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f*** I spent so long on another topic I forgot I came here to ask for some help.

I'm trying to find a simple maths technique that will reduce any number to an even single digit number. So far the best I've got is 'Think of a number, times it by 9 and minus 1. If there are two digits add them together, if there are still two digits add them together again.'

No matter what they pick they will end up with 8.

This is fine but multiplying something by nine can be tricky for some people to do in their heads and I would rather it could be any single digit even number, while still being simple enough for anyone to do quickly in their head, and quick enough to not seem that important.

This is based around a numerology presentation so if it's less 'maths like' the better. Ideally someone thinks of a random number and then just ads the digits together until they get a single digit number.

Any help would be appreciated.

What about 'divide it by 2, if you dont get a whole number just round up / round down, now if your number has more than 1 digit, we'll just use the last one'.

I was going to suggest something similar to Relish but instead of divide, multiply the number by 2 and use the last digit. If you multiply then you don't need to worry about it being a number that doesn't go into 2.

Perfect exactly what I needed, thanks both of you.

I know you've got this sorted already, but going back to your original method - you could teach your spectator to do the nine times table on their fingers...
http://gwydir.demon.co.uk/jo/numbers/fi ... ltiply.htm

Alright I'm going to need some help again, I really really suck at maths and I have tried to look this up and work it out myself but no success.

I'm trying to work out permutations for a routine I'm doing. It will be a combination of three out of six shapes i.e:

Square Circle Triangle

Square Circle Cross

Triangle Square Star

etc

The shapes will be Square, Circle, Triangle, Star, Cross, Heart.

Thanks.

Is the question how many permutations are there? If so it's 120
(6*5*4)

Or if you allow for repeated selections if the same shape its 216
(6*6*6)

Lawrence wrote:Is the question how many permutations are there? If so it's 120
(6*5*4)

Or if you allow for repeated selections if the same shape its 216
(6*6*6)

Totally agree Lawrence

mr invisible wrote:

Lawrence wrote:Is the question how many permutations are there? If so it's 120
(6*5*4)

Or if you allow for repeated selections if the same shape its 216
(6*6*6)

Totally agree Lawrence

Glad to see my maths degree is completely going to waste!

Right if I understand this, if there were seven objects with no repeating it would be 7 times 6 times 5. If there were four spaces open with six objects no repeating it would be 6 times 5 times 4 times 3, right?

Yes

Good good, now what are these A-B-C's I'm meant to learn?

Jean wrote:Alright I'm going to need some help again, I really really suck at maths and I have tried to look this up and work it out myself but no success.

I'm trying to work out permutations for a routine I'm doing. It will be a combination of three out of six shapes i.e:

Square Circle Triangle

Square Circle Cross

Triangle Square Star

etc

The shapes will be Square, Circle, Triangle, Star, Cross, Heart.

Thanks.

Just to clarify, does the order of the objects matter ??

i.e. Is "square circle cross" considered the same as "circle cross square" or is this considered a different selection ??

Fro the purpose of the presentation, the order does matter.

Jean wrote:Fro the purpose of the presentation, the order does matter.

..... in that case I agree with Lawrence's answer .

If the order didn't matter, there would only be 20 permutations ( 120/(3x2x1) ) ...... I think .

This is fairly obvious, as for every selection of A, B and C, there are 6 permutations that would be considered the same; ABC, ACB, BCA, BAC, CAB and CBA.