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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!
* l; G$ C# m* Y. U首先此方法是在号码对应码的基础上创立的!9 r4 H" @7 ^' f( R1 s' G
差5对应码:05 16 27 38 496 s3 \$ j1 `! z( D# T% H1 x% B
和9对应码:09 18 27 36 45
3 `" A- |7 @! L q* |* K! F1 V断组方法中出现的几种形态及解决方法:
2 p# H' Y( q$ `+ i/ u" k, ^形态一:奖号中3个号码无对应码关系
H# Q( a" J2 } _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。
* P- H+ c) z. F% M9 c* `断组步骤:
- B7 ]+ g" d8 s! R/ L6 B8 _! a! H! R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。5 q! D, ?% s9 C
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。
: L" J$ a9 U1 [: b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。8 d; _, d% J, ~% a& ?: T j) a7 l/ T
综合起来:864-291-0357! J6 C$ J: e/ U
2009251期开奖结果为587,断去第二组号码291正确!!1 t! A) Z' x8 V3 E/ |' A X, S" E. I
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)
# m9 j- _! d4 J; z7 m3 x( F: q例如2009242期开奖结果为156,号码中16出现差5对应关系!, f* }8 d; m: p, ^& n2 T
断组步骤:& `% b/ ^: n: c. L1 ~) `
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。 [% v- W/ B& ~+ ?- f; m
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$ a- B" e9 `8 F
第三组为剩下号码!$ n' }1 m" `9 U
综合起来:106-954-2378
: ^* x d8 u& J( H1 R2009243期开奖结果为695,断去第三组号码2378正确!!' C1 [% B+ t( z0 T. ?* C4 @
形态三:奖号中出现对子的情况(也就是组三形态)4 R/ y* B7 S. y0 }) [9 ~$ l$ i
例如2009253期开奖结果为707,组三形态。
7 e4 o: h2 Y' ^ t( s/ f. i6 M4 Y断组步骤:
& e) H. S* j- \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。
+ X2 v. A( ^. V& b% Q% c p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。, a0 b" v' W+ g. I. [
第三组为剩下号码!
- M8 u: e0 P& B2 [4 |综合起来:057-943-12686 q9 v0 W* A5 P; \: D/ h" o
2009254期开奖结果为008,断去第二组号码943正确!" s- K m( _. H" b+ C& U
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