TheRob wrote:

How many five digit numbers can be formed using digits 0,1,2,3,4,5, Which are divisible by 3, whithout any of the digits repeating?

A. 15

B. 96

C. 120

D. 181

E. 216

Key Knowledge:

(Divisibility Rules, Number Theory)

math-number-theory-88376.htmlFor a number to be divisible by 3, the sum of it's digits must be a multiple of 3

Case Method:

Case#1:\(\sum 0,1,2,4,5 = 12\)

\(5!=120\) = Number of arrangements of 5 digits

Less:

\(4!=24\) = Number of arrangements of 4 digits

(Removes arrangements where 0 is the leading digit (i.e. fix 0 as leading, arrange the remaining 4 digits 4! ways))

\(5!-4!=96\)

Case #2:\(\sum 1,2,3,4,5 = 15\)

\(5!=120\) = Number of arrangements of 5 digits

\(120+96=216\)