Free Facebook Likes XMinds Education: 3 Algebra Tricks That Will Even Fool Yourself !!!

3 Algebra Tricks That Will Even Fool Yourself !!!



 Trick 1.The Brain Control Experiment


This is how the trick will actually look to your audience.

Here is an experiment that looks at the power of the human mind to control distant events. Try it with your own deck of cards first before you look at the explanation - you will even fool yourself! You need one deck of playing cards and one brain.

This is where the actual trick itself is explained.
Give the cards a good shuffle. Spread them the table faces down. Now think of the color RED and select any 13 cards. Shuffle again if you wish, then think of the color BLACK and select another 13 cards at random. Take both sets of selected cards and turn them face up in a single pile, keeping the rest of the cards in a face down spread.
Now the remote control starts. Concentrate. You are going to separate the cards you selected (and that are now in your face-up pile) into two piles, a RED pile and a BLACK pile, in the following way.
Go through your face-up cards one at a time. If the card is RED, put it in the RED pile. For each RED card you put in your RED pile think RED and select a random card from the face-down cards on the table without looking at it. Put this random card in a pile face-down in front of your RED pile.
Similarly, if you come to a card which is BLACK, put it face-up on your BLACK pile. Think BLACK and select a random face-down card. Put this face-down card in a pile in front of your BLACK pile. Go through this procedure until you run out of face-up cards.
You now have the following: a RED pile and in front of that a pile containing the same number of face-down cards you selected while thinking RED. You also have a BLACK pile in front of which is a pile of random cards you selected while thinking BLACK.
Interestingly your thoughts have influenced you choice of random cards! Don’t believe me? Look at the pile of random cards you chose and put in front of your RED pile. Count the number of RED cards in this pile. Remember that number.
Now look at the random cards in front of your BLACK pile, and count the number of BLACK cards you selected. They are the same! Rather than two random numbers of RED and BLACK, you selected the same number of RED and BLACK cards totally at random because your brain was controlled!
This is a final proof that your sub-conscious mind can make you choose random cards to balance those numbers!  So you have done our remote control brain experiment. Is mind control a reality?
Of course it’s not mind control. Its mathematics, but you knew that didn’t you? But how does this mind-reading miracle work? Well, it’s all down to Abracadabra algebra...


Mathematical Principles Behind The Trick.

 Let’s call the number of cards you dealt R1 for the RED pile and B1 for the BLACK pile. The two other piles in front of these contain a random mixture of red and black, so let’s say that the pile in front of R1 contains R2 reds and B2 blacks, and the pile in front of B1 contains R3 reds and B3 blacks. See the picture.
In the first part of the experiment, what you are actually doing is dividing the pack in half. You dealt two piles of 13 cards each, 2 altogether, half a pack of 2 cards.
Now we also know that in a full pack of cards half are red, and the other half are black, so all your red piles must add up to 2cards and similarly for the blacks. Writing this mathematically we get:
Equation (1): R1 + R2 + R3 = 2Equation (2): B1 + B2 + B3 = 2
We also know the number of cards in the RED pile R1 is the same as the number of face down cards placed in front of it (made up of R2 red cards and B2 black cards). So together R2 and B2 must have the same number of cards as R1. Similar reasoning holds for the cards in front of the BLACK pile. So:
Equation (3): R1 = R2 + B2 Equation (4): B1 = R3 + B3
So if we substitute Equation (3) in Equation (1), we can eliminate R1 to get:
Equation (): (R2 + B2) + R2 + R3 = 2
Similarly if we substitute Equation (4) in Equation (2), we can eliminate B1 to get:
Equation (): (R3 + B3) + B2 + B3 = 2
As Equations () and () both add up to 2, we get:
(2xR2) + B2 + R3 = 2 = R3 + (2xB3) + B2
We can subtract R3 and B2 from each side. That leaves:
2xR2 = 2xB3
Finally, we can divide both sides by 2 giving:
R2 = B3
So what the maths shows is that the number of RED cards (R2) in front of the RED pile must always be equal to the number of BLACK cards (B3) in front of the BLACK (B1) pile. That is how the magic works. Maths.
In the card trick, we don’t know how many cards will be in the piles. Yet by using letters to represent the numbers (these letters are called variables because, well, they can be variable) we can work through the maths, simplifying the expressions to find the answer. The algebra proves the numbers R2 and B3 will always be the same: so long as you follow the instructions for the trick it will always work. The rest of the trick is just as we have done before, presentational flim-flam ... but don’t tell anyone how it works!

Trick 2. The Number of Matches Prediction


This is how the trick will actually look to your audience.
Let’s do a trick with coins or matches or even cards if you don’t have anything else. In this effect, even with your back turned, you can reveal the exact number of things that a spectator chooses freely from a pile in a stunning three phase prediction.

This is where the actual trick itself is explained.
Build this up. You have a stunning three stage prediction, each stage seemingly more impossible than the last. Take time to build up, chat a bit to misdirect the audience so they forget the exact details of the earlier part of the prediction, and, if you act amazed at each step they will react the same. This trick is known in some magical circles as The Trick that Fooled Einstein. If you present it well, it will fool anyone!

Let’s assume you have a pile of matches. Ask the spectator to grab a small random number of matches from the pile and hide them, while your back is turned. They don’t know how many and so, of course, you can’t know either. You then turn round and take a bunch of matches, secretly making sure you take more than the spectator. You told them to take a small number so you take a large number.

You then count your matches and work out what your target number should be. Choose a small number to be your difference number to take away from your total match count. Say for example you have taken 17 matches. You choose say 3 to take away from this total, to reach your target number which in this instance will be 14.

You would turn to your spectator and announce, or write down on a bit of paper, “I’ve got as many matches as you, then an extra 3 more and then just enough left to make your number up to 14.”With the prediction made, it’s time for the spectator to count their matches. You then count yours out and your prediction is proven true. Say they have seven matches. From your pile, you count out seven matches and put them aside. Part one of your prediction is true: you have as many matches as them.

You then count off your difference number of matches off to the side from what’s left of your pile (in this example, it would be three). That’s Part two of the prediction correct - you have three more matches than they had.

In your pile you now have 17-7-3 =7 matches left. Count your remaining seven matches onto their pile of seven matches, starting counting, “8, 9, 10 …” up to the total of your stated target number which was 14. Part three of your amazing prediction is correct.

Mathematical Principles Behind The Trick.
The trick here is the three part prediction - it camouflages the simple maths going on. Say you choose X matches, the spectator chooses Y matches and you decide the difference number should be Z.

Part One of your prediction is that you have more matches than them, i.e. X>Y. This is true because they took a small number of matches and you took a large number.

Part Two was that you had a particular number Z more matches than them. If you chose your difference number sensibly in the first place, you should have Z to spare, leaving you with X-Y-Z matches.

Part Three is the clever bit. You now add the matches you have left to their Y matches. So you count Y+X-Y-Z = X-Z – exactly the target number
you promised!

Algebra wins again, hidden in your prediction. It’s really quite simple and will always be true.

Trick 3 Algebra and Addition The Amazing Coincidence

This is how the trick will actually look to your audience.
You claim that some things are simply meant to be, and set about proving it. You need your friend to help you create a ‘random’ target number. But in fact, you can predict what the target number will be, whatever numbers your friend gives you. The total is... 1089.

This is where the actual trick itself is explained.
You get your friend to secretly write down any three-digit number. To make it ‘more impossible’, you say the digits must all be different and the biggest digit must be at the front. You say it’s still too easy. Now you want them to jumble things up a bit. You get them to reverse their selected number and write it underneath the first number. They should then subtract this lower number from their first number. Finally, to make it even harder still, have them write their answer backwards and add this new reversed number to their answer. After all of this, you now have a random number even your friend couldn’t have predicted in advance. But you can, as you reveal your previous prediction of this amazing coincidence!

As you will see, the maths shows that the trick will always work if the spectator does what’s expected of them and follows the rules. Mathematically we call the things the spectator must do, like choose different digits and put them in the right order, the constraints. In this trick it’s the presentation (where we say ‘lets make it harder’ and so on) that makes sure these constraints are followed. If we let things go astray and, for example, let the spectator choose digits that were the same we would have a constraint violation and it would upset the maths. Not a good idea!

Mathematical Principles Behind The Trick.
This trick blends simple arithmetic and a bit of algebra. Let’s represent the chosen number as letters ABC (that’s A hundreds, B tens and C units). So, for example, if the chosen number is 2 7 that’s A=2, B=5 and C=7.

Following the instructions your spectator has to reverse this number, so they will get CBA. Next we ask them to subtract CBA from ABC, which we can write like this:

Hundreds                     Tens                 Units
  A                                  B                      C
- C                                  B                      A


Now you do something sneaky. If you add 100 to ABC and then take away 100, then you’re still left with the same number, ABC right? Well, now imagine the second 100 which you added on was made up from
nine tens and ten units (ie 90+10). It makes no difference to the final value of ABC, but it changes our table to

Hundreds                     Tens                             Units
(A-1)                            (B+9)                           (C+10)
- C                                    B                                A

Doing the subtraction, we get:

(A-1)-C                         (B+9)-B                        (C+10)-A

So here is the first bit of magical prediction: the centre number will always, always be a 9! But you can also see why it’s important to have all the digits different and with the largest digit first. It’s not to make it harder - that’s just presentation - it’s to make the maths work.

But there is more mathematical mystery to follow. When you get them to write this number backwards and add, what you’re really doing is adding
A-1-C                           9                                  10+C-A
+ 10+C-A                     9                                  A-1-C  
9                                  18                                    9

We can simplify this. That 18 in the middle means 180 (remember it’s in the tens column) so we can take the 100 from 180 and shift it to the hundreds column, leaving us 8 in the middle. So the final sum is:

10                                8                                    9

Which of course is 10 hundreds, 8 tens and 9 units or 1089 - the number you had predicted all along!


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