling. Group theory concerns itself with understanding sets and properties preserved by operations on those sets. For instance, the set of all configurations of a deck of 52 cards forms a group, and a shuffle is an operation on that group.

Fun With Maths - Red-Black Pairs Card Trick

Red-Black Pairs Card Trick

Here's a terrific mathematical card trick that will impress your friends. When you do this trick, the effect of the card trick will look like this: You have a deck of cards, and you ask for a volunteer who knows how to do a riffle shuffle. You then cut the deck and then give the volunteer the halves of the deck and ask him to do one riffle shuffle and return the deck to you. Now say "There's no way I could know anything about the deck right now, right? Well, I was born with the amazing ability to feel the redness and blackness of cards with my fingertips. However, my talent is not that refined. I can only feel red and black cards in pairs." As you say this, put the deck of cards behind your back (so that you cannot see them) and then, at regular intervals, you fish around in the deck and pull out pairs of cards and show them to the audience. These pairs will all have exactly one black and one red card! Presentation Suggestions: Before performing the trick, order the deck alternating colors, all the way through, red-black-red-black-... etc. (When you flash the deck before their eyes, they really won't notice this pattern if you do it quickly.) After this, there is really only one thing you need to remember to ensure that the trick works: you must cut the deck (not the spectator), and you must do it in such a way that the bottom of each half of the deck is a different color. Then, no matter how the spectator riffle shuffles the deck, the cards will always drop in red-black or black-red pairs. See below for explanation. Then, all you have to do after the deck is returned and you put it behind your back is to pull out the top 2 cards. It will be either red-black or black-red! Then pull out the next 2 cards, which again will be red-black or black-red. You can continue in this fashion to the end of the deck, if you like! Of course, you should make it look as if you are trying really hard to find the cards (even though what you are really doing is very easy). Spectators will wonder if you are pulling one card off the top and one card off the bottom; but you can pull the deck out and show them that this is not the case. The Math Behind the Fact: The reason the trick works at the point of the riffle shuffle is both simple and stunning: if you cut the deck so that the cards at the bottom of each half are different colors, then the first card that gets "dropped" in the shuffle will be a different color then the second card that gets dropped, no matter which half of the deck they come from. As an example, if the first card that gets dropped is black, then after that both halves will have red cards at the bottom, so no matter which card falls next it will be red! After this, both halves again have different colored cards at bottom and we are back to the situation at start. So all the cards will fall off in either red-black or black-red pairs. This amazing fact is a special case of something known as the Gilbreath principle. The message of this trick is that one shuffle is not enough to randomize a deck of cards-- you really can know something about the deck after one shuffle... but only if you stack the deck in a particular way first!

Fun With Math - Magic 1089

Magic 1089

Here's a cool mathematical magic trick. Write down a three-digit number whose digits are decreasing. Then reverse the digits to create a new number, and subtract this number from the original number. With the resulting number, add it to the reverse of itself. The number you will get is 1089! For example, if you start with 532 (three digits, decreasing order), then the reverse is 235. Subtract 532-235 to get 297. Now add 297 and its reverse 792, and you will get 1089! Presentation Suggestions: You might ask your students to see if they can explain this magic trick using a little algebra. The Math Behind the Fact: If we let a, b, c denote the three digits of the original number, then the three-digit number is 100a+10b+c. The reverse is 100c+10b+a. Subtract: (100a+10b+c)-(100c+10b+a) to get 99(a-c). Since the digits were decreasing, (a-c) is at least 2 and no greater than 9, so the result must be one of 198, 297, 396, 495, 594, 693, 792, or 891. When you add any one of those numbers to the reverse of itself, you get 1089!

Fun With Maths - Birthday Problem

Birthday Problem

How many people do you need in a group to ensure at least a 50 percent probability that 2 people in the group share a birthday? Let's take a show of hands. How many people think 30 people is enough? 60? 90? 180? 360? Surprisingly, the answer is only 23 people to have at least a 50 percent chance of a match. This goes up to 70 percent for 30 people, 90 percent for 41 people, 95 percent for 47 people. With 57 people there is better than a 99 percent chance of a birthday match! Presentation Suggestions: If you have a large class, it is fun to try to take a poll of birthdays: have people call out their birthdays. But of course, whether or not you have a match proves nothing... The Math Behind the Fact: Most people find this result surprising because they are tempted to calculate the probability of a birthday match with one particular person. But the calculation should be done over all pairs of people. Here is a trick that makes the calculation easier. To calculate the probability of a match, calculate the probability of no match and subtract from 1. But the probability of no match among n people is just (365/365)(364/365)(363/365)(362/365)...((366-n)/365), where the k-th term in the product arises from considering the probability that the k-th person in the group doesn't have a birthday match with the (k-1) people before her. If you want to do this calculation quickly, you can use an approximation: note that for i much smaller than 365, the term (1-i/365) can be approximated by EXP(-i/365). Hence, for n much smaller than 365, the probability of no match is close to EXP( - SUMi=1 to (n-1) i/365) = EXP( - n(n-1)/(2*365)). When n=23, this evaluates to 0.499998 for the probability of no match. The probability of at least one match is thus 1 minus this quantity. For still more fun, if you know some probability: to find the probability that in a given set of n people there are exactly M matches, you can use a Poisson approximation. The Poisson distribution is usually used to model a random variable that counts a number of "rare events", each independent and identically distributed and with average frequency lambda. Here, the probability of a match in a given pair is 1/365. The matches can be considered to be approximately independent. The frequency lambda is the product of the number of pairs times the probability of a match in a pair: (n choose 2)/365. Then the approximate probability that there are exactly M matches is: (lambda)M * EXP(-lambda) / M! which gives the same formula as above when M=0 and n=-365

Fun With Maths - Mind Reading Number Trick

Mind-Reading Number Trick

Think of a number, any positive integer (but keep it small so you can do computations in your head). 1. Square it. 2. Add the result to your original number. 3. Divide by your original number. 4. Add, oh, how about 17. 5. Subtract your original number. 6. Divide by 6. The number you are thinking of now is 3! How did I do this? Presentation Suggestions: Ham it up with magician's patter. Step 4 could be anything you want---someone's age, or their favorite number--- just ask the crowd for suggestions. (This will change the final outcome of Step 6, but see below for how.) The Math Behind the Fact: Clearly no matter what you start with, the answer should come out the same (zero wasn't allowed because of Step 3). We can see why this trick works by using a little bit of high school algebra! If you follow the instructions starting with the variable X instead of an actual number, you will see that X is eliminated by Step 5. Using this idea, you can make up your own mental math trick right on the spot! (Just don't do anything too obvious, like tell people to add 5, subtract their original number, and say "the number you are thinking of is 5".)