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Ricky wrote:

Given a and b (not just a/b), there is no way to relate them.

I wanted to say given all of L1, L2, L3, a, b, except 1 of them.

LQ wrote:

Listen to me, is there anything outside even though knowing everything is boring?

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.

knowing everything is boring...

(At least bores me )

So here's the BIG BANG - you already KNOW EVERYTHING, but you just don't know that you know it

the problem was rather trivial after all

Which problem?

ohh, I understand now, you're talking about the exponents! But then these functions will no longer be polynomilas. I don't have any clue about the structure of such numbers, and it's likely that there's no theory of these at all, because before even thinking about the structure of the set of roots of such all such functions, you need some information about these roots - bounds, number (i suspect there may be infinite number of roots of such a "polynomial" with "finite" degree).

I don't think dealing in general with numbers like

But when we take the exponents to be rational, you may take some common divisor of the rationalexponents and after a suitable substitution, you're again in the good old algebraic field.

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

the solution of

is given by:

hm interesting, but i'm stuck.

first of all, notice this (it may actually be the key to the problem if someone uses some strange trigonometric identities):

now, let:

0<y<1 (i like the open intervals)

the equation bacomes

Now, this has a root in the interval we wanted (because of the signs at 0 and 1), but I don't know how to find it analytically...

Actually, plying whit it a little, I noticed that z = -1/4 is a solution to

blah, surely the solution will be some involving tirgonommetric identities.

OK, here's the whole thruth.

Take the the sums of squares. The sequence starts like this:

0,1,5,14,30...

And we want to determine how is this sequence generated. There's a method, maybe it's called the difference method, which may help us, but ONLY if the generator of the sequence is a polynomial.

We put the numbers in rows like this:

```
0 1 5 14 30
1 4 9 16
3 5 7
2 2
```

where under two numbers is their difference. If the generator is polinomial, this process will eventually lead us to a row in which all the numbers are the same (or, which is the same, to a row in which all the numbers are 0)

Call the furst number on the first row

, the first number on the second - , and so on till the first number of the nth row - . Then we use these coefficients in the "magic" formula:We just have to keep in mind that the sequence is starting at index null, so the first term in our sequence corresopnds to .

In the case of sums of squares, we get:

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:

We may also take advantage of it by factoring our number into prime divisors. In this case,

56=7*8

From this and the formula, we may deduce

, which, accidently leads to solution instantly, else you may play arranging the divisors in some way.

JaneFairfax wrote:

krassi_holmz wrote:4|150

It does???!!!

You got me!

So... Since 1500 = 37.40 + 20, then

Blah, and since i don't want to multiply by 9 3times, I'll use not Euler, but Carmichael theorem, first find:

And by Carmichael's theorem, that is:

Hence, the last 2 digits are 01, because

Jane, you ruined my first proof. Happy now!;)

It would've been just fine if 100 were a prime...

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

Now, Euler theorem gives you:

, and 4|150, so the last two digits are 01.

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

2310809578111909272693109431184832846484968454396283812529115413319435556973292122101720139716262409469889717513948376726158501486182636383531313869623735751595188198743086350673413093292498741784795660389148326466211372843337723144590341000538769498117226562808444716579845071408688720747381120135479199680471885180482121554984483377022066232113498842614313541610752653690490275184816645775562870080222550532658199952189433551842563700110684899935346509404097452154895288699360903786484615575470777362920177718027703180305669825349020152868844727956235156059960772077214312800556301983539901820176796454299860300131486822074451633519214994291242241850066416567586776526981799888092096865174444248800546127994730384671200620078154936387315118122040519117349396356198197315367766646921156257797160661163192060277301722871797101013527586327674333920080776435765228230385762165404932957246335621254520607306174400378047735376342518713628466946614321497738427647167939078993913147690702992638955965837451910388196619417991842556404018337740923222175380153877003983519741457506625666074023573361964063311318755920947209571433341645962479076131201964416276406387762072361807631460445372486964777059883706070699922193176468574966898852772974365576953262531708962699524486008324366541931862849776343129934761158587798436540362085500749517170402734545433993099089688531543345393342814833105525138762128016370025483881594201769798453765660001

And in neater form:

Look at the picture. It is sufficient to prove that

.To show this, notice AO=DO=a half of the diagonal. The angle AOD is 90degs, so we end up with 2 congruent triangles, namely APO and ODQ. That means that OQ = AP, and by Pythagorean theorem,

mathsyperson wrote:

(Except in the boring case when A=0 and nothing happens ever)

No, you don't mean this - the zero exceptions give color to maths. A space becomes funny when we find some distinguishable elements of it. The zero case often is called trivial, but it's not trivial at all - it's foundamental.

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

*relative* difference between the two sticks - the numerator and denominator). The last thing is that just before the sticks become with same length, the one is twice the other.

Interesting. Like it.

I'll give a hunt.

For a

to be an integer it must be that {x,y} is a solution to the diopantine equaitonFirst let's investigate this diophantine: Let

So the general solutions of (1) in integers are:

Try to continue from here.

I don't think that should be theorem - just a result...

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

Are these r children given flu jabs?