alpha_plus_gamma wrote:

1+2+3+4+5 is divisible by 3, we can form 5! numbers from 1...5.

Considering 0 as one of the digits, the remaining 4 digits must be 5,4,2,1 in order to the number be divisible by 3

for this, first digit cant be 0,

so 4*4*3*2*1= 96 possibilities

thus total numbers that can be formed are 5! + 96 = 216 possibilities.

E

Perfect. I missed the possiblity of 0 as one of the digits (excluding the first one) and ended up with the answer as 5!

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