Math Is Fun Forum

It doesn't matter if there's anything outside, because it don't fall into "EVERYTHING" column, so we just can't know what's outside the box, we can only make guesses without strong arguments, and this is not knowledge anymore.

hm, it makes sence - but I have other plans - imagine that everything around us is just not real, but it's some mental simulation of the real world, so that everybody is actually a simulation inside yourself. But then you and your mental imgination stimulate the science, and without knowing, YOU discover the scientifical results, because the people are controlled by YOU.
So here's the BIG  BANG - you already KNOW EVERYTHING, but you just don't know that you know it

Which problem?

Noop. Plugging in gives 11/46656, not 1/46656, but this is something useful:
the solution of

So, the algebraic number field is algebraically closed, this is a foundamental property of the field, and means exactly that every root of a polynomial with algebraic coefficients is algebraic too. This comes from basic factorisation properties of plynomials and using the fact that every first degree polynomial over the algebraic numbers has an algebraic solution.

First, the binomial coefficients have their combinatorial significance, so actually, what you're doing may seem only algebraical, but it's combinatorial. For small numbers (like 56) it's really most prefferable to use trial and error, because it will lead to solution faster. I can think of some kind of optimisation, where you don't actually try all the numbers. It may be used to tell if some number is binomial. There's a formula for the order (the maximal degree of a prime number p) which divides n!, that is:

You got me!
So... Since 1500 = 37.40 + 20, then

To get the last 2 digits you need the remainder modulo 100. First find the phi of 100:

And if you don't beleive, the number is actually
2310809578111909272693109431184832846484968454396283812529115413319435556973292122101720139716262409469889717513948376726158501486182636383531313869623735751595188198743086350673413093292498741784795660389148326466211372843337723144590341000538769498117226562808444716579845071408688720747381120135479199680471885180482121554984483377022066232113498842614313541610752653690490275184816645775562870080222550532658199952189433551842563700110684899935346509404097452154895288699360903786484615575470777362920177718027703180305669825349020152868844727956235156059960772077214312800556301983539901820176796454299860300131486822074451633519214994291242241850066416567586776526981799888092096865174444248800546127994730384671200620078154936387315118122040519117349396356198197315367766646921156257797160661163192060277301722871797101013527586327674333920080776435765228230385762165404932957246335621254520607306174400378047735376342518713628466946614321497738427647167939078993913147690702992638955965837451910388196619417991842556404018337740923222175380153877003983519741457506625666074023573361964063311318755920947209571433341645962479076131201964416276406387762072361807631460445372486964777059883706070699922193176468574966898852772974365576953262531708962699524486008324366541931862849776343129934761158587798436540362085500749517170402734545433993099089688531543345393342814833105525138762128016370025483881594201769798453765660001

Look at the picture. It is sufficient to prove that

What does "with a replacement" means? Is it you take a ball, observe it and put it back, or what?

Interesting. Like it.
I'll give a hunt.

For a

yes! the answer is 8000000, but that's not permutations!~