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相信大家对断组不陌生吧!断组形态3-3-4,也就是说每期开奖号会开在一组或两组中,断去其中的一组或两组,通常大多数情况下是断去一组!
/ `1 _+ T$ Z, o首先此方法是在号码对应码的基础上创立的!
% I/ c2 P) e5 \3 O5 o& W差5对应码:05 16 27 38 49
0 ?# B5 L& r7 g4 Z+ t/ X) B2 w y和9对应码:09 18 27 36 45% F, {% J$ f& Z7 o
断组方法中出现的几种形态及解决方法:$ v5 U; J. X: _$ F. z
形态一:奖号中3个号码无对应码关系
3 M3 Z* r' `% a' {% O/ E: M例如2009250期开奖结果为319,号码中3个数字无对应码关系,这种形态断组比较简单。
' R, |1 x; [, P( f: @断组步骤:% j" s [6 w5 a6 a U4 D4 i. l5 _4 c
第一组用319的对应号码组成!3对应8,1对应6,9对应4,得到第一组号码为864。, r# e$ r! ?& {# l8 ~; f4 P3 V
第二组号码用第一组号码进行和10,8+2=10,6+4=10,4+6=10,得到第二组号码246,但是这里问题出现了46两号已经在第一组中出现过,那么这里我们就使用46的对应码9和1,最终得到第二组号码291。 i+ q* J* }0 m/ g
第三组号码简单了,除去第一、第二组的号码剩下的四个号码为第四组,得到号码0357。
4 b) W# [6 y8 F" Z/ E综合起来:864-291-0357
3 y! M8 A; u7 i j: G* m) \- U2009251期开奖结果为587,断去第二组号码291正确!!
. x) B, N- M4 m4 c: N3 p: |形态二:奖号中3个号码中有2个号码出现对应码关系(这里对应码关系我们只考虑差5对应关系,和9不考虑)- L# \2 ^1 K5 F! ~
例如2009242期开奖结果为156,号码中16出现差5对应关系! @3 R4 W& P8 ?/ I5 x
断组步骤:3 N" v) H6 ]9 |" w6 O
第一组保留奖号中的对应码16加上5的对应码0组成第一组号码,得第一组号码为106。
9 Q* a3 q l% O第二组用第一组号码进行和10,1+9=10,遇0使用0的对应码,6+4=10,得到第二组号码954。9 l' ]9 T5 s4 w( F6 w1 G
第三组为剩下号码!
" O k- z7 B3 l Y4 O( b K综合起来:106-954-2378
! m- [: }0 {8 `1 D6 h) R2 |4 R2009243期开奖结果为695,断去第三组号码2378正确!!
1 H2 }8 p4 ^2 V$ q t; _形态三:奖号中出现对子的情况(也就是组三形态): x) N) b, T: q4 ~/ l7 w
例如2009253期开奖结果为707,组三形态。7 d- k4 }* Q* \8 \
断组步骤:" R$ B* D ~! H$ b/ o n; D% h% G
第一组保留对子中出现的号码7,加上0的对应码5,得到第一组号码为057。- M# Q- Z' \6 D7 g& h
第二组用第一组进行和10,遇0时用0的对应码,这里问题出现了0的对应码5在第一组中已经出现过,这里我们使用0的和9对应码也就是9,5+5=10,而5在第一组中也出现了,那么用5的对应码0,而0在第一组中也出现了那就只能用5的和9对应码4了,7+3=10,得到第二组号码为943。
' F' \( @) y4 A1 p0 c0 h7 W第三组为剩下号码!
# a. O* t3 D/ i9 w) a( }综合起来:057-943-1268
; ~/ K% d- h! _0 ~2009254期开奖结果为008,断去第二组号码943正确!
2 u# ^8 W) P4 m S1 Z |
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