plese solve these

[sanjay]

Section 1 : Series Transformation
1) If 102101->210212 then 1112112->?

2) if 102101-> 200111 then 1122112->?

3) If 102101->101201 then 1122112->?

[keerthi]

I think the answers are
Section 1 : Series Transformation
1) If 102101->210212 then 1112112->2220220

2) if 102101-> 200111 then 1122112->2221111

3) If 102101->101201 then 1122112->2112211
Edited by: keerthi

[keerthi]

Hi sanjay ,


As per ur request i m posting the explanations.If iam wrong please tell me .


1) If 102101->210212 then 1112112->2220220

if 1=2 , 0=1, 2=0 then 102101 = 210212


similarly 1112112 = 2220220





2) if 102101-> 200111 then 1122112->2221111


I think series is written in increasing order based on no. of times the number is repeated.


In 102101 has one-2 , two-0's and three-1's,


so the increasing order = 200111


similarly 1122112 has four-1's and three-2's


so increasing order = 2221111

[keerthi]

3) If 102101->101201 then 1122112->2112211


here divide the series into two equal parts


102101 = 102 and 101


then reverse both of them 201 and 101


now reverse their positions 101201 is the answer.


similarly 1122112 =112 ,2 & 112 here there is extra no in the middle.


reverse them 211 ,2 , 211


interchange their positions


2112211 is the answer

[keerthi]

I have doubt regarding the procedure . if u know the answers plz post them.