varun

• Jul 3rd, 2005

 

38664

ROHAN

• Mar 7th, 2006

 

starting with 1,2 & 3 series of no.

1230   2031  3012

1320   2013  3102       TOTAL SUM= 38664

1023   2103  3201

1032   2130  3021

1302   2310  3120

1203   2301  3210

santhosh_sandy

 Profile Answers by santhosh_sandy 

• Jul 4th, 2011

 

in addition to them
we have to add
0123
0321
0132
0213
0312
0231


these are also digits dude!!! so i think  your answer (38664) must be added with 1332
which leads to 39996 would be the right answer!!!  
what do u say guys??

superstardlx

 Profile Answers by superstardlx 

• Jul 4th, 2011

 

total number of such numbers = 18 

sum of all digits in units place :

0*2 ( 3 in thousands place)

1*2

2*2

0*2 (2 in thousands place)

1*2

3*2

0*2 (1 in thousands place)

2*2

3*2

------

24

4 in units place...2 carry

now tens place :

((1 +2) + (0 + 2) + (0 + 1)) + ((1+3) + (0+3) + (0 + 1)) + ((0+1) + (0+2) + (1+2)) 

----------------------------------      --------------------------------    -------------------------------

3 in thousands place                  2 in thousands place             1 in thousands place

=6 + 8  +6 = 20

20 + carry = 22

2 in tens place 2 carry

hundreds place :

this is similar to tens place

=20

20+ carry=22

2 in hundreds place...2 carry

thousands place :

6 combinations for each digit in thousands place (as 3! = 6)

6*3 + 6*2 + 6*1 =36

36 + carry = 38

therefore, sum = 38224