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picp3 patterns

Topic closed. 4 replies. Last post 2 years ago by dr san.

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bgonçalves
Brasil
Member #92564
June 9, 2010
2133 Posts
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Posted: December 16, 2014, 6:18 am - IP Logged

hello, now pick 3 was in patterns of letters
aaa
Saab
aac
aad
aae
aBB
acc
add
aee
aBC
abe
ace
bbb
bbc
bcc
bcd
bce
bde
ccc
cee
cde
cdd
dee
dde
ddc
ddd
eee
table
A = 0-1
B = 2-3
C = 4-5
6-7 = d
= 8-9 and
you have put the draws in ascending example last sweepstakes gave 913 = 139 has put in ascending order to see the patterns of the letters = 913 = 139 = = growing abe
only work when Colcar the draw in ascending order ex = 736 = 367
need to put in excel to convert, of course then to assemble bet will be positional example predicted the pattern = ddc = 6-7
the advantage that the quantity of letters and digits amount of
example (dee) = means a type in (6-7) and two digits in the (8-9)
course has to register in the draw for faces of all possible tables above = advantage of these patterns, in ascending order, finds the equivalent letters and the number of digits by letter

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    bgonçalves
    Brasil
    Member #92564
    June 9, 2010
    2133 Posts
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    Posted: December 16, 2014, 6:42 am - IP Logged

    ex=

    000=aaa

    111=aaa

    010= aaa

    101= aaa

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      bgonçalves
      Brasil
      Member #92564
      June 9, 2010
      2133 Posts
      Offline
      Posted: December 16, 2014, 6:44 am - IP Logged

      saab= error     corret= aab

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        bgonçalves
        Brasil
        Member #92564
        June 9, 2010
        2133 Posts
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        Posted: December 16, 2014, 8:37 am - IP Logged

        Hello example draw
        315 = 135 = = rising standard abc
        Abc = has the digits =a= 0.1,  b= 2.3,   c =4.5
        6 digits closing 3 =0,1,2,3,4,5
        123
        124
        125
        134
        135
        145
        234
        235
        245
        345
        Of course in ascending order

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          bgonçalves
          Brasil
          Member #92564
          June 9, 2010
          2133 Posts
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          Posted: December 20, 2014, 6:24 am - IP Logged
           Sub NumberToLetterPattern()  Dim X As Long, Z As Long, Num As String, Numbers As Variant  Numbers = Range("A1", Cells(Rows.Count, "A").End(xlUp))  For X = 1 To UBound(Numbers)    For Z = 1 To 3      Num = Format(Numbers(X, 1), "000")      Mid(Num, Z) = Chr(Int((InStr("0123456789", Mid(Num, Z, 1)) - 1) / 2) + 97)      Numbers(X, 1) = Num    Next  Next  Columns("B").Clear  Range("B1:B" & UBound(Numbers)) = NumbersEnd Sub