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 Four Digits (Posted on 2013-11-05)
You have a number where any digit appears at most twice. The sum of all neighboring four digits in this number is a square number.

What is the maximum possible value for this number?

Example: 205290 is such a number because no digit appears more than twice and 2+0+5+2, 0+5+2+9, and 5+2+9+0 are square numbers.

 No Solution Yet Submitted by Danish Ahmed Khan Rating: 4.0000 (1 votes)

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 computer solution | Comment 2 of 4 |

5   dim Used(9),Digits(20)
10   for Strtbase=1 to 9
20     Nbr=Strtbase
30     Digits(1)=Nbr
40     Used(Nbr)=1
70     Trial4=Nbr
90     L=1
135     Used(Nbr)=0
140   next
150   end
160
180   local I,Did1,Sr,Svtrial4
190   Did1=0
200   for I=0 to 9
210      if Used(I)<2 then
220        :inc Used(I)
225        :Svtrial4=Trial4
230        :Trial4=Trial4+I
235        :if L>3 then Trial4=Trial4-Digits(L-3):endif
240        :Sr=int(sqrt(Trial4)+0.5)
250        :if Sr*Sr=Trial4 or L<3 then
260            :Nbr=Nbr*10+I
265            :Digits(L+1)=I
270            :Did1=1
275            :inc L
290            :Nbr=Nbr\10
295            :dec L
300        :endif
310        :dec Used(I)
315        :Trial4=Svtrial4
320   next I
330   if Did1=0 and Nbr>Max then print Nbr:Max=Nbr
340   return

looks for answers in lexicographic order and when it gets a new high number, reports it. The results are:

10031573
10036793679
10126082698
10961026892
11026802689
11026892680
11026892689
12068206199
17531003679
178010358935
1978100367936
3976300187918
5398530108719
63976300187918
81978100367936

so the highest is 81978100367936.

The sample given, 205290, does not appear, as when the numbers starting with 2 were being done, the numbers starting with 1 had already reached 13 digits.

 Posted by Charlie on 2013-11-05 13:17:32

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