A^5+B^5+C^5+D^5+E^5=S
Given that the concatenation ABCDE digits is a 9-digit zero-less pandigital number
and so is the sum S - there is only one sum S to make the above equation true.
Find it.
DefDbl A-Z
Dim crlf$, used(9)
Private Sub Form_Load()
ChDir "C:\VB5 projects\flooble"
Text1.Text = ""
crlf$ = Chr(13) + Chr(10)
Form1.Visible = True
DoEvents
Open "1-9scram.txt" For Output As #2
For a1 = 1 To 5
used(a1) = 1
a$ = LTrim(Str(a1))
For b1 = a1 + 1 To 6
b$ = LTrim(Str(b1))
used(b1) = 1
For c1 = b1 + 1 To 7
c$ = LTrim(Str(c1))
used(c1) = 1
For d1 = c1 + 1 To 8
d$ = LTrim(Str(d1))
used(d1) = 1
For e1 = d1 + 1 To 9
e$ = LTrim(Str(e1))
used(e1) = 1
r$ = ""
For i = 1 To 9
If used(i) = 0 Then
r$ = r$ + Str$(i) ' adds a space and a digit as a string
End If
Next
h$ = r$
Do
pix = 0: prt = 1: ix = 0
pix = ix + 1
ix = InStr(pix, r, " "): If ix = 0 Then ix = Len(r) + 1
aval = Val(a$ + Mid(r, pix, ix - pix))
pix = ix + 1
ix = InStr(pix, r, " "): If ix = 0 Then ix = Len(r) + 1
bval = Val(b$ + Mid(r, pix, ix - pix))
pix = ix + 1
ix = InStr(pix, r, " "): If ix = 0 Then ix = Len(r) + 1
cval = Val(c$ + Mid(r, pix, ix - pix))
pix = ix + 1
ix = InStr(pix, r, " "): If ix = 0 Then ix = Len(r) + 1
dval = Val(d$ + Mid(r, pix, ix - pix))
pix = ix + 1
ix = InStr(pix, r, " "): If ix = 0 Then ix = Len(r) + 1
eval = Val(e$ + Mid(r, pix, ix - pix))
Print #2, aval
Print #2, bval
Print #2, cval
Print #2, dval
Print #2, eval
permute r$
Loop Until r$ = h$
used(e1) = 0
Next
used(d1) = 0
Next
used(c1) = 0
Next
used(b1) = 0
Next
used(a1) = 0
Next
Close 2
Text1.Text = Text1.Text & "done"
End Sub
was used to find all possible arrangements of the digits 1 throught 9 into 5 groups with at least one digit in each group. The groups are arranged in lexicographic order so that a number beginning with a 1 would appear before any beginning with any other digit; those beginning with 2 would appear before any beginning with any higher digit, etc.
It creates a file that feeds into a UBASIC program which does the raising to powers and addition, and compares the sum to see if it is a 9-digit pandigital with no zeros.
The file that feeds into the UBASIC program starts:
1
26
37
48
59
1
26
37
489
5
1
26
37
49
58
1
26
37
498
5
1
26
378
4
59
1
26
378
49
5
1
26
3789
4
5
1
26
379
4
58
1
26
379
48
5
1
26
3798
4
5
1
26
38
4
579
1
26
38
4
597
1
26
38
47
59
1
26
38
479
5
1
26
38
49
57
1
26
38
497
5
1
26
387
4
59
1
26
387
49
5
1
26
3879
4
5
1
26
389
4
57
1
26
389
47
5
1
26
3897
4
5
1
26
39
4
578
1
26
39
4
587
1
26
39
47
58
1
26
39
478
5
1
26
39
48
57
1
26
39
487
5
1
26
397
4
58
1
26
397
48
5
1
26
3978
4
5
1
26
398
4
57
1
26
398
47
5
1
26
3987
4
5
1
267
3
4
589
1
267
3
4
598
1
267
3
48
59
1
267
3
489
5
1
267
3
49
58
1
267
3
498
5
1
267
38
4
59
1
267
38
49
5
1
267
389
4
5
1
267
39
4
58
...
and ends
...
5
61
7
8423
9
5
61
7
843
92
5
61
7
8432
9
5
61
72
8
934
5
61
72
8
943
They are grouped by 5 as every set has 5 numbers. The lowest comes first, so the highest low digit is 5.
The UBASIC program that reads the file and produces the result is:
10 open "1-9scram.txt" for input as #1
20 repeat
30 input #1,A,B,C,D,E
35 a=val(a):b=val(b):c=val(c):d=val(d):e=val(e)
40 v=cutspc(str(a^5+b^5+c^5+d^5+e^5))
50 if len(v)=9 then
60 :good=1
70 :for i=1 to 8
80 :if instr(i+1,v,mid(v,i,1)) > 0 then good=0:cancel for:goto 200:endif
90 :next i
100 :if instr(v,"0")>0 then good=0:endif
110 :if good then
120 :print a,b,c,d,e,v
200 rem
240 until eof(1)
250 close
and it finds
17 26 43 58 9 816725493
meaning 17^5 + 26^5 + 43^5 + 58^5 + 9^5 = 816725493.
Of course permutations of A, B, C, D and E will work also.
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Posted by Charlie
on 2015-03-19 15:29:27 |
computer solution
