• 3 Posts
  • 12 Comments
Joined 1 year ago
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Cake day: June 14th, 2023

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  • God! Which one to choose?

    The electrical company that let a drunk work because “he had 20+ years of experience”, which was rather dubious…

    Or furniture retailers that had delivery drivers…read that again delivery drivers, drive drunk/stoned/coked out, and they were well aware they were.

    Or people trying (Eh, let me be a little more specific…People certifying drivers) to drive in careers where people with mental issues should not be driving…(One guy I knew drove onto the shoulder chasing an oversize load with two bicyclers. He missed them, but god damn)…

    Or my entire fucking family who don’t realize I’m getting bad at driving, and expect me to drive regardless that I might get somebody hurt…(I get a lot of flack for that).


  • I did find that it can be done arbitrarily. Mind is definitely not into writing about it, though, but here’s the gp code I wrote to look it over.

    /*
        There may exist a 0<=t<s such that
        s divides both x and (x+(x%d)*(t*d-1))/d.
    
    
        To show this for solving for divisibility of 7 in 
        any natural number x.
    
        g(35,5,10) = 28
        g(28,5,10) = 42
        g(42,5,10) = 14
        g(14,5,10) = 21
        g(21,5,10) =  7
    */
    
    g(x,t,d)=(x+(x%d)*(t*d-1))/d;
    
    /* Find_t( x = Any natural number that is divisible by s,
               s = The divisor the search is being done for,
               d = The modulus restriction ).
    
        Returns all possible t values.
    */
    
    Find_t(x,s, d) = {
        V=List();
        
        for(t=2,d-1,
            C = factor(g(x,t,d));
            for(i=1,matsize(C)[1],if(C[i,1]==s, listput(V,t))));
            
        return(V);
    }   
    

    One thing that I noticed almost right away, regardless what d is, it seems to always work when s is prime, but not when s is composite.

    Too tired…Pains too much…Have to stop…But still…interesting.



  • Not sure, (“Older and a lot more decrepit” doesn’t mean “younger an a lot more mentally sound”, heh. Do wish I could change that, but meh, I can’t).

    Anyway, I did find a method similar to what you wrote, so I can redefine it in your terms.

    A base 20 number is divisible by 7 if the difference between 8 times the last digit and the remaining digits is divisible by 7.

    Ok, a little description on a base 20 number (Think Mayan and Nahuatl/Aztec numbers). 0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19 should be considered single digits. So a base 10 number, 7*17 = 119 (1*10^2+1*10+9), would be 7*17 = 5:19 in a base 20 system (5*20+19).

    • is 1:8 divisible by 7? (28 in base 10). 8*8 = 3:4. 3:4-1 = 3:3
      • is 3:3 divisible by 7? 8 * 3 = 1:4. 1:4 - 3 = 1:1 (1*20+1 = 21).
    • is 9:2 divisible by 7? (182). 2*8 = 16. 16-9 = 7 Check.

    I’ll just leave that there. So a long weird way of saying, yes, that’s pretty much my reasoning, but not exactly at the same time. As the first message included the base 20 numbers divisible by the base 20 single digits 7, 13, and 17. (Hopefully that came off a little better).

    (Note: Saying “base 20 number[s]” is not important overall. Just being overly descriptive to differentiate between base 10 digits and base 20 digits).