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Smallest Perfect Cube Puzzle (Posted on 2015-07-15) Difficulty: 3 of 5
Determine the smallest positive perfect cube whose first three digits (reading from the left) are 201 (in this order) and the last digit is 5.

No Solution Yet Submitted by K Sengupta    
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Solution computer assisted solution | Comment 1 of 2
The program takes the ceiling of the cube root of 201 followed by P zeros, and finds any solutions (cubes with the appropriate characteristics) that have the same number of digits as that number. Due to the way the cube roots are formed, if any exist with that number of digits, they will come first, before larger numbers.

    3   kill "smalcube.txt"
    5   open "smalcube.txt" for output as #2
   10   for P=1 to 999
   20       N=201*10^P
   30       Cr=-int(-(N^(1/3)))
   40       Dv5=-int(-Cr/5)
   50       if Dv5@2=0 then Dv5=Dv5+1
   55       loop
   60        Cr=5*Dv5
   70        N=Cr*Cr*Cr
   80        Ns=cutspc(str(N))
   90        if left(Ns,3)="201" and right(Ns,1)="5" then print N
   91        if left(Ns,3)="201" and right(Ns,1)="5" then print #2,N:inc Ct:if Ct>12 then end
   92        if left(Ns,3)>"201" then goto 100
   93        Dv5=Dv5+2
   95       endloop
  100   next

finds these numbers that begin with 201 and are the smallest perfect cubes of their length and also end with 5. The first is 201745589625.

 201745589625 
 2012306640625 
 2017092147875 
 20112552439875 
 20134747640125 
 20156959163375 
 20179187015625 
 201075567351625 
 201178550994875 
 201281569795125 
 201384623758375 
 201487712890625 
 201590837197875 


  Posted by Charlie on 2015-07-15 16:03:18
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