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Pi repeats (Posted on 2012-02-29) Difficulty: 3 of 5
The Feynman point is a pattern in pi where the digit 9 repeats six times in a row. The 762nd to 767th digits are all 9. Here is pi to the Feynman point. Each row has 50 digits.

3.

14159265358979323846264338327950288419716939937510
58209749445923078164062862089986280348253421170679

82148086513282306647093844609550582231725359408128
48111745028410270193852110555964462294895493038196

44288109756659334461284756482337867831652712019091
45648566923460348610454326648213393607260249141273

72458700660631558817488152092096282925409171536436
78925903600113305305488204665213841469519415116094

33057270365759591953092186117381932611793105118548
07446237996274956735188575272489122793818301194912

98336733624406566430860213949463952247371907021798
60943702770539217176293176752384674818467669405132

00056812714526356082778577134275778960917363717872
14684409012249534301465495853710507922796892589235

42019956112129021960864034418159813629774771309960
51870721134999999...

1. Find three times where a digit repeats three times consecutively before the Feynman point.

2. The number 999 repeats twice consecutively in 999999. Find two 3-digit numbers that repeat twice consecutively before the Feynman point.

3. Find two 5-digit numbers that appear twice in different places before the Feynman point.

4. In one of the first 7 blocks of 100 digits, there is a 5-digit number ABCDE where ABCDE, ABCD, and BCDE all appear in different places in the same block. Which block of 100 digits is it, and what is ABCDE?

5. The number 999999 is of the form AABBAB, where A=B=9. Find a number of the form AABBAB before the Feynman point where A≠B.

6. Find a 6-digit number ABCDEF where ABCDEF and ABCDDEF both appear at different places before the Feynman point.

See The Solution Submitted by Math Man    
Rating: 5.0000 (1 votes)

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Solution computer solution | Comment 1 of 3

   10   point 165
   20   Pi=cutspc(str(#pi))
   30   Blocks=mid(Pi,3,800)
   40   print Blocks
   50   Feynman=instr(Blocks,"999999"):print Feynman
   55   print "Part 1:"
   60   for Dig=0 to 9
   70    S=cutspc(str(Dig)):S=S+S+S
   75    Ix=0
   80    repeat
   90        Ix=instr(Ix+1,Blocks,S)
  100        if Ix>0 and Ix<Feynman then print S,Ix
  110    until Ix=0
  120   next Dig
  130   print:print "Part 2:"
  140    for St=1 to Feynman-6
  150       S=mid(Blocks,St,3)
  160       if mid(Blocks,St+3,3)=S then print S+S,St
  170    next St
  180   print:print "Part 3:"
  190    for St=1 to Feynman-5
  200       S=mid(Blocks,St,5)
  210       if instr(St+1,Blocks,S)>0 then print S,St,instr(St+1,Blocks,S)
  220    next St
  230   print:print "Part 4:"
  240   for St=1 to 696
  250      S5=mid(Blocks,St,5)
  260      BlockNo=int((St-1)/100)+1
  265      B=mid(Blocks,(BlockNo-1)*100+1,100)
  270      Abcd=left(S5,4):Bcde=mid(S5,2,*)
  275      St2=St-(BlockNo-1)*100
  280      Tst=left(B,St2-1)+" "+mid(B,St2+5,*)
  290      if instr(Tst,Abcd)>0 and instr(Tst,Bcde)>0 then print S5;St:print mid(Blocks,(BlockNo-1)*100+1,100)
  300    next
  310   print:print "Part 5:"
  320   for St=1 to Feynman-6
  330      if mid(Blocks,St,1)=mid(Blocks,St+1,1) and mid(Blocks,St,1)=mid(Blocks,St+4,1) then
  340      :if mid(Blocks,St+2,1)=mid(Blocks,St+3,1) and mid(Blocks,St+2,1)=mid(Blocks,St+5,1) then
  350      :if mid(Blocks,St+1,1)<>mid(Blocks,St+2,1) then print mid(Blocks,St,6),St
  360   next
  370   print:print "Part 6:"
  380   for St=1 to Feynman-6
  390     Abcddef=mid(Blocks,St,4)+mid(Blocks,St+3,3)
  400     if instr(Blocks,Abcddef)>0 and instr(Blocks,Abcddef)<Feynman then print mid(Blocks,St,6),St,Abcddef,instr(Blocks,Abcddef)
  410   next

finds

Part 1:
000      601
111      153
555      177

Part 2:
209209   326
305305   365

Part 3:
23846    16      579
60943    397     551

Part 4:
56482 225
4428810975665933446128475648233786783165271201909145648566923460348610454326648213393607260249141273

Part 5:
006606   307

Part 6:
502841   158    5028841  31

So,

In part 1,
000 begins at 601
111 begins at 153
555 begins at 177

Part 2 is similarly interpreted.

For part 3 both locations are listed for each string.

For part 4, string 56482 begins at 225. The block itself is shown below it, so you can see the 5648 and 6482 in their other locations.

Parts 5 and 6 show the respective locations of the shown strings of digits.


  Posted by Charlie on 2012-02-29 15:35:48
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