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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!# g7 W, A% G; A0 N8 j" p
首先此方法是在号码对应码的基础上创立的!' y \1 u5 i' h) A3 g6 C
差5对应码:05 16 27 38 49( l+ }: k& v4 m
和9对应码:09 18 27 36 45! N6 u2 c" P6 {% }
断组方法中出现的几种形态及解决方法:' B5 L" A6 {2 l- Z; z# w5 m2 r
形态一:奖号中3个号码无对应码关系0 d( F/ K$ e3 q6 g7 ~% i
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。* L) }' Z$ N; ]: d
断组步骤:
% q" J: d9 s" v) [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。
" w: Z6 [& y7 K" u& b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。
: `9 h2 Q3 f4 @0 w" P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。
) _- a3 F4 {8 D \& ^: z综合起来:864-291-0357" R; S" A9 x2 P% `$ q
2009251期开奖结果为587,断去第二组号码291正确!!
' O3 o; f, |. }! X' e1 A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)) Z2 f+ n u' u9 e' }
例如2009242期开奖结果为156,号码中16出现差5对应关系!: _; H$ O2 t/ {: e* u
断组步骤:) u* V* u) O, F( A: n$ X8 N, Y
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。
Q) b# m1 I( _) @% n t& O/ F' l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。: t$ }( O8 E# s' Z. d
第三组为剩下号码!4 ]' V# I& q' m
综合起来:106-954-2378& w- C/ s/ Y! f, A2 r' e, P
2009243期开奖结果为695,断去第三组号码2378正确!!8 _; A! t1 X# z1 {( Q
形态三:奖号中出现对子的情况(也就是组三形态)
/ K9 ~7 f5 G. d$ N1 a( b例如2009253期开奖结果为707,组三形态。' m5 Z" I; A2 A/ t2 b! V7 a5 |# S
断组步骤:) h3 v A4 n: F: `4 w
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。
; O0 R0 j8 j3 t; b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。
5 P3 M+ f' ]7 F- I" J0 A第三组为剩下号码!, G C0 ~/ _* C1 b
综合起来:057-943-1268
2 D ]) v& H; x4 y5 ~1 T2009254期开奖结果为008,断去第二组号码943正确!4 j4 {; }* N! Y1 }2 S, x
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