Welcome Guest
You last visited January 16, 2017, 5:37 am
All times shown are
Eastern Time (GMT-5:00)

# picp3 patterns

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

 Page 1 of 1
bgonÃ§alves
Brasil
Member #92564
June 9, 2010
2133 Posts
Offline
 Posted: December 16, 2014, 6:18 am - IP Logged

hello, now pick 3 was in patterns of letters
aaa
Saab
aac
aae
aBB
acc
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

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

ex=

000=aaa

111=aaa

010= aaa

101= aaa

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

saab= error     corret= aab

bgonÃ§alves
Brasil
Member #92564
June 9, 2010
2133 Posts
Offline
 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

bgonÃ§alves
Brasil
Member #92564
June 9, 2010
2133 Posts
Offline
 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`

 Page 1 of 1