Problem 365

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JiminP
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Problem 365

Post by JiminP »

Problem 365 (View Problem)

Would anyone verify these?

(Removed. Sorry for posting sub-results..)
Last edited by JiminP on Sat Jan 14, 2012 6:08 pm, edited 2 times in total.
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Marcus_Andrews
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Re: Problem 365

Post by Marcus_Andrews »

EDIT: Please disregard whatever I said in this post pre-edit.
Last edited by Marcus_Andrews on Sat Jan 14, 2012 6:10 pm, edited 7 times in total.
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Eternalcode
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Re: Problem 365

Post by Eternalcode »

I am curious.
What does it mean by not to post any results?
viewtopic.php?f=50&t=1356
Does it mean that I should not post the final result and it is fine by posting intermediate results?

Thanks
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hk
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Re: Problem 365

Post by hk »

Don't post any results means that you're not supposed to post results at all.
So you're not supposed to post intermediate results (and of course no final result)
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rockstome
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Re: Problem 365

Post by rockstome »

i figured out how to calculate C(m,n)mod(p*q*r) where p,q,r are prime but my sum is wrong.
can anybody confirm that are XXXXXXXX triplets (p,q,r) : p,q,r prime and 1000<p<q<r<5000.
my result is from 10^17 to 10^18
Last edited by rockstome on Tue Jan 17, 2012 8:28 pm, edited 1 time in total.
Thanks for reply
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hk
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Re: Problem 365

Post by hk »

@Rocksome: don't post intermediate results.
Please edit them away.
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Hossein
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Problem 365

Post by Hossein »

The following is the problem:
"The binomial coefficient C(10ˆ18,10ˆ9) is a number with more than 9 billion (9×109) digits.

Let M(n,k,m) denote the binomial coefficient C(n,k) modulo m.

Calculate ∑M(10ˆ18,10ˆ9,p*q*r) for 1000<p<q<r<5000 and p,q,r prime."

1-What are the index, lower bound and upper bound of the summation?
2- Maybe p,q and r , on these two conditions are unique, but if not , are they arbitrary?

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mpiotte
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Re: Problem 365

Post by mpiotte »

Hossein wrote:The following is the problem:
"..."
1-What are the index, lower bound and upper bound of the summation?
2- Maybe p,q and r , on these two conditions are unique, but if not , are they arbitrary?
A) Do not create a new topic when one already exist.
B) The summation is over all triplets (p, q, r) that satisfy the conditions. So you are asked to sum over all possible combinations of valid p, q and r.

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Oliver1978
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Re: Problem 365

Post by Oliver1978 »

I suspect there are 20[...]50 triplets to process. Is that somewhat close?
49.157.5694.1125

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Oliver1978
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Re: Problem 365

Post by Oliver1978 »

It is :wink:
49.157.5694.1125

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