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Old 2022-10-31, 07:23   #1
tgan
 
Jul 2015

2616 Posts
Default November 2022

https://research.ibm.com/haifa/ponde...ember2022.html
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Old 2022-10-31, 08:25   #2
SuikaPredator
 
Aug 2022
China

25 Posts
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.

Last fiddled with by SuikaPredator on 2022-10-31 at 08:26 Reason: deleted
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Old 2022-10-31, 08:50   #3
LaurV
Romulan Interpreter
 
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Jun 2011
Thailand

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Not really. While, with the current computing power of a good CPU today, a single line in Pari incrementing the enumerator and denominator alternatively (the silly-stupid way) will give you the EP100 solution in about half hour or so (didn't have the patience and the time, stopped it after few seconds), here (by replacing 2 with the constant they give) it will not work, due to the size of the numbers, you will get quite old going one by one for 100 digits , and will have to do binary search, or come with a more clever algorithm.
Code:
gp > m=1; n=1; while(1, if((zn=(2*n*(n-1)))<(zm=(m*(m-1))),n++,if(zn>zm,m++,print(n", "m);n++)))
1, 1
3, 4
15, 21
85, 120
493, 697
2871, 4060
16731, 23661
97513, 137904
568345, 803761
3312555, 4684660
19306983, 27304197
112529341, 159140520
edit: Hey buddy, grrr, I just found out that you deleted the post and let mine hanging !

Last fiddled with by LaurV on 2022-10-31 at 15:02
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Old 2022-11-01, 03:15   #4
Rubiksmath
 
Sep 2022

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a must be one of the following forms mod 487085 (i think), but considering the fact we are dealing with 100 digit numbers it hardly helps:
0,1,79850,114985,194835,292251,372101,407236.

Last fiddled with by Rubiksmath on 2022-11-01 at 03:22
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Old 2022-11-01, 05:58   #5
Dieter
 
Oct 2017

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Quote:
Originally Posted by Rubiksmath View Post
a must be one of the following forms mod 487085 (i think), but considering the fact we are dealing with 100 digit numbers it hardly helps:
0,1,79850,114985,194835,292251,372101,407236.
You wanted to say "b must be ..." ?
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Old 2022-11-01, 06:06   #6
Rubiksmath
 
Sep 2022

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Quote:
Originally Posted by Dieter View Post
You wanted to say "b must be ..." ?
yes you are right.
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Old 2022-11-01, 08:21   #7
SuikaPredator
 
Aug 2022
China

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Quote:
Originally Posted by LaurV View Post
edit: Hey buddy, grrr, I just found out that you deleted the post and let mine hanging !
I'm sorry, but I discovered my mistake immediately after posting that one. Actually I misread the fraction 1/974170 in the problem statement into 1/2. Maybe the actual problem could also be solved by continued fractions but with some more complex deductions than the 1/2 version.
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Old 2022-11-02, 06:52   #8
tgan
 
Jul 2015

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Default Please explain

"Your goal: find a,b such that the probability for two comfortable socks is exactly \frac{1}{974170} , and such that this is the minimal solution with b having at least 100 digits (minimal with respect to the size of b )"

My understanding is b should be bigger than 10^99 and smaller than any other potential b?
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Old 2022-11-02, 13:28   #9
Dr Sardonicus
 
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Feb 2017
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Quote:
Originally Posted by tgan View Post
"Your goal: find a,b such that the probability for two comfortable socks is exactly \frac{1}{974170} , and such that this is the minimal solution with b having at least 100 digits (minimal with respect to the size of b )"

My understanding is b should be bigger than 10^99 and smaller than any other potential b?
No, just that b is the smallest solution that is greater than or equal to 10^99.

If I did my sums correctly, the smallest possible b is 114985, with a = 117.

(I don't think this is giving away anything useful for answering the stated question.)
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