Math Is Fun Forum

many more to come:D

PS sorry about that last one I worked it out myself (the otehrs I got from a book years ago) I'm no good at factorising so if someone could do that for me I would be most greatful

One of my obsessions with puzzles is to extend them and make harder versions

like this puzzle from the site:

Dropping BallsImagine that you had 3 balls and 1000 storey building

What would be the solution now?

(Everything else is the same.)

The Riemann Zeta function is defined as

It's trivial zeros are said be negative even numbers,

However if you put a negative whole number (even or odd) the two negatives (from the minus sign in the function and the value of "s") the result is a whole number and positive power.

So the terms of the sequence are all whole numbers, which indicates an infinite sum, so why do all sources that I've read about the Riemann Hypothisis say that the trivial zeros (values of s for which the sum is 0) are negative even numbers?

Either:

I've gone wrong somewhere (In which case I would really appreciate it if someone would tell me:D)

I don't know the true definition of a "zero" (in a function)(In which case ot would be useful to know what one is)

or something else is happening???

but this has been bothering me for a while.

thank you:D

A common misconception (I think) is that if you've got a set of permuatuations (which are open to variation that could mean that they're not all distinct) the simple way to find out how many there are (ignoring what they are) is to take the number of permuations P (where variation are classed as different) and divide by the number of variations V :these could be "rotations and reflections" (for shapes) cyclic combinations of colours (for couloured patturns etc)

but this doesn't always work Proof:

Lets say that there are P permuations and V variations then if you take "P/V" distinct possibilities and then write them down in a list

1a,1b,1c ... 1V
2a,2b,2c ... 2V

            .
            .
            .

Ra,Rb,Rc, ... RV (where R=P/V)

then if all these possibilities are different! then the number of possibilties is R X V (=P)

but if some have variations which are the same (such as them having isometry themselves which matches the isometry of the shape for e.g.) then some permuations will be mapped on to themselves (instead of another permuation) this means that this list (containing P possibilties in total) has duplicates, once filtered there are less than P (and that must mean there's more distinct possibilties to find)

Conclusion:

The number of dinstinct possibilities can end up being more than "P/V"

this can apply to a lot of things:

Polyforms
Coloured grids
and much more:D

this problem has been driving me mental for such a long time and I just give up on it!

Basically if you have a number "n" and one of it's factors "r" (neither "r" nor" any of it's roots if any can have powers which are also factors of "n")

e.g. if you take 40, 2 as a factor 4 (it's square is a factor) or 4 (2 it's square root has 8 as a factor) so you could use 8 but not 2 or 4

if you take this function:

HCF(x,n) for x= 1,2,3,...n-1 this set (I think) forms a group "multiplication mod n"

I can prove the: inverse, associative and closure property but for the life of me I can't prve the indentity property,

I've tried finding a formula for it (it's not always the factor that is the identity element)
e.g. for 12 and it's factor 3 there are only two elements 3 and 9. 9 is the identity

I tried looking at it backwards given r consider it multiples (to account for the same problem as above don't multiply by another multiple of "r")

Has this result been proven (or disproven)? as if so what is the proof (disproof) is there a formula for the identity?

any help is most appreciated:D