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Palindromic sum (Posted on 2017-08-08) Difficulty: 3 of 5
Express 314159265358979323846 as a sum of positive palindromes, the idea being to use as few summands as you can.

No Solution Yet Submitted by Danish Ahmed Khan    
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Solution The "grunt work" left to the computer (spoiler) | Comment 1 of 3
Take the first half of the original number and append that half's reverse to that half to get a palindrome. Include the middle digit separately between the forward and reverse if an odd number of digits. Lower the middle digit or two of that created palindrome if necessary to make sure the palindrome is less than or equal to the number. Use that palindrome, and subtract from the original number; then continue this process working on what remains each time, until there is nothing left:


    5   open "palinsum.txt" for output as #2
   10   dim Pal(30)
   20   Num$="314159265358979323846"
   30   N=val(Num$)
   40    gosub *Palinate(1)
   50    print Pal(1)+Pal(2)+Pal(3)+Pal(4)+Pal(5)
   60   close #2
  999   end
 1010   *Palinate(Wh)
 1015   P$=left(Num$,ceil(len(Num$)/2)):P1$=P$
 1020   for I=floor(len(Num$)/2) to 1 step -1
 1030      P$=P$+mid(Num$,I,1)
 1040   next
 1050   if val(Num$)-val(P$)<0 then
 1060    :P$=cutspc(str(val(P1$)-1))
 1070    :for I=floor(len(Num$)/2) to 1 step -1
 1080       :P$=P$+mid(Num$,I,1)
 1090    :next
 1100   :endif
 1150   Pal(Wh)=val(P$)
 1160   Num$=cutspc(str(val(Num$)-val(P$)))
 1170   if val(Num$)=0 then
 1180     :for I=1 to Wh
 1190        :print Pal(I):print #2,Pal(I)
 1200     :next
 1210   :else
 1220     :gosub *Palinate(Wh+1)
 1230   :endif
 1260   return

  Posted by Charlie on 2017-08-08 10:13:42
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