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\[\sum_{k = 1}^{1000} 2^{(k!)!} \equiv n\pmod{1000}\]\(n\) is a three-digit number.

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@amistre64 Phi function?
Well, \((k!)!\) would be divisible by all positive numbers under \((k!)!\)
How can I proceed with that information?

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Other answers:

well, the sum of mods reduces each term .. might be useful. 23 = k (mod 3) k = 2; since 3(7) + 2 = 23 23 = 3+3+3+3+3+3+3+2 3+3+3+3+3+3+3+2 = k (mod 3) 0+0+0+0+0+0+0+2 = k (mod 3) k = 2
im not sure how applicable the phi function might be, is it stated in the question by chance? or was that just an idea of yours?
Yup, know that property. :-)
Idea of mine.
Or would it be recognizing some pattern?
it might be patternable ... if your mod reductions simplify down to a repetition what are the first 10 terms .. or 5 terms?
1!, 2!, 6!, 24!, 120!, 720!, .... that can be a nightmare, but might be reduceable by phis
a1 = 2 = 0 (mod 1000) a2 = 4 = 4 (mod 1000) a3 = 2^(720!) = O_O (mod 2)
a3 = 2^(6!) = 2^(720) sorry
2^(1!) = 2 = 2 (mod 1000) 2^(2!) = 4 = 4 (mod 1000) 2^(3!) = 64 = 64 (mod 1000) 2^(4!) = 216 (mod 1000) Hmm
Do you see any pattern? o_O
not yet ... when does 2^n equal a multiple of 1000? i know 2^10 = 1024, which is 24 mod 1000
2^2 = 4 4^3 = 64 64^4 has the last three digits 216 216^5 has last three digits equal to 216
no, 576.
lol, dont look at this:
Super-exponential function! :-O
i see a pollard factorization method that might be useful ... but might not
Dag nab those fancy names.
2^(p-1)=1 (mod p) per fermat little thrm.
if k! = (p-1)q for some interger q, then that reduces to 1 mod p
Hmm... :-|
Is there anything to do now? :-\
Oh.\[k! \equiv 2^{(p - 1)} \pmod{p}\]Is it so?
thats fermats little thrm i believe
Were you able to solve it then?
Gotta sleep, or rather: gotta sleep on the problem. ;-) If you have any solution, feel free to post.
Thanks for your time!
me? no, i got no idea what the solution would be for this. number theory is too far behind and too little used for me to think clearly about it.
Thanks anyway sire!
ive got my elementary number theory textbook that im looking thru, and might be able to determine something in about a day :)
@experimentX Any ideas on this one?
I don't know ... but let me see.
What does Mathematica compute?
overflow :(((
Oh . . . what do you mean by overflow? It isn't able to compute it due to the hugeness?
Yep ...
Darn. I thought Mathematica was smart.
Let me try finding the sum of mods first
Yeah, try it one-by-one, maybe we can find some pattern. Actually we did above, but let's see what we get for the further values.
Sum[Mod[2^((i!)!), 1000], {i, 1, 1000}]
returns overflow in computation ... let me try on matlab.
looks like application of Euler totient function. but let me compute it numerically first.
Yup, that.
But it would get harder and harder to compute \(\phi((n!)!)\) as \(n\) increases. :-|
trust me .. I don't know that ... I am least experience with number theory.
Me too :-(
damn!! even sum of these numbers is huge digit ... returns NaN
@Callisto Hi, any ideas?
Try calculating modulo 1000? Just an idea...
bug on my code .. :(((
Well, it's more like pattern recognition, I guarantee you that.
yeah i know ... my code was correct ,,, just 2^(,,(..)) is such a huge number. i guess Bigger than grahms number.
But still lesser than Googol.
Or is it?
Grahams number is bigger than googol
numerically it's a fiasco ... let me look for totient function.
Fiasco indeed.
I never used Totient function on my entire life. ... looks like worth it.
freaking group theory is again here.
up until now I managed to calculate 2^1000! mod 1000 numerically.
A very interesting property \[ a^{\phi (n)} = a^{mn} = k \mod n \] So the problem is asking, for what value of \( k \ge n \), \((k!)! = 0 \mod \phi(1000)\)
The answer is 454 ... Amazing, Apart for Euler theorem I managed to learn Euclid algorithm, System of linear congruence, Chinese Remainder theorem, ...
Is there a CRT here too?!
oh ... just step for understanding Euler's theorem. I have never done number theory in my entire life.
I see, number theory and contest mathematics combined is pretty hard!
quite hard ... i often have difficulty doing problems on most of them.

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