I wonder if anyone can help me with this.

I want to be able to force the number nine but so that it seems that the outcome could be any number between say 1 and 100. I am looking for some kind of math based force where the spectator thinks of a number or two numbers and does some kind of fair seeming math and arrives at nine.

If someone can point me in the direction of such a force, or a book with such a force in it I'd appreciate it.

One - fairly simple - method would be to:

1. ask the spectator to choose any two digit number (say 65)
2. have them add the two digits together (6+5 = 11)
3. tell them to take away the total of the two digits from the original number (65 - 11 = 54)
4. this new number will always be a multiple of nine, so if they repeat step 2 they will find they come to the number nine (5+4 = 9)

Of course you will need to add some kind of rationale for all of this... adn some strong scripting to hide the method. I would suggest to them that they will be dooing some simple maths with the number so that it will create a new 'random' number - this, you say, is to eliminate any thoughts that things might have been pre-arranged...

Any help?

Jack

I like your method Jack but I add a little more to it to really throw the spectator off. After I force the number nine on them, I tell them that I have the coin value in my hand that equals the number they picked... They know they got you because the answer would be 4 and a half and there's no suff coin. They now are more interested in proving me wrong than in figuring out the trick. Well, surprise, surprise.. I open my hand and let 4 pennies fall to the table along with another penny that I have cut in half (4 1/2 cents)... It makes for a good illusion, it throws them off, and it always gets a laugh..

May not be what you want but ...

1. any four digit number (all different numerals)
2. Jumble up the 1st number to make a 2nd four digit number
3. Subtract smaller from larger
4. Add the digits of the new number
5. If total is a 2 digit number, add again
6. Answer will always be nine

Sorry, I left something out - changed the post!!

After I force the number nine on them, I tell them to divide their total in half (divide by 2). Then I tell them I have the coin value in my hand that equals the total they now have... They know they got you because the answer would be 4 and a half and there's no such coin. They now are more interested in proving me wrong than in figuring out the trick. Well, surprise, surprise.. I open my hand and let 4 pennies fall to the table along with another penny that I have cut in half (4 1/2 cents)... It makes for a good illusion, it throws them off, and it always gets a laugh..

91 ~ 28 (mod 63)

Strange. This was meant for ANOTHER post. Hmmm.

1. Get a 3 digit number (all digits different, ex: 123)
2. Flip it around (ex: 321)
3. subtract the smaller from the bigger number (ex.: 321-123=198)
4. middle digit is always 9
btw: the outcome is a multiple of 99...

(not so different from what paulmagic had to offer)
just my 2 cents
Heinz

Oce you get a root # 9, any # that you multiply it by will have a root of 9. Any # with a root of 9 can be simplified to 9 by adding digits.
Example: 3*3=9... 9*7=63... 6+3=9.
Example: 9*176=1584... 1+5+8+4=18... 8+1=9

Practical example:How many cylinders does your car have/
4
Multiple that by any #
7
7*4=28
Multiply that # by 9
9*28=252(root 9)

Also, any # reversed and subtracted from the original will give a root # of 9.
Knowing the secret of 9's can lead to miracles.
Good luck.

Dear dcobbs,
This may interest you:
Multiply any whole number by nine (except zero of course) then repeatedly
add the digits of the answer till you get a SINGLE digit, and that digit will always
be the number NINE.
1,357 x 9= 12,213
(1+2+2+1+3=9)

8,642 x 9= 77,778
(7+7+7+7+8= 36; 3+6=9)
Marilyn Vos Savant, Parade, page 17, 23 Sep 2018
Sincerely,
miistermagico

..and 14 years later still no TY from the OP.

At just over two posts per year, he can hardly afford to waste them on "thank you" messages.

I'm shocked (SHOCKED I say!) that nobody has mentioned two really good resources for approaches like this:

http://www.richardbusch.com/mentalism/pr......lease-2/
Richard Busch's "Number...Please". I think his website is password protected, so go here first if you can't get in through that link, he thoroughly explains the password: http://www.richardbusch.com/mentalism/
In any case, here is an excerpt from his ad copy of this, see if it intrigues you:

My next recommendation is "Mathematical Wizardry" by Harry Lorayne. Anything by Mr. Lorayne really, but for this discussion, that's the book: http://www.harryloraynemagic.com/store/p......ok).html
I bought the physical book when it came out and I will be honest, I was disappointed that it is now in PDF form (I want it all to myself!). But, as they say, if you want to hide something, put it in a book. It'd be impossible to even list all the ways you could force the #9 from the book but I will allude to one example:
Look at miistermagico's post. Then imagine you don't have to do it, you can just have a participant pull out their own calculator and just multiple random number after random number (you never touch the calculator), and the result is the same. Mr. Lorayne discusses it.

If anyone has either of those resources and wants to talk about it, post here and I'll PM you.

Best wishes!

As we are on the subject of forcing '9', I would be grateful if somebody would kindly PM me the procedure by L.Vosburgh Lyons called simply "nine"?

It's published in "The Phoenix" #53. Just a variation of the fact that a number minus its digit sum is divisible by 9.
I'll send you a PM with some more details.

Cheers saxonia, and thx for the PM.

[quote]On Jul 3, 2019, Chris K wrote

Look at miistermagico's post. Then imagine you don't have to do it, you can just have a participant pull out their own calculator and just multiple random number after random number (you never touch the calculator), and the result is the same. Mr. Lorayne discusses it.


Exactly! Here is an idea for a presentation using such as approach.

Print off nine different images taken from the internet onto card and laminate. You will need three duplicates of each image so that makes twenty-seven images in total. Use your imagination. Think of a connecting theme such as victims of Jack the Ripper or super heroes or whatever.....

You start by taking out all the images from a black velvet bag.

Now have your volunteer multiply different single digits randomly together onto a calculator until they obtain a six digit total. Emphasise that the numbers should be "as different as possible and as random as possible". Your back is turned whilst they do this. You do not see anything that is going on.

All the images are numbered one to nine discretely on the back. Ignoring any noughts in their total you can now ask them to create their number total using the image numbers to represent the digits within the total and when the face up lineup is complete to mix all the images about on the table. They are then to clear the total from the calculator and to put it upside down so that the display can not be seen and so that the calculator can not be inferfered with.

They are then instructed to place all the unused left over images into the black bag which which they originally came. These images are thus also out of sight.

Finally they are to take any one of the images away from the others, to place it face down, and to also to put their hand over it to hide it completely.

Only then do you turn around to face them.

And then you reveal gradually -bit by bit- the exact image that they are thinking of!

Martin Gardners MMM has a great section on this ...