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#1 |
Jul 2015
110002 Posts |
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#2 |
Jan 2017
3×29 Posts |
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Has anyone been able to solve this? I wrote a program to find all numbers with the given number of relative primes, but found no exact match for the divisor sum among those. Closest I got to the target sum in absolute value was this:
Code:
sage: x=7943551318529981932577436079331984148138176 sage: factor(x) 2^6 * 7 * 13 * 29 * 1223 * 144323 * 586543 * 7695239 * 59035244685044657 sage: euler_phi(x) 3031634148236289733373855928919180891127808 sage: sum(divisors(x))-x 12142617410592093155511288408870474224367424 By the way the phrasing of the question talks about "natural numbers smaller than 278" etc, but seems to exclude 0 - this would be ambiguous without the example values. |
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#3 | ||
"Rashid Naimi"
Oct 2015
Remote to Here/There
111110000002 Posts |
![]() Quote:
Pari-gp code: Code:
counter=0 for(i=2,278,{ if(gcd(278,i)<2,counter=counter+1); }) counter 1 is not a prime number (at this point in history) and thus not coprime to any number. and 0 is not coprime to any number: gcd(278,0) = 278 It looks like the Puzzle-Master is considering 1 as coprime to the solution. I fail to see how 0 is being relevant. ETA: Quote:
So 1 is a coprime by convention/definition. But 0 is still irrelevant as far as I can see. ![]() corrections are appreciated. Last fiddled with by a1call on 2020-06-01 at 21:56 |
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#4 | |
Jan 2017
3×29 Posts |
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But yeah 0 is not relevant, I wasn't really thinking when writing that... Last fiddled with by uau on 2020-06-01 at 21:54 |
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#5 |
"Rashid Naimi"
Oct 2015
Remote to Here/There
26·31 Posts |
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We cross posted. I stand corrected on 1. But, I still think it's a matter of convention and not logic.
![]() ETA: In the same way 1 was considered to be a prime number at some point until someone with authority decided it was not. And Retina has his own logic. ETA II: 1 is the same number it has always been. ETA III: Needless to say that I still consider Pluto to be a planet. It's the same heavenly body it was before someone decided to demote it. Last fiddled with by a1call on 2020-06-01 at 22:06 |
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#6 | |
Romulan Interpreter
Jun 2011
Thailand
52×7×53 Posts |
![]() Quote:
![]() ![]() ![]() A lot of work you did there, man... ![]() gcd, if, for,... grrrr (edit: even if not talking about the futile accolade) haha Code:
gp>eulerphi(278) %1=138 Last fiddled with by LaurV on 2020-06-02 at 16:14 |
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#7 |
"Rashid Naimi"
Oct 2015
Remote to Here/There
198410 Posts |
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Thank you LaurV, But I actually new about that function.
One day in elementary school my geometry teacher asked if anyone could prove some particular theorem. I raised my hand described a valid proof which took me about 20 minutes to describe in the 45 minutes class. At this point the teacher replied, that is correct but it is equivalent to turning your arm behind your head and then taking the spoon to your mouth when eating. Afterwards he proved the theorem with 3 sentences. We grow up on the outside but we are the same person we were when we were kids. Kind of like Pluto. ![]() |
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#8 | |||
"Rashid Naimi"
Oct 2015
Remote to Here/There
26·31 Posts |
![]() Quote:
![]() Google >> define coprime numbers >> Click-Top-Result >> Quote:
Google >> define prime numbers >> Click-Top-Result >> Quote:
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#9 |
Jan 2017
3·29 Posts |
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#10 |
Oct 2017
103 Posts |
![]() Big numbers, not many solvers. Can anyone check the following - itβs not the solution, of course: x = 2**7 * 5 * 23 * 127 * 659 * 53323 * 1876187 * 97544836889 * 665320793909 = 7998766649128898059663516612687535453720960 euler_phi(x) = the wanted value sum(divisors(x)) - x = 12142697391577851168337274092012830083559040 I would like to know, if at least one prime of the solution > 2**64...π |
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#11 | |
"Ben"
Feb 2007
3,371 Posts |
![]() Quote:
Code:
>> x=2^7* 5 * 23 * 127 * 659 * 53323 * 1876187 * 97544836889 * 665320793909 7998766649128898059663516612687535453720960 >> totient(x) 3031634148236289733373855928919180891127808 >> sigma(x,1)-x 12142697391577851168337274092012830083559040 >> |
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