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

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ParthKohli
 3 years ago
Best ResponseYou've already chosen the best response.0@amistre64 Phi function?

ParthKohli
 3 years ago
Best ResponseYou've already chosen the best response.0Well, \((k!)!\) would be divisible by all positive numbers under \((k!)!\)

ParthKohli
 3 years ago
Best ResponseYou've already chosen the best response.0How can I proceed with that information?

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.0well, 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

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.0im 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?

ParthKohli
 3 years ago
Best ResponseYou've already chosen the best response.0Yup, know that property. :)

ParthKohli
 3 years ago
Best ResponseYou've already chosen the best response.0Or would it be recognizing some pattern?

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.0it might be patternable ... if your mod reductions simplify down to a repetition what are the first 10 terms .. or 5 terms?

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.01!, 2!, 6!, 24!, 120!, 720!, .... that can be a nightmare, but might be reduceable by phis

ParthKohli
 3 years ago
Best ResponseYou've already chosen the best response.0a1 = 2 = 0 (mod 1000) a2 = 4 = 4 (mod 1000) a3 = 2^(720!) = O_O (mod 2)

ParthKohli
 3 years ago
Best ResponseYou've already chosen the best response.0a3 = 2^(6!) = 2^(720) sorry

ParthKohli
 3 years ago
Best ResponseYou've already chosen the best response.02^(1!) = 2 = 2 (mod 1000) 2^(2!) = 4 = 4 (mod 1000) 2^(3!) = 64 = 64 (mod 1000) 2^(4!) = 216 (mod 1000) Hmm

ParthKohli
 3 years ago
Best ResponseYou've already chosen the best response.0Do you see any pattern? o_O

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.0not yet ... when does 2^n equal a multiple of 1000? i know 2^10 = 1024, which is 24 mod 1000

ParthKohli
 3 years ago
Best ResponseYou've already chosen the best response.02^2 = 4 4^3 = 64 64^4 has the last three digits 216 216^5 has last three digits equal to 216

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.0lol, dont look at this: http://www.wolframalpha.com/input/?i=2%5E%28n%21%29%21%2C+n%3D1..3

ParthKohli
 3 years ago
Best ResponseYou've already chosen the best response.0Superexponential function! :O

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.0i see a pollard factorization method that might be useful ... but might not

ParthKohli
 3 years ago
Best ResponseYou've already chosen the best response.0Dag nab those fancy names.

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.02^(p1)=1 (mod p) per fermat little thrm.

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.0if k! = (p1)q for some interger q, then that reduces to 1 mod p

ParthKohli
 3 years ago
Best ResponseYou've already chosen the best response.0Is there anything to do now? :\

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

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.0thats fermats little thrm i believe

ParthKohli
 3 years ago
Best ResponseYou've already chosen the best response.0Were you able to solve it then?

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

ParthKohli
 3 years ago
Best ResponseYou've already chosen the best response.0Thanks for your time!

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.0me? 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.

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.0ive got my elementary number theory textbook that im looking thru, and might be able to determine something in about a day :)

ParthKohli
 3 years ago
Best ResponseYou've already chosen the best response.0@experimentX Any ideas on this one?

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.2I don't know ... but let me see.

ParthKohli
 3 years ago
Best ResponseYou've already chosen the best response.0What does Mathematica compute?

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

ParthKohli
 3 years ago
Best ResponseYou've already chosen the best response.0Darn. I thought Mathematica was smart.

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.2Let me try finding the sum of mods first

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

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.2Sum[Mod[2^((i!)!), 1000], {i, 1, 1000}]

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.2returns overflow in computation ... let me try on matlab.

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.2looks like application of Euler totient function. but let me compute it numerically first.

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

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.2trust me .. I don't know that ... I am least experience with number theory.

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.2damn!! even sum of these numbers is huge digit ... returns NaN

ParthKohli
 3 years ago
Best ResponseYou've already chosen the best response.0@Callisto Hi, any ideas?

ParthKohli
 3 years ago
Best ResponseYou've already chosen the best response.0Try calculating modulo 1000? Just an idea...

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.2bug on my code .. :(((

ParthKohli
 3 years ago
Best ResponseYou've already chosen the best response.0Well, it's more like pattern recognition, I guarantee you that.

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

ParthKohli
 3 years ago
Best ResponseYou've already chosen the best response.0But still lesser than Googol.

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.2Grahams number is bigger than googol

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.2numerically it's a fiasco ... let me look for totient function.

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.2I never used Totient function on my entire life. ... looks like worth it.

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.2freaking group theory is again here.

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.2up until now I managed to calculate 2^1000! mod 1000 numerically.

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

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

ParthKohli
 3 years ago
Best ResponseYou've already chosen the best response.0Is there a CRT here too?!

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

ParthKohli
 3 years ago
Best ResponseYou've already chosen the best response.0I see, number theory and contest mathematics combined is pretty hard!

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.2quite hard ... i often have difficulty doing problems on most of them.
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