OG problem solving #176 (number property-Hard)

[gooddice]

1234

1243

1324

.....

....

+4321

 

 

The addition problem above shows four of the 24 different integers that can be formed by using each of the digits 1,2,3, and 4 excactly once in each integer. What is the sum of these 24 integers?

 

Please explain your method.

 

a)24,000

b)26,664

c)40,440

d)60,000

e)66,660

[racsun_81]

IMO : e

[mydreamGMAT]

All the units didits will have the 6 sets of each digit as 1,2,3,4,1,2,3,4,1,2,3,4,1,2,3,4,1,2,3,4,1,2,3,4

 

Hence the sum of units side =1+2+3+4+1+2+3+4+1+2+3+4+1+2+3+4+1+2+3+4+1+2+3+4=60

Same goes for the Tens, hundred and the Thousands Unit

 

Hence the sum wil be 66660 (e)

[visgmat]

Total number of such combinations = 24.

 

Each digit (i.e. 1,2, 3 or 4) would occupy the unit's, ten's, hundred,s or thousand's place 6 times.

 

Hence start with unit's place: 6(4+3+2+1) = 60

Likewise 60 would be the total in the unit's place, hundred's place and thousand's place.

 

If you were to start the addition process, the total would be 66,660.

 

Let me know if the explanation looks ok :)

[gooddice]

OA is E

[serena1]

most significant digit can be 1,2,3 or 4 so 1xxx,2xxx,3xxx,4xxx

 

6 numbers in each category so sum has to be > (6 + 6*2+6*3+6*4)1000

[pawanjsvn]

E it is.