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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!
" `& `2 L3 ?2 H7 ^& g首先此方法是在号码对应码的基础上创立的!
2 R+ n7 {. }5 K$ N差5对应码:05 16 27 38 490 @' p j; x- c6 L' p3 _4 K
和9对应码:09 18 27 36 45
* V& @& p! D2 B断组方法中出现的几种形态及解决方法:
- g0 x* P% ~- D/ M; ^* e* N形态一:奖号中3个号码无对应码关系
H8 N( c! a, m! F& x( F+ G/ z. L+ f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。5 R% q; Z; }: n+ y6 U7 ^9 I
断组步骤:
/ a, n* n. @; y, O3 ]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。) X8 S( r, Y/ [9 h4 E8 L& e" m! Y
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。
% ~% u: W( }6 k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。+ w5 N8 i: n) d, J7 i
综合起来:864-291-0357
" R4 ^" l4 ^( n1 Z) y/ o: X. P2009251期开奖结果为587,断去第二组号码291正确!!
T( G2 @8 U. Q0 o, D6 _% _* D! y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)7 w- [& ~! U5 \, ^) }7 S' m
例如2009242期开奖结果为156,号码中16出现差5对应关系!3 z& s) n* b! g. v f9 m) V, m
断组步骤:
$ L5 @. |+ K/ R- z- p8 ~# T; r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。4 \9 N( P2 N i4 Z( G; j8 k
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。/ O1 g: W( y1 c' Z7 D' T1 L
第三组为剩下号码!
* ~+ [. Z- S& ~# B' m# o: T综合起来:106-954-2378
9 ]4 g: ]: z7 y$ J; {8 L! k2009243期开奖结果为695,断去第三组号码2378正确!!
! a3 V, L! K0 U9 |形态三:奖号中出现对子的情况(也就是组三形态)1 d6 b1 A/ Z* [0 H& V
例如2009253期开奖结果为707,组三形态。; B: B4 p* Y/ w8 J
断组步骤:- i* Q7 ]7 v# \* {3 @2 e
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。: J& R3 g0 F0 _+ \
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。
. b4 B& ` `( ~* Z# }第三组为剩下号码!
; h; c4 l8 H* t# _' D8 b综合起来:057-943-1268
& |: B1 T' k8 a w. h* q2009254期开奖结果为008,断去第二组号码943正确!
4 H1 q" G L* K3 a8 D4 ? T |
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