How many 10 digit multiples of 99 are pandigital? Remember, no leading zeros.
All the pandigitals are multiples of 9, so the question is how many are multiples of 11.
If the even digits are as indicated below for those cases where the sum of the even and that of the odd digits differ by a multiple of 11, the sum of the even digits and that of the odd digits are as shown on the same line:
0 1 2 5 9 17 28
0 1 2 6 8 17 28
0 1 3 4 9 17 28
0 1 3 5 8 17 28
0 1 3 6 7 17 28
0 1 4 5 7 17 28
0 2 3 4 8 17 28
0 2 3 5 7 17 28
0 2 4 5 6 17 28
0 4 7 8 9 28 17
0 5 6 8 9 28 17
1 2 3 4 7 17 28
1 2 3 5 6 17 28
1 3 7 8 9 28 17
1 4 6 8 9 28 17
1 5 6 7 9 28 17
2 3 6 8 9 28 17
2 4 5 8 9 28 17
2 4 6 7 9 28 17
2 5 6 7 8 28 17
3 4 5 7 9 28 17
3 4 6 7 8 28 17
There are 22 such cases.
In 11 of the cases there is a zero in the odd positions, including the first position. In each of those cases, there are 5! - 4! = 96 valid permutations of the odd digits and 5! = 120 of the even digits for 96*120=11520 valid permutations of all the digits.
In the other 11 cases, there are 5! valid permutations of each. Then, 120^2 = 14400 is the number of valid permutations of all the digits.
In each of these two sets of cases, there are actually 11 cases, so:
That's (11520 + 14400) * 11 = 285,120 possibilities altogether.
Table produced by:
DefDbl A-Z
Dim crlf$
Private Sub Form_Load()
Form1.Visible = True
Text1.Text = ""
crlf = Chr$(13) + Chr$(10)
For a = 0 To 5
For b = a + 1 To 6
For c = b + 1 To 7
For d = c + 1 To 8
For e = d + 1 To 9
lhs = a + b + c + d + e
rhs = 45 - lhs
If Abs(rhs - lhs) Mod 11 = 0 Then
Text1.Text = Text1.Text & Str(a) & Str(b) & Str(c) & Str(d) & Str(e)
Text1.Text = Text1.Text & Str(lhs) & Str(rhs) & crlf
ct = ct + 1
End If
Next
Next
Next
Next
Next
Text1.Text = Text1.Text & crlf & ct & " done"
End Sub
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Posted by Charlie
on 2016-08-03 14:19:15 |
solution
