divisibility

AndrewTom

How many of the three-digit numbers which can be formed using different digits from $0, 1, 2, 3$ and $5$ are divisible by $6$ ?

**Posted:** Fri Nov 06, 2009 1:55 am

Gen8 · Poincare Conjecture Offline Joined: 26 May 2008 Posts: 241

Well, I guess the answer is: $12$ .

The total number of three-digit numbers which can be formed by using the digits $0, 1, 2 , 3, 5$ are: $100$ . Barring the hundredths place--which can only be filled in 4 ways, the remaining two places can be filled in 5 ways each. (4*5*5)

For a 3 digit number to be divisible by $6$ , the 3 digit number should be divisible by both $3$ and $2$ . I surmise, you know the divisibility rules of both $3$ and $2$ . For a number (3 digit or any number) to be divisible by $3$ -- the sum of the digits should be divisible by $3$ and for a number to be divisible by $2$ -- the units place of the 3 digit number should be either $0, 2, 4, 6, 8$ .

After calculating, I observed that the only possible cases are as given as follows: $120, 210, 300, 150, 510, 330, 102, 222, 522, 252, 312, 132$ .

**Posted:** Fri Nov 06, 2009 3:49 am

AndrewTom

Yes, but only seven of the twelve use different digits.

**Posted:** Fri Nov 06, 2009 5:22 am