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# of 5-digit numbers

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by mehrasa » Wed Oct 12, 2011 5:25 am
How many five-digit numbers can be formed using the digits 0, 1, 2, 3, 4 and 5 which are divisible by 3, without repeating the digits?
(A) 15
(B) 96
(C) 120
(D) 181
(E) 216


OA: E
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by CappyAA » Wed Oct 12, 2011 5:39 am
For all numbers that are divisible by 3, their sums must be divisible by 3. The only two combinations of 5 numbers from the set that are divisible by 3 are:

5, 4, 3, 2, 1 --> sum = 15
5, 4, 2, 1, 0 --> sum = 12

According to the rule, all combinations of the above numbers must be divisible by 3 because their sums are divisible by 3.

In the first group, there are 5!, or 120 combinations.

In the second group, 0 cannot be in the first slot. So there are 4*4!, or 96 combinations.

96 + 120 = 216

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by sam2304 » Wed Oct 12, 2011 5:48 am
For the number to be a multiple of 3 the sum of digits should be a multiple of 3

Out of 6 digits we can remove 0 or 3 for the nos to be multiple of 3. Any other combination would not result in a multiple of 3.

W/o 0 as part of the number - sum of 1 to 5 = 15 be it any order

_ _ _ _ _
5 4 3 2 1 = 120 ways

With 0 as part of the number - sum of the digits = 12

_ _ _ _ _
4 4 3 2 1 = 96

The first digit cannot take a 0 so it has only 4 ways.

total = 120 + 96 = 216 ways.
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