IMO E

Any number is divisible by 3 if the sum of the digits of that number is a multiple of 3.

The digits are 0,1,2,3,4,5

Now we can two possible situations were the sum of the digits is a multiple of 3

1) 1,2,3,4,5
The total number of ways of arranging these digits = 5! = 120

2) 0,1,2,4,5
The total number of ways of arranging these digits = 4*4*3*2*1 = 96
We cannot take 0 as the first digit since no number starts with 0. Hence only 4 options for the first place.
Sine we have already used up one digit we are left with only 4 digits. Hence we have only 4 options for the 2nd place.

Hence total number of arrangements = 120+96 = 216

Whats the OA??

Answer is 216.

Digits: 12345 and 01234

12345: 5! arrangements

01234: 4*4! arrangements

Total: 5! + 4*4! = 9*4! = 9*24 = 216

yes ans is 216

mals24 wrote:IMO E

Any number is divisible by 3 if the sum of the digits of that number is a multiple of 3.

The digits are 0,1,2,3,4,5

Now we can two possible situations were the sum of the digits is a multiple of 3

1) 1,2,3,4,5
The total number of ways of arranging these digits = 5! = 120

2) 0,1,2,4,5
The total number of ways of arranging these digits = 4*4*3*2*1 = 96
We cannot take 0 as the first digit since no number starts with 0. Hence only 4 options for the first place.
Sine we have already used up one digit we are left with only 4 digits. Hence we have only 4 options for the 2nd place.

Hence total number of arrangements = 120+96 = 216

Whats the OA??