Looking for "math-miracles" :o

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MagicalGerman

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Oct 9, 2007
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Hello,

Okay, I suppose the Subject doesnt really explain what I want.
Well, I'm gonna tell the whole story :).

I'm a conjuror/magician and at the moment I'm working on Mentalism and effects with cards. Now for a "mind-reading" effect I actually need some sort of way to let the spectator randomly pick 4 cards out of the whole deck and add these numbers up to find a "page"... I'll find a way to know the line then.

But I really tried all I can to find a way to know the spectators number without actually looking at the cards... You know I thought of some sort of math-thing like, for instance the one where you pick 4 random numbers, alter the order of those, substract the smaller one from the larger one and then add up the digits and always get nine.

Its really important I hope yours can help me :).

Cheers and Thanks already,

Nick
 
We're volunteers who help math students with math questions from their math homework. Whoever told you that you could get free consultation or work-product here was mistaken. Some one or another might be familiar with an old card trick, but, other than that, we're unlikely to be terribly helpful. Sorry. :oops:

To develop tricks, you might want to do some research into your industry, so that you can create your own work-product. If you are wanting to obtain "canned" product, then please contract with a supplier in your industry. :idea:

My best wishes to you in your endeavors. :D

Eliz.
 
MagicalGerman said:
I'm a conjuror/magician and at the moment I'm working on Mentalism and effects with cards. Now for a "mind-reading" effect I actually need some sort of way to let the spectator randomly pick 4 cards out of the whole deck and add these numbers up to find a "page"... I'll find a way to know the line then.

But I really tried all I can to find a way to know the spectators number without actually looking at the cards... You know I thought of some sort of math-thing like, for instance the one where you pick 4 random numbers, alter the order of those, substract the smaller one from the larger one and then add up the digits and always get nine.
Click to expand...
If a number from 1 to 10 will suffice, try the following:

Prepare the top 12 cards of a deck of cards as follows from top to bottom, face down:
B-W-B-B-W-W-W-W-B-W-B-B (B = black card and R = red card)
Place them on top of the reainder of the face down deck.
Deal the 12 cards out on the table in 4 rows of 3 cards each.
You should end with the following array of cards:
B----R----B

B----R----R

R----R----B

R----B----B

Proceed to tell him that you are going to try to predict the card he will end up on.
Write the name of the middle card of the bottom row on a piece of paper, fold it up, and place it in your pocket or on the table.
Tell your subject to think of one of the 12 cards and to keep his eye on that card only.
Next, tell him to move his eye up or down to the nearest red card.
Next, tell him to move his eye left or right too the nearest black card.
Next, tell him to move his eye diagonally to the nearest red card.
Finally, tell him to move his eye to the nearest black card.
Tell him to remove the card that his eye ends up on from the array of cards.
Have him open the piece of paper you wrote on earlier.

APPLAUSE!!!

Rather than deal out the cards from an unshuffled deck by the subject, have the subject shuffle the cards and you deal them onto the table in the desired pattern. This might lead to some suspicion but at lest you are dealing the cards from a shuffled deck and tends to eliminate the thoughts of a prepared deck.

Since the spectator will always end up on the same card position, you will know the value of the card he ends up on.

You can have him multiply the number by some number, add some number, divide by some number, etc., with you always knowing the result which you can then use for whatever purpose you want..
 
Here is another way to get a number.

An entertainer has an ordinary deck of cards. He gives them to his subject turns his back and tells her to make some piles according to these instructions:

First, the top card of the deck is turned over. If it is a picture card, put it back in the deck. Beginning with the first number card that turns up, she starts counting to herself until she gets to 12. (aces are 1) With each count, she takes one card from top of deck and places it face up on top of pile she is creating. When she reaches 12 she turns the pile over so card started with is face down on top. For example: She intitially turns up an 8, she places a card on top of the 8, and silently counts "9". The she places a card on top of the pile and silently counts "10". Then another card and counts "11" and another card and counts "12". Then she turns over the pile with the 8 face down on top. In this example the pile would have 5 cards all together.

Once a pile is complete, repeat the steps with the remaining cards until she runs out of cards. If she runs out of cards while trying to complete a pile, she picks up all the cards in that incomplete pile. When she is done, the entertainer asks her to give him the leftover cards. He sees that she has given him five cards but he does not look to see which cards they are. He also sees that she has made 6 complete piles.

He then tells her to take the top cards from the piles and add them together. Without her showing him the cards he tells her the sum. She gets the same answer. Figure out what the sum was and how he figured it out. Is there some kind of rule so that one can figure out how the entertainer did it and what the sum was that he guessed.

Some sample data:
>Leftover / # Piles / Sum
>4 6 30
>5 7 44
>0 7 39
>8 6 34
>1 7 40
>1 6 27
>3 6 29
>1 8 53
>4 7 43
>2 6 28
>4 8 56
>1 5 14
>1 6 27
>0 9 65
>2 7 41
>7 5 20
>3 9 68
>5 6 31

Let S = the sum, P = the number of piles and L = the number left over.
If you examine the data carefully, you will see that the differences in sums is 13 between adjacent pile quantities for any given leftover quantity
.
L = 4 .. P......S
...........6......30
...........7......43
...........8......56

Therefore, the P & S data can be expressed as y = mx + b or S = mP + L for each leftover quantity.

For zero leftover with 7 and 9 piles, the slope m = 26/2 = 13.
Then, using one of the sets of data, 65 = 13(9) + b making b = -52. The sum of any situation where there are no cards leftover but with differing numbers of piles is therefore, S = 13P - 52.

If you do the same thing with all the other sets of data you will end up with a set of equations as follows:
1 leftover - S = 13P - 51
2 leftover - S = 13P - 50
3 leftover - S = 13P - 49
4 leftover - S = 13P - 48
5 leftover - S = 13P - 47 and so on

Therefore, the entire spectrum of possibilities is defined by S = 13P - (52 - L). The sum of the cards sought is given by 13 times the number of remaining piles minus the difference between 52 and the number of remaining cards.

Another way of looking at it is to let R = the number of remaining cards, D = the number of cards dealt out, F = the sum of the first card values, A = the number of additional cards dealt out, and n = the number of piles.

Then, F + A = 12n, A = 52 -- R - n, and F = 12n - 52 + R + n = 13n - 52 + R.

ood luck
 
Here is another method you can use to reach a number you want. Just put any set of random numbers on 15 cards, arranged as shown below,face down on the table, with your desired end result placed in the center.

Place the numbered cards face down so that at the end, only the subject can see, and read the final card number to you.

In each box, write the following numbers, or any set of numbers you want.

.1.....2.....4.....3....5

.6.....8.....7....10....9

11....13....15....12...14

(The numbers shown are merely for illustrative purposes. You can put any different numbers you want as long as the odd-even relationship to the boxes is retained. In other words, the odd numbers must be in the boxes containing the odd numbers in the example and the even numbers must be in the boxes containing the even numbers in the example.)

Tell the subject that you are going to do an experiment in mental telpathy (thought transferance).(You are really getting a page number that you want to use in some way.

In your case, have the subject pick up the final card and read it to you

Tell the subject to sselect and focus on any odd number he wishes.

Next instruct him to move his eyes left or right to the nearest even number.

Next instruct him to move his eyes up or down to the nearest odd number.

Now tell him that you believe that his eyes arenot on the 1, 6, 11, 5, 9 or 14 so remove them from the table.

Next instruct him to move his eyes one space right or down moving only to the cards still on the table.

Now tell him that you believe his eyes are not on the 2, 4 or 3 so remove them from the table

Next tell him to move his eyes one space up, down, right or left, again only to the nearest remaing card

Now tell him that his eyes are not on the 8 so remove it.

Next tell him to move his eyes three spaces up, down, right or left, again only to remaining cards.

Now tell him that his eyes are not on the 1 or 2 so remove them.

Next tell him to move his eyes two times up, down, right or left, again only to remaining cards.

Now tell him that his eyes are not on the 13 so remove it.

Next tell to move his eyes one space up, down, right or left.

Now tell him that his eyes are not on the 10 or 15 so remove them.

The remaing card, the 7, can be used in any manner you desire.

Regardless of the numbers in the boxes or the number first selected by each person, they will end up in the middle box.

Good luck.
 
Just remembered a golden oldie:

Write the number 5 on a piece of paper, fold it up, and give it to your friend. Tell him to
1--Pick a number from 1 to whatever you wish, usually 10 to 20.
2--Double it
3--Add 10
4--Divide by 2
5--Subtract the number you started with
6--Look at the paper

Wow!!! His answer is 5. I'm sure you will be able to understand the reason why by writing out the steps.

1--Pick a number from 1 to whatever you wish, usually 10 to 20. Lets say 12 or N.
2--Double it - 24 or 2N
3--Add 10 - 34 or 2N + 10
4--Divide by 2 - 17 or N + 5
5--Subtract the number you started with - 12 or 5
6--Look at the paper - Answer = 5.

The cute thing about this one is that you can continue doing it for a while when they say do it again. Why? The secret is in the step 2 and 3. If you use "double it" you must divide by 2 in step 4. You can add any number you wish in step 3 as long as it is evenly divisible by the number used in step 2. The answer is always the added number divided by the multiplier. So you have the multiplier and the added number to vary every time you do the trick. For instance

1-Number 2-times 3-add 4-divide by 5-subtract 6-answer
......9.........2=18.....16=34.....2=17..........9=8..........8 = 16/2
.....13........3=39.....21=60.....3=20.........13=7..........7=21/3
......7.........4=28.....16=44....4=11...........7=4..........4=16/4
.....12........5=60.....15=75....5=15..........12=3.........3=15/5

Just remember, pick your multiplier first, add a number that is evenly divisible by the multiplier, and use the same number to divide as you used to multiply.
 
Pick a card, any card.

Show it to everyone present.

Put it back in the deck.

Shuffle the cards.

What is your age?

Count that many card from the bottom of the deck.

Is this your card?

Since it is guaranteed to be, APPLAUSE please. :))
 
Thanks alot :). I'm more or less not that good in math that why I'm asked here its not like a "product" I'm young so I'm not really into the "buisness".

Thanks alot!

Nick
 
TchrWill said:
Pick a card, any card.

Show it to everyone present.

Put it back in the deck.

Shuffle the cards.

What is your age?

Count that many card from the bottom of the deck.

Is this your card?

Since it is guaranteed to be, APPLAUSE please. :))
Click to expand...

What if you're over 52?
 
daon said:
TchrWill said:
Pick a card, any card.

Show it to everyone present.

Put it back in the deck.

Shuffle the cards.

What is your age?

Count that many card from the bottom of the deck.

Is this your card?

Since it is guaranteed to be, APPLAUSE please. :))
Click to expand...
Click to expand...

What if you're over 52?

OUCH!!

The trick is obviously best played on younger people, preferably your children or grandchildren.
 
Nick, I like this "trick":

Get participant to shuffle deck, then put 10 cards face up on table.
Left to right: card1, card2, ....., card9, card10
Then tell him to touch one of them.
Then tell him to give you any 3 of the cards.
You then rearrange these (not necessarily all 3) and pass them
to your "partner" say under the door (partner in adjoining room).
Your partner then announces the cardnumber that was touched.

Works like this:

none = 1 (pass no cards means card1 was touched)
one = 2 (pass 1 card = card2)
12 = 3 (pass 2 cards in ascending order = card3)
21 = 4 (pass 2 cards in descending order = card4)
123 = 5 **
132 = 6
213 = 7
231 = 8
312 = 9
321 = 10

** lowest middle highest = card 5 ; and so on

If you happen to be given a pair or 3 same, suit determines
order: like heart > diamond > spade > clubs.

I play that with my daughter as partner...sure fun to have
the in-laws beg to know how we do it!

Plus the "indicator" order can be changed at will:
none = 10
one = 9
12 = 8
21 = 7
123 = 6
132 = 5
213 = 4
231 = 3
312 = 2
321 = 1
(in case someone notices a pattern!)
 
Denis said:
Nick,

If you happen to be given a pair or 3 same, suit determines
order: like heart > diamond > spade > clubs.

quote]

Neat trick ...but you must not play bridge (in that land spade > heart > diamond > club)
Click to expand...
 
Subhotosh Khan said:
Denis said:
Nick,
If you happen to be given a pair or 3 same, suit determines
order: like heart > diamond > spade > clubs.
quote]
Neat trick ...but you must not play bridge (in that land spade > heart > diamond > club)
Click to expand...
Click to expand...
No Subhotosh: if I use the bridge manner, then bridge players may notice!
 
Hello, Nick!

There are various methods for forcing a number.
Most of them are quite elaborate and may distract from the effect itself.

Here's one that takes forever, but is an impressive effect by itself.

Give the spectator a pad of lined paper, a pen, and a calculator.
Prepare the top sheet with a column of numbers at the left side of the page,
. . numbered from 1 to 20.

Have him choose a one-digit number (mentally) and write it on line 1.
Have him choose a two-digit number and write in on line 2.
Have him add these two numbers and write the sum on line 3.

Instruct him to add the last two numbers on the list and write them on the next line.
Have him continue this process until he has filled the twenty lines.

When he's done, have him pick any number from the last five on the list
. . and divide by the preceding number.

Punchline: You already know the first five digits of his final answer, 16180.


~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~


This is a Fibonacci-type sequence.
The ratio of consecutive terms approaches \(\displaystyle \phi\), the Golden Mean:
. . \(\displaystyle \frac{1\,+\,\sqrt{5}}{2} \:=\:1.618033989\,\cdots\)


Try it yourself.
Suppose your first two numbers were 8 and 37.

The list looks like this:

\(\displaystyle \begin{array}{cc}1 & 8 \\ 2 & 37 \\ 3 & 45 \\ 4 & 82 \\ 5 & 127 \\ 6 & 209 \\ \vdots & \vdots \\16 & 25586 \\ 17 & 41399 \\ 18 & 66985 \\ 19 & 108384 \\ 20 & 175369\end{array}\)

Now divide #18 by #17: \(\displaystyle \:66985 \div 41399 \:=\:1.618034...\)

I'll let someone else explain why this always works.
 
soroban said:
I'll let someone else explain why this always works.
Click to expand...

Let a and b be ANY two integers. The numbers themselves are not important, as the nature of the sequence defines the outcome. (edit: oops, please mind the nonsense choices such as a=b=0).

Let the sequence \(\displaystyle a_n\) be defined as follows:

\(\displaystyle a_1 = a; \,\, a_2=b\)
\(\displaystyle a_{n+1} = a_{n-1}+a_{n}\)

You're looking to show \(\displaystyle lim_{n \rightarrow \infty} \frac{a_{n+1}}{a_n} \,\, = \,\, \frac{1+\sqrt{5}}{2}\).

\(\displaystyle \L L =lim \frac{a_{n+1}}{a_n} = lim \frac{a_{n}}{a_{n-1}}\) (1)
and, \(\displaystyle \L L = lim \frac{ a_{n-1}+a_{n}}{a_{n}} = lim \frac{a_{n-1}}{a_n} + 1\) (2)

Using (1) and (2),

\(\displaystyle \L L = lim \frac{a_{n-1}}{a_n} + 1 = \frac{1}{L}+1\)

Finally giving the famous quadratic equation:

\(\displaystyle \L L^2-L-1 = 0\), of which \(\displaystyle \phi\) is one of the roots.
 
Lovely proof, Daon!


Hello, MagicalGerman!

Here's a trick you may be able to modify to your purposes.

Give the volunteer a pad of paper, a pen and a calcultor.

Tell him to write down any three-digit number.
Have him repeat the number ... so he has a six-digit number.

From across the room, you divine that the number is divisible by 7.
Have him divide by7.

Tell him to divide the result by his original three-digit number.

The answer is always \(\displaystyle \L143\).


If (somehow) you can get him to factor it, we get \(\displaystyle 11\) and \(\displaystyle 13\).
And you can send him to page 11, word 13.

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

The explanation is quite simple.

The repeated number is of the form: \(\displaystyle \,"ABCABC"\)

And: \(\displaystyle \:"ABCABC" \:=\:"ABC"\,\times\,1001\;=\;"ABC"\,\times\,(7\,\times\,11\,\times\,13)\)

 
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