Firstly I must say that I found this principle in a novel, and then came up with the use. Im not massively versed in mathematical magic, and so if this is already in print I'm happy to take it down.

Crowd force (threes a crowd)

Effect
The spec picks any number between 1-100. You then write down the next two numbers next to it. All the numbers are added together until the spec is left with a single digit.

This number can then be

Revealed as a prediction
Used for a chapter / page / word in a book test
Used to count down cards in a deck
Used to force an object out of a numbered selection of items


Principle
This is a mathematical principle which relies on the following rule:

The total of any 3 sequential numbers, where the last one is divisible by three, will be 6

So for 1+2+3 this is easy to see. Any larger numbers will mean that the digits are added together e.g.

7+8+9 = 24, 2+4=6

Foe even bigger numbers, there are two ways this works
Method 1 - add the numbers then the digits
22+23+24 = 69, 6+9 = 15, 1+5 = 6

Or
Method 2 - add the digits
22 23 24 = 2+2+2+3+2+4 =15, 1+5=6



Presentation
This would begin by having a piece of paper with three lines on it for the numbers to be written on, eg

- - -

Then ask the spec to tell you any number between 1-100 (you could use a larger or smaller number depending on your maths ability). The spec writes this nimber down on the paper, and you explain that I is very dificult to get someone to choose a random number, as eveyone tends to remember ages, door numbers, favourite or least favourite numbers, so you ask them to add some of the next numbers in the sequence.

Depending on the number they say, there are 3 different outcomes. each time you will ask them to write their number in one of the spaces, and two other numbers, so that all 3 numbers are in sequence and the last is divisible by 3, so that it will total 6.

As it is the 3 times table, the specs number will either be a multiple of 3, one less than a multiple of 3, or two less than a multiple of three, and this is true for any number.

Eg

3,6,9,12,15,18 etc - multiples of 3
2,5,8,11,14,17 etc - one less than the multiple
1,4,7,10,13,16 etc - two less than the multiple


so:
1. If their number is divisble by 3, you ask them to write it on the right hand line, and write the two numbers before it to make the number 'more random'. These will then be added together to get 6.

2.if their number is 1 less than the multiple of 3 (it might be easier to think of it as being 1 less than the three times table) then they write the selected number in the middle and write the number before it in the space to the left, and the number after it in the space to the right.

3. If their number is two less than the multiple of 3, then they write the number in the left hand space and put the two numbers after it in the last two spaces.

It spunds quite difficult, but really isn't with practice (as long as you know your three times table).


Notes

1) For larger numbers, I find it easier to minus groups of 30 from it, to see whats left. This means the maths never gets more difficult than the 3 times table.

Eg 77 is selected
77-30-30= 17
18 is a multiple of 3 so this is one less
The number 77 would then go on the middle line, with 76 before it and 78 after

76+77+78= 42, 4+2 = 6 (adding digits)
76+77+78=231, 2+3+1 = 6 (adding numbers)

2) in my opinion, this would work best as a prediction style effect, rather than mind reading, even with a book test. The drawback with presenting it as mind reading would be that you need to know the specs first number. A spectator could think that you are quickly doing the maths in you head before revealing the answer. If the prediction is already on display, produced from a pocket for the reveal or even posted to someone before the show, then this removes this possible explanation.

3) You could do an example addition, where the last number isn't divisible by 3 (or ask other people to repeat the process) to produce a random number to show that its fair, but I don't think this is necessary.

Anyones thoughts or opinions would be much appreciated

Rich

You can add up the digits in any larger number and tell if the number is divisible by 3. In your example of 77, 7+7=14, which is one less than the nearest multiple of 3 (15). This method may be easier than subtracting 30 for some. I will try this as a card force with some of my students and see how they do. Had never heard of this one.

Thanks.

Interesting principle, Rich!

Between 1 and 100, there are only 33 multiples of 3, so the simplest solution is making sure you know all the multiples of 3 by heart.

When you know that, there's a simple way to remember which numbers to use. Think of the multiples of 3 as the "ceiling", and just remind yourself to always work up to the ceiling.

For example, someone says 75, you recognize that as a multiple of 3, and that's already a ceiling, so you would work up to 75 (73, 74, 75).

How about 38? OK, 38...39...wait! 39 is a multiple of 3, so 39 is the ceiling, so the sequence needs to be 37, 38, and 39. Get the idea?

---

Bonus idea: Always work DOWN to the multiple of 3 as the "floor" instead, and the sum of the digits total will be 3 instead.

65? Ok, 65, 64, 63...aha! That's a multiple of 3, so you're going to use 63, 64, and 65. Add them up, and you get 192, 1+9+2=12, and 1+2=3!

Another bonus: if you keep the multiple of three in the middle, the digital root of the sum is 9.

Good call Greerj, to easier check the reminder.

/Tomas

And it seems as though the digital root (thanks for letting me know the name Tomas) of keeping the number first is a 3

so we have

3,9,6

this means it could be used for repeat performances with multiple different outcomes.
Ive messed with series of 2 or 4 numbers and it only seems to work with three.

In the following blog post, the author points out that you can have someone write down any 5 positive integers, and you'll always be able to pick out 3 of those integers which will total a multiple of 3.
http://mindyourdecisions.com/blog/2013/0......le-by-3/

In this case, you wouldn't know ahead of time whether the digital root of the number will be 3, 6, or 9, so m*****le o**s would be needed.

I'm wonder if there's some sort of P**EO-ish or eq****ue-ish process to guarantee the selection of 3 such numbers.

Hello guys, I found this topic very interesting! I ve been thinkin about it and here s my thought on it:

If the 1st number of the (3) suit is a multiple of 3, the sum will be 3
If the number in the middle of the suit is a multiple of 3, the sum will be 9
If the number is the last of the suit and is a multiple of 3, the sum will be 6.

Knowing that, you can make a prediction in advance (3, 6 or 9) and lead the spectator to the predicted number.

Here s an example with 74
If I want to predict a 3, I ll lead the spectator to sum up 72-73-74 (by saying "please add 2 numbers preceding the 74")
If I want to predict a 6, the spectator will sum up 73-74-75 (by saying "74 is in the middle of 2 numbers...73 and 75 right? Let s add them together!)
If I want to predict a 9, simple ask the spectator to add 74-75-76 (by saying "please add the 2 numbers fallowing 74..")

Your attitude is everything, but I m confident it would work.

I hope it is clear for you.

Have a nice day!

Hi

Can this principle be applied to create a 4 th force number . That way with 4 people each will end up with a single digit which can represent together bank PIN number or the entry code on my phone

Andrew

Hi all,

I've been thinking about this method since this thread started and I wanted to share the way I used it for the past days.

I wanted a version where I don't know which numbers are thought of by the spectators, so I could use the result as a force, rather than a prediction.

I do this with 3 spectators: I give the first spectator a piece of paper and ask him to think of a 2 digit number.

Then, as an after thought, I tell him "you know what, let's go for a bigger number: double that number... No! Triple it! Use you cellphone's calculator if you have to". (if they don't have a cell phone or calculator, I as them to think of a number between 1 and 10 instead of a 2 digit number).

Then I tell him to write that number on the upper part of the paper and hand it to the second spectator, all that without me seeing the paper.

I tell the second spectator "would you rather add one or remove one from the number on the paper?"

He either says "add" or "remove" and writes the result below the first number and hand the paper to spectator number three.

If the second spectator added, I tell the third: "now, you can either go the oposite direction and remove from the first number or you can go the same direction and add to the second number, which you prefer?".

If the second spectator removed, I tell the third: "now, you can either go the oposite direction and add to the first number or you can go the same direction and remove from the second number, which you prefer?"

Third spectator answers and writes down his number.

I then have him add all the numbers up and add the digits. The result is either 3, 6 or 9.

If they both added, the result is 3.
If either one added and the other one removed, the result is 9.
If they both removed, the result is 6.

Now, in their mind, they arrived at a completely random number, but I actually know what number that is and I can use this information anyway I want: I can simply reveal the number or, if I'm at a bar, I usually have them look on the menu what drink is at their number and guess it. I can also use it as the starting point of a one ahead routine, etc.

That's great, it would even work with 1 person as long as they told you their decisions.

love it

edit:
ive just tried this out with 28, and it didn't work

adding twice would be 28+29+30 = 87 = 15 = 6
one of each would be 27+28+29 = 84 = 12 = 3
subtracting twicw would be 26+27+28 = 81 = 9

Hi,

The key is that the spectator has to think of a number and then multiply ir by three. Only then you can do the adding/removing procedure.

28 is not a multiple of three. If they think of 28, then they have to multiply by 3. 28x3 = 84.

Adding twice: 84+85+86=255. 2+5+5=12. 1+2 = 3.
One of each: 83+84+85=252. 2+5+2 = 9.
Subtracting twice: 82+83+84 = 249. 2+4+9 = 15. 1+5=6.

Ah, I see, apologies

You don't have to apologize, my friend. In fact I should be thanking you for sharing this cool principle. I've been using it a lot lately and I like it more and more. So, thank YOU!

How about his as food-for-thought: using this principle as a month divination?

Give a spectator a business card and, in the guise of doing a somewhat strange numerology reading, ask him to write the month he was born on the top.

Then, do the add/remove procedure: tell him to choose wether he wants to add one or subtract one from his month. Whatever he choses, he writes the result below his month.

Then, ask him one more time if he wants to add (to the second number) or subtract (from the first number). Whatever he choses, he writes the result below the second number.

Ask him to add everything up and, if the result has more than one digit, add the digits. Lets call this number the FIRST RESULT.

At this point, if he were to tell you his first result, you could narrow his month to four possibilities:

If he added twice and the result is 3, the possible months are 3, 6, 9 and 12.
If he added twice and the result is 6, the possible months are 1, 4, 7 and 10.
If he added twice and the result is 9, the possible months are 2, 5, 8 and 11.

If he subtracted twice and the result is 3, the possible months are 2, 5, 8 and 11.
If he subtracted twice and the result is 6, the possible months are 3, 6, 9 and 12.
If he subtracted twice and the result is 9, the possible months are 1, 4, 7 and 10.

If he added once and subtracted the other and the result is 3, the possible months are 1, 4, 7 and 10.
If he added once and subtracted the other and the result is 6, the possible months are 2, 5, 8 and 11.
If he added once and subtracted the other and the result is 9, the possible months are 3, 6, 9 and 12.

But, to narrow it down even more (and to add another seemingly random choice), tell the spectator that, if he was born on an odd month, he must chose if he wants to add one of subtract one from his first result. This one he doesn't need to tell wether he is adding or subtracting. Let's call this number the SECOND RESULT.

Ask him what his second result is. Besides the choices of adding or subtracting, this is the only information you get from the spectator.

If his second result is NOT 3, 6 or 9, it means he was born on an odd month.

But, from the second result you can easily figure out the first result: 2 or 4 means the first result was 3; 5 or 7 means the first result was 6; 8 or 10 means his first result was 9.

So, for instance, if the final result is 8 and he added twice, we know he was born on an odd month and that his first result was 9. So, he was born on either the month 5 or 11.

After a quick numerology reading, you can say "You weren't born on May, were you?". If they say yes, you say "I thought so", if he says no, you say "I thought so, because although you do have some traits of someone born near the begging of the year, I would say you definitively were born on November, right?".

With so many choices and the fact that there are a lot of other legitimate results, the spectator will really believe that the end result had absolutely nothing to do with his month. For instance, I was born on March and, with this procedure, I could end up with 2, 4, 5, 7, 8 or 10.

I'll have to read in the morning to make sure it all sinks in. The principles sound amazing!

Here's my 2 cents :]

I believe the biggest strength of the crowd force is it can be used for ANY 3 consecutive numbers thought of, and this could be a really pure and direct mentalism premise in itself.

How about just write down a number 6 on a small piece of paper, show the prediction if the digits add up to 6. Rotate it around to show 9 if the digits add up to 9. There is 2 out of 3 chances you will be spot on. that's 66.6%!

How about apply this principle to a clock face? if the spectator randomly add up any 3 consecutive hours on a clock face, there is 50% chances the added digit would be 6. Specifically, if the starting hour of the 3 consecutive hour is 4, 7, 10, 11, 12 , 1, the final added digit would be 6. By asking the spectator to think of a not-so-early hour, will increase the probability.

If you combine the 6/9 rotation prediction with the clock face idea, there will be 9 out of 12 chances that you will be spot on! that's 75%!

I have a few other ideas to share later :> off for lunch now!

Hey Rich, this is one of the best ideas I have come around in mathemagic! I absolutely love this! Thank you so much for your contribution!

Love,
Perl

Looking at this clock, all the circled hour(and the subsequent addition of the next 2 hours) will give you the results of either 6 or 9. Asking a spectator to randomly drawing in an hour hand to the clock, will easily force them to point to an hour other than the 3,6 and 9. This is because the position of 3,6,9 is too common, straightforward and will be consider not-so-random in human's mind.

Or, they can draw in the hour hand behind their back? Will it help?

Hope the diagram helps the understanding of the ideas.

Now, I really have to go for my yummy lunch :>

Love,
Perl

Click here to view attached image.

Using this as a force with a stacked deck in which the 1st and 4th cards are jokers and the others are indifferent, all face down, take off the top card and put it by the spec's elbow.
Then ask them to choose a number from 1 to 10 and triple it (smaller numbers are easier for people to work with, even on paper or with a calculator). Then they add the next 2 numbers after that result to the result and then add the digits of the final answer. You have forced the number 3.

Ask them to count down to that card in the deck. You haven't been told anything by the spec. The card will be the joker, matching the one (face down) by their elbow. All of the other cards may be examined and are found to be non-jokers.

Then, while doing some math-babble about digital roots, slip in another 3 cards at the top and reset the deck (leaving the face up joker by the spec's elbow). Now do this with the next spec. After tripling the random number add the 2 BEFORE the result they just got, add the digits and have them count down in the reset, face down deck (you've forced a 6). Since you slipped in 3 extra cards the 6th one will be a joker.

Any thoughts on the above routine?

Personally, I wouldn't use this force with cards. The reason is, although the maths involved are simple, I can't ask a spectator to to them without paper or a calculator.

If they pick 8, they have to do: 8x3=24. Then 24+25=49. Then 49+26=75. Then 7+5=12. Then 1+2=3.

This is way too much to ask of someone who is "on the spot". They might mess it up so many times I don't want to risk it.

So, because of this, I prefer to hand them a piece of paper and ask them to think of a number, them multiply by three and write it down. They write the next two numbers below the first and add them up at the end. By the time they write the result, they forgot they even multiplied their first number by three (actually, this happened to Pixelated, who read my post then tried it and it didn't work because he forgot to multiply, which goes to prove the spectator WILL forget that step afterwards )

The reason I wouldn't do your routine is that I would have to hand them a piece of paper and have them do some maths, then hand them the deck and do some dealing. That seems to involve to many props for my personal taste, but that's just me

Another idea would be to tell them, after they get to the final number (3, 6 or 9): "If your number is odd, think of Black. If your number is even, think of Red. If your number if 5 or less, think of Clubs or Diamonds, depending on your color, and if your number is 6 or more, think of Spades or Hearts. Imagine a card with that suit and the value of your number".

This limits the possible cards to 3 of Clubs, 6 of Hearts and 9 of Spades. But after they write down the numbers and before they add them up, you already know which card they will pick.

You could have only three cards in your pocket and after they imagine their card, you tell them "my lucky card is this one" and remove their card from your pocket. Or you could just remove a card from the deck after they add up the numbers.

I prefer giving them the choice of adding the next number or the previous one (as I described in a previous post), but if you don't give them that choice, you can limit the possible cards to only one.

Hi Adrien,
I tried the trick 2 days ago just as I described it, with a smart high school student (getting the 3) and an adult getting the 6 immediately after (resetting the deck and slipping in the 3 extra cards at the top). It went over fine but I think that random audiences would not find the trick interesting. It would depend on the group you were working with.