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38s and Bunches of 1s (Posted on 2017-06-30) Difficulty: 3 of 5
Prove every number in the sequence 38, 381, 3811, 38111, 381111, ... is composite.

See The Solution Submitted by Brian Smith    
Rating: 4.5000 (2 votes)

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Some Thoughts No proof but a pattern found | Comment 1 of 7
 Number and its prime factors:

 38 2 19
 381  3 127
 3811  37 103
 38111  23 1657
 381111  3 127037
 3811111  17 37 73 83
 38111111  233 163567
 381111111  3 3 42345679
 3811111111  37 113 613 1487
 38111111111  31 2333 526957
 381111111111  3 2399 52954163
 3811111111111  37 103003003003
 38111111111111  23333 1633356667
 381111111111111  3 73 1740233384069
 3811111111111111  37 2287 45038479669
 38111111111111111  353 661 163333566667
 381111111111111111  3 3 131 323249458109509
 3811111111111111111  37 114346289 900798827
 38111111111111111111  19 227 541 2857 5716953331
 381111111111111111111  3 879449 1140233 126685261
 
     4   word 255:kill "38and1s.txt":open "38and1s.txt" for output as #2
     5   N=38
    10   for I=2 to 20
    20     print N;:Fctors=fnFactor(N):N=N*10+1
    25     print #2,N;
    30   next
    40   close #2
   999   end
  1000   fnFactor(Num)
  1005   local Fctrs,S,N,I
  1010     N=abs(Num):Ct=0:Fctrs=0:S=""
  1020     if N>0 then Limit=sqrt(N):else Limit=0
  1030     if Limit<>int(Limit) then Limit=int(Limit+1)
  1040     Dv=2:gosub *DivideIt
  1050     Dv=3:gosub *DivideIt
  1060     Dv=5:gosub *DivideIt
  1070     Dv=7
  1080     loop
  1090      if Dv>Limit then goto *Afterloop
  1100      gosub *DivideIt:Dv=Dv+4 '11
  1110      gosub *DivideIt:Dv=Dv+2 '13
  1120      gosub *DivideIt:Dv=Dv+4 '17
  1130      gosub *DivideIt:Dv=Dv+2 '19
  1140      gosub *DivideIt:Dv=Dv+4 '23
  1150      gosub *DivideIt:Dv=Dv+6 '29
  1160      gosub *DivideIt:Dv=Dv+2 '31
  1170      gosub *DivideIt:Dv=Dv+6 '37
  1180      if inkey=chr(27) then S$=chr(27):end
  1190    endloop
  1200    *Afterloop
  1210    if N>1 then S=str(S)+str(N):Ct=Ct+1
  1220   print S
  1221   print #2,S
  1230   return(Ct)
  1240  
  1250   *DivideIt
  1255   Did=0
  1260    loop
  1270     Q=int(N/Dv)
  1280     if Q*Dv=N and N>0 then
  1290       :N=Q:Fctrs=Fctrs+1:S=str(S)+str(Dv):Ct=Ct+1
  1291       :Did=Did+1
  1300       :if N>0 then Limit=sqrt(N):else Limit=0:endif
  1310       :if Limit<>int(Limit) then Limit=int(Limit+1):endif
  1320      :else
  1330      :goto *Afterdo
  1340     :endif
  1350    endloop
  1360    *Afterdo
  1370    return
  
  Wolfram Alpha supplies:
  
   3811111111111111111111 37 393380951 261840342653
   38111111111111111111111 17 73 1372549 22374429543379
   381111111111111111111111 3 55711 158823283 14357381249
   
   Every third one seems to be a multiple of three, and the one after that is a multiple of 37. The one after than would seem to be unpredicatable, but is actually always divisible by 2333..., with one more 3 each occurrence. it's disquised by the fact that 233333 = 353*661 and 2333333 = 19*227*541 and 23333333 =  17*1372549.


  Posted by Charlie on 2017-06-30 09:04:10
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