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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!
* m& n6 S$ v) C7 Q- s首先此方法是在号码对应码的基础上创立的!: j* K8 ^' u5 ^" b7 v! J
差5对应码:05 16 27 38 49
( s& o" {% d8 X- @ P' N# J和9对应码:09 18 27 36 45: v) e9 n: h; G% A6 i( W
断组方法中出现的几种形态及解决方法:
: U6 q' b% y* O o& v) R+ z形态一:奖号中3个号码无对应码关系6 ^8 w! w1 r# T+ a; w
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。
/ d9 J1 f* S/ n5 d0 j$ v/ U$ a! N2 o2 x断组步骤:
& D4 T- }6 c1 G# l( n: P9 ?0 S: P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& c$ W+ V8 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。
* }5 Q" b' w: Y/ [ h; \, 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。
' j5 ]/ k: s r) G* j综合起来:864-291-03578 s9 |7 ^7 b- R$ U! }% ~5 i
2009251期开奖结果为587,断去第二组号码291正确!!
' z* I" r6 W! ]; v8 Z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)
- s) c& |- B9 @3 \0 Q* x例如2009242期开奖结果为156,号码中16出现差5对应关系!: Z h- y7 Z l: ` {; t
断组步骤:
/ [- E: i. D# N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。
3 e/ u7 K' ]3 U& W7 y ?5 U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。
7 ^9 Z: c. C Y' k1 o第三组为剩下号码!8 } G4 u+ r) L* Y: l3 q4 V* L
综合起来:106-954-2378; S1 @( z7 q( T2 @, k1 Q' K
2009243期开奖结果为695,断去第三组号码2378正确!!
% N; W' n, o9 m1 s, ?/ Z+ n4 O形态三:奖号中出现对子的情况(也就是组三形态)
" H( N8 l& {2 i2 m5 ~例如2009253期开奖结果为707,组三形态。
* C5 c$ c2 \6 |6 ]: f$ r9 {断组步骤:4 J \0 R' [' G1 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。) s7 L/ D. p8 l2 h$ R
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。$ g& S L n+ C% l/ r$ i
第三组为剩下号码!
* m2 v+ K& L% o! S综合起来:057-943-1268
+ O0 m" E& `- i- r2009254期开奖结果为008,断去第二组号码943正确!
# y2 ^( S8 M6 Q* R; A. q3 Z7 F3 G |
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