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\[\sum_{k = 1}^{1000} 2^{(k!)!} \equiv n\pmod{1000}\]\(n\) is a threedigit number.
 one year ago
 one year ago
\[\sum_{k = 1}^{1000} 2^{(k!)!} \equiv n\pmod{1000}\]\(n\) is a threedigit number.
 one year ago
 one year ago

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ParthKohliBest ResponseYou've already chosen the best response.0
@amistre64 Phi function?
 one year ago

ParthKohliBest ResponseYou've already chosen the best response.0
Well, \((k!)!\) would be divisible by all positive numbers under \((k!)!\)
 one year ago

ParthKohliBest ResponseYou've already chosen the best response.0
How can I proceed with that information?
 one year ago

amistre64Best ResponseYou've already chosen the best response.0
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
 one year ago

amistre64Best ResponseYou've already chosen the best response.0
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?
 one year ago

ParthKohliBest ResponseYou've already chosen the best response.0
Yup, know that property. :)
 one year ago

ParthKohliBest ResponseYou've already chosen the best response.0
Or would it be recognizing some pattern?
 one year ago

amistre64Best ResponseYou've already chosen the best response.0
it might be patternable ... if your mod reductions simplify down to a repetition what are the first 10 terms .. or 5 terms?
 one year ago

amistre64Best ResponseYou've already chosen the best response.0
1!, 2!, 6!, 24!, 120!, 720!, .... that can be a nightmare, but might be reduceable by phis
 one year ago

ParthKohliBest ResponseYou've already chosen the best response.0
a1 = 2 = 0 (mod 1000) a2 = 4 = 4 (mod 1000) a3 = 2^(720!) = O_O (mod 2)
 one year ago

ParthKohliBest ResponseYou've already chosen the best response.0
a3 = 2^(6!) = 2^(720) sorry
 one year ago

ParthKohliBest ResponseYou've already chosen the best response.0
2^(1!) = 2 = 2 (mod 1000) 2^(2!) = 4 = 4 (mod 1000) 2^(3!) = 64 = 64 (mod 1000) 2^(4!) = 216 (mod 1000) Hmm
 one year ago

ParthKohliBest ResponseYou've already chosen the best response.0
Do you see any pattern? o_O
 one year ago

amistre64Best ResponseYou've already chosen the best response.0
not yet ... when does 2^n equal a multiple of 1000? i know 2^10 = 1024, which is 24 mod 1000
 one year ago

ParthKohliBest ResponseYou've already chosen the best response.0
2^2 = 4 4^3 = 64 64^4 has the last three digits 216 216^5 has last three digits equal to 216
 one year ago

amistre64Best ResponseYou've already chosen the best response.0
lol, dont look at this: http://www.wolframalpha.com/input/?i=2%5E%28n%21%29%21%2C+n%3D1..3
 one year ago

ParthKohliBest ResponseYou've already chosen the best response.0
Superexponential function! :O
 one year ago

amistre64Best ResponseYou've already chosen the best response.0
i see a pollard factorization method that might be useful ... but might not
 one year ago

ParthKohliBest ResponseYou've already chosen the best response.0
Dag nab those fancy names.
 one year ago

amistre64Best ResponseYou've already chosen the best response.0
2^(p1)=1 (mod p) per fermat little thrm.
 one year ago

amistre64Best ResponseYou've already chosen the best response.0
if k! = (p1)q for some interger q, then that reduces to 1 mod p
 one year ago

ParthKohliBest ResponseYou've already chosen the best response.0
Is there anything to do now? :\
 one year ago

ParthKohliBest ResponseYou've already chosen the best response.0
Oh.\[k! \equiv 2^{(p  1)} \pmod{p}\]Is it so?
 one year ago

amistre64Best ResponseYou've already chosen the best response.0
thats fermats little thrm i believe
 one year ago

ParthKohliBest ResponseYou've already chosen the best response.0
Were you able to solve it then?
 one year ago

ParthKohliBest ResponseYou've already chosen the best response.0
Gotta sleep, or rather: gotta sleep on the problem. ;) If you have any solution, feel free to post.
 one year ago

ParthKohliBest ResponseYou've already chosen the best response.0
Thanks for your time!
 one year ago

amistre64Best ResponseYou've already chosen the best response.0
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.
 one year ago

ParthKohliBest ResponseYou've already chosen the best response.0
Thanks anyway sire!
 one year ago

amistre64Best ResponseYou've already chosen the best response.0
ive got my elementary number theory textbook that im looking thru, and might be able to determine something in about a day :)
 one year ago

ParthKohliBest ResponseYou've already chosen the best response.0
@experimentX Any ideas on this one?
 one year ago

experimentXBest ResponseYou've already chosen the best response.2
I don't know ... but let me see.
 one year ago

ParthKohliBest ResponseYou've already chosen the best response.0
What does Mathematica compute?
 one year ago

ParthKohliBest ResponseYou've already chosen the best response.0
Oh . . . what do you mean by overflow? It isn't able to compute it due to the hugeness?
 one year ago

ParthKohliBest ResponseYou've already chosen the best response.0
Darn. I thought Mathematica was smart.
 one year ago

experimentXBest ResponseYou've already chosen the best response.2
Let me try finding the sum of mods first
 one year ago

ParthKohliBest ResponseYou've already chosen the best response.0
Yeah, try it onebyone, maybe we can find some pattern. Actually we did above, but let's see what we get for the further values.
 one year ago

experimentXBest ResponseYou've already chosen the best response.2
Sum[Mod[2^((i!)!), 1000], {i, 1, 1000}]
 one year ago

experimentXBest ResponseYou've already chosen the best response.2
returns overflow in computation ... let me try on matlab.
 one year ago

experimentXBest ResponseYou've already chosen the best response.2
looks like application of Euler totient function. but let me compute it numerically first.
 one year ago

ParthKohliBest ResponseYou've already chosen the best response.0
But it would get harder and harder to compute \(\phi((n!)!)\) as \(n\) increases. :
 one year ago

experimentXBest ResponseYou've already chosen the best response.2
trust me .. I don't know that ... I am least experience with number theory.
 one year ago

experimentXBest ResponseYou've already chosen the best response.2
damn!! even sum of these numbers is huge digit ... returns NaN
 one year ago

ParthKohliBest ResponseYou've already chosen the best response.0
@Callisto Hi, any ideas?
 one year ago

ParthKohliBest ResponseYou've already chosen the best response.0
Try calculating modulo 1000? Just an idea...
 one year ago

experimentXBest ResponseYou've already chosen the best response.2
bug on my code .. :(((
 one year ago

ParthKohliBest ResponseYou've already chosen the best response.0
Well, it's more like pattern recognition, I guarantee you that.
 one year ago

experimentXBest ResponseYou've already chosen the best response.2
yeah i know ... my code was correct ,,, just 2^(,,(..)) is such a huge number. i guess Bigger than grahms number.
 one year ago

ParthKohliBest ResponseYou've already chosen the best response.0
But still lesser than Googol.
 one year ago

experimentXBest ResponseYou've already chosen the best response.2
Grahams number is bigger than googol
 one year ago

experimentXBest ResponseYou've already chosen the best response.2
numerically it's a fiasco ... let me look for totient function.
 one year ago

experimentXBest ResponseYou've already chosen the best response.2
I never used Totient function on my entire life. ... looks like worth it.
 one year ago

experimentXBest ResponseYou've already chosen the best response.2
freaking group theory is again here.
 one year ago

experimentXBest ResponseYou've already chosen the best response.2
up until now I managed to calculate 2^1000! mod 1000 numerically.
 one year ago

experimentXBest ResponseYou've already chosen the best response.2
A very interesting property \[ a^{\phi (n)} = a^{mn} = k \mod n \] So the problem is asking, for what value of \( k \ge n \), \((k!)! = 0 \mod \phi(1000)\)
 one year ago

experimentXBest ResponseYou've already chosen the best response.2
The answer is 454 ... Amazing, Apart for Euler theorem I managed to learn Euclid algorithm, System of linear congruence, Chinese Remainder theorem, ...
 one year ago

ParthKohliBest ResponseYou've already chosen the best response.0
Is there a CRT here too?!
 one year ago

experimentXBest ResponseYou've already chosen the best response.2
oh ... just step for understanding Euler's theorem. I have never done number theory in my entire life.
 one year ago

ParthKohliBest ResponseYou've already chosen the best response.0
I see, number theory and contest mathematics combined is pretty hard!
 one year ago

experimentXBest ResponseYou've already chosen the best response.2
quite hard ... i often have difficulty doing problems on most of them.
 one year ago
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