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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!
; r: b8 R/ [6 Y6 @首先此方法是在号码对应码的基础上创立的!6 r8 m3 W; I/ ^8 i* y1 z
差5对应码:05 16 27 38 49
1 K* u4 f" O3 }" g w& ^' ]和9对应码:09 18 27 36 45
. k1 j7 s7 }- ~, A& [; Z- Y断组方法中出现的几种形态及解决方法:
4 J6 n; d2 A5 B. P( y$ W- o z# m) t/ U形态一:奖号中3个号码无对应码关系
* c. X+ T9 S/ w! p0 d- ?8 }& G% y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。
+ C( d& H" c# b" x$ u- S4 Q断组步骤:
8 ~6 E2 a. S2 @ U+ H& h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。' n6 a$ F0 O7 _. m9 q8 _2 Q
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。: k3 l3 q5 r8 E3 p: ]" N
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。
% U- n. q0 W0 k0 A/ s# ?6 h( R综合起来:864-291-0357
5 Q* z1 G' L- f- I6 d6 n( v6 F$ s2009251期开奖结果为587,断去第二组号码291正确!!" C) h" @+ A; l, 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)
! }. x+ o) |& ^* u* X/ M例如2009242期开奖结果为156,号码中16出现差5对应关系!( L+ S1 V! Y9 j4 O3 G
断组步骤:) X3 X7 o' r& v$ s4 [2 L
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。
z+ {4 I' R7 w7 ^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。6 A- @: [7 p7 _: R* ?/ \
第三组为剩下号码!' T* x: Y; ^- q' I& u$ d# e
综合起来:106-954-2378- `0 V0 k; Y3 c- |) v- s. Z
2009243期开奖结果为695,断去第三组号码2378正确!!
$ d- e) M! b5 I; }形态三:奖号中出现对子的情况(也就是组三形态)/ F/ x; T% n6 A. I
例如2009253期开奖结果为707,组三形态。
9 ~8 I1 E8 ]7 n, V# h断组步骤: h. a3 f' Z" m9 ^" D( O8 z
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。2 m4 N% c) I: s/ r# W
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 V0 ^! g4 [9 R8 p* `8 ~2 ^/ R
第三组为剩下号码!
* O+ g) C$ F- M( H ^综合起来:057-943-1268- B+ Q0 E- g, ~ V
2009254期开奖结果为008,断去第二组号码943正确!4 |: V) I/ [9 H" N1 O
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