At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga.
Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus.
Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get our expert's

answer on brainly

SEE EXPERT ANSWER

Get your **free** account and access **expert** answers to this and **thousands** of other questions.

See more answers at brainly.com

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your **free** account and access **expert** answers to this and **thousands** of other questions

@amistre64 Phi function?

Well, \((k!)!\) would be divisible by all positive numbers under \((k!)!\)

How can I proceed with that information?

Yup, know that property. :-)

Idea of mine.

Or would it be recognizing some pattern?

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

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: http://www.wolframalpha.com/input/?i=2%5E%28n%21%29%21%2C+n%3D1..3

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.

Yup.

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!

Thanks anyway sire!

@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

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.

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.

Ditto!

freaking group theory is again here.

up until now I managed to calculate 2^1000! mod 1000 numerically.

Is there a CRT here too?!

I see, number theory and contest mathematics combined is pretty hard!

quite hard ... i often have difficulty doing problems on most of them.