An integer is divisible by 3 if the sum of its digits is divisible by 3. The sum of 0,1,2,3,4,5 is 15. There are six digits, so we have to choose one digit to omit from the list to form a 5 digit number. To make it a multiple of 3, we can only omit 0 or 3.

Omitting 0:

1,2,3,4,5 can be arranged 5!=120 ways.

Omitting 3:

0,1,2,4,5

Any digit can come first except 0. Using slot method: 4*4*3*2*1=96

[spoiler] 120+96=216 Ans: C[/spoiler]

Pete,
A slight variation to this problem.

How many five digit numbers can be formed using the digits 0, 1, 2, 3, 4 and 5 which are divisible by 4, without repeating the digits?

Plz help!!

Here is one option along the same lines:

Any integer whose last two digits are divisible by 4 is divisible by 4 itself. For example 53124=53100+24. 4 is a factor of 53100 because 100 is a multiple of 4, and 4 is obviously a factor of 24.

First, find the two digit pairs that are divisible by 4: 52, 40, 32, 24, 12, 04

52, 32, 24, 12:

First three digits can be selected in 3*3*2=18 ways. any of them can be paired with any of those 4 endings so 18*4=72 numbers in this case

40, 04:

4*3*2=24 ways to choose the first three digits. 24*2=48 numbers in this case

72+48=120 numbers

Double check that, but something along those lines should work.

To make the numbers divisible by 4, the last 2 digits of the number must be divisible by 4.

We can have, 12, 24, 32, 52 (set 1) as the last 2 digits. We are left with 4 digits (including a zero) , therefore, 3 * 3 * 3 * 2 are the number of ways in which this can be done. 18 for each combination listed in set 1. Total1 = 18 * 4 = 72.

We can also have 04, 40 (set 2) as the ending. We now have 4 digits which can be arranged in 4 * 3* 2 ways. (0, 4) can be arranged in 2 ways. Total2 = 4 * 3 * 3 * 2 = 48.

We can have 20 as the ending digits (set 3). That leaves us with 4 digits which can be arranged in 4 * 3 * 2 ways = 24.

Total = 72 + 48 + 24 = 144.

Whoops. forgot 20. thanks, bhavsk9.