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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!$ G1 Y$ A. L4 L3 O* Y0 v
首先此方法是在号码对应码的基础上创立的!
- Z+ P1 f% y8 I- ^差5对应码:05 16 27 38 49
! M- V3 L; x" w和9对应码:09 18 27 36 45
, v9 Q @) W9 O断组方法中出现的几种形态及解决方法:' x2 ?# `4 R" Z4 V" V$ @% R9 p
形态一:奖号中3个号码无对应码关系: C# ^* R; ]2 z- {) N
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。
& X2 z# W; y: n断组步骤:5 ^& v! v0 w- n3 q/ x1 b* A
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 ]& k9 g" \+ V; T( n, b3 J) V( 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。
H0 q( r) W, L3 _( G/ b7 E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。5 ~8 W, t( a8 _- l4 h
综合起来:864-291-0357
; ^1 D3 u6 W w6 {1 m, O& Q2009251期开奖结果为587,断去第二组号码291正确!!% Y6 W; U- D1 k# K' J1 d& O- l1 }8 ~
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): p% F; @" o/ n; ^
例如2009242期开奖结果为156,号码中16出现差5对应关系!5 E# S% t- P, G" w8 ?
断组步骤:
. n" g. X% A! R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。
& i; a& f4 O& Y% ~, M5 T( l; s) \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。2 [2 R/ f: T! x! i; S& r5 B% ~9 `
第三组为剩下号码!
[4 v4 ~% [7 `+ i3 F综合起来:106-954-2378
8 T7 Y3 i% E6 p2009243期开奖结果为695,断去第三组号码2378正确!!
4 @0 H& ?. J5 D D形态三:奖号中出现对子的情况(也就是组三形态)
0 h1 _7 L7 x2 }例如2009253期开奖结果为707,组三形态。4 s+ P* n7 E' N0 k
断组步骤:
, B, H' U2 a# J' q/ w3 J: x. F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, P6 Z* K$ N 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。
8 K9 o+ A8 z! _ c* ]第三组为剩下号码!
1 `7 z1 a5 O' L1 Z) T) f" c综合起来:057-943-1268) r1 s. n, w+ B [* P0 k
2009254期开奖结果为008,断去第二组号码943正确!
0 J. y& Z# b: S( z! }" G! i5 V- F |
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