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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!
, X7 k2 t1 b3 k7 u# ^首先此方法是在号码对应码的基础上创立的!
5 r% w! ?# P$ l" T9 D) s" N差5对应码:05 16 27 38 49 g4 C' ^" v+ a6 G4 V- | p
和9对应码:09 18 27 36 45' I, |0 B* V: w/ p7 `8 P9 s9 x
断组方法中出现的几种形态及解决方法:
$ K4 D, b& I. F- S6 J/ L2 C形态一:奖号中3个号码无对应码关系
: y4 d j7 M8 Q! E; I" j4 @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。
$ B# B; I, G- U7 A/ g断组步骤:0 c0 ~# V( m' f- r+ Z
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。
/ [) F, D4 v- t( t* B1 t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。2 T5 I) h9 H& I3 k6 Y
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。
& i; u8 h/ ?/ I综合起来:864-291-0357
0 K+ M. E6 b; Z2 O+ p" P$ ]2009251期开奖结果为587,断去第二组号码291正确!!
8 B5 t6 Q8 A x2 `, b- y+ K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)4 u) t$ x1 F$ G/ O- y3 ~9 f
例如2009242期开奖结果为156,号码中16出现差5对应关系!1 g' u+ |* S1 z8 L ~
断组步骤:
$ G' |9 K' i# {: J1 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。! [' g! E- e* r: 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。
. o. e. n) V6 C5 L- f& _. z6 U5 J第三组为剩下号码!
/ m" \, x1 }% {# u, R综合起来:106-954-23784 {5 K: A8 X0 s: e: k: A2 C4 R5 Q
2009243期开奖结果为695,断去第三组号码2378正确!!
F, l7 l G1 T- n7 E形态三:奖号中出现对子的情况(也就是组三形态)
" y% o4 K" ~; n( P7 v" R+ |例如2009253期开奖结果为707,组三形态。 ]& V* r4 e* e& ^
断组步骤:! {) q# ^/ [# ^. |0 C
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。! u! ~6 G% P& i6 n6 j/ _
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。
" J( q ?. f8 h3 a; O6 V第三组为剩下号码!+ V, ^' I0 y& X5 ^
综合起来:057-943-1268$ k, U- O i/ J Z' g0 f7 S! N! w
2009254期开奖结果为008,断去第二组号码943正确!" t1 t* x6 F! u0 {/ l9 o0 O2 t
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