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A five-digit number divisible by 3 is to be formed using

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A five-digit number divisible by 3 is to be formed using [#permalink] New post 06 Aug 2012, 08:23
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A five-digit number divisible by 3 is to be formed using numerical 0, 1, 2, 3, 4 and 5 without repetition. The total number of ways this can be done is:

A. 122
B. 210
C. 216
D. 217
E. 225
[Reveal] Spoiler: OA
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Re: A five-digit number divisible by 3 is to be formed using [#permalink] New post 06 Aug 2012, 08:30
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miteshsholay wrote:
A five-digit number divisible by 3 is to be formed using numerical 0, 1, 2, 3, 4 and 5 without repetition. The total number of ways this can be done is:

A. 122
B. 210
C. 216
D. 217
E. 225


First step:

We should determine which 5 digits from given 6, would form the 5 digit number divisible by 3.

We have six digits: 0, 1, 2, 3, 4, 5. Their sum=15.

For a number to be divisible by 3 the sum of the digits must be divisible by 3. As the sum of the six given numbers is 15 (divisible by 3) only 5 digits good to form our 5 digit number would be 15-0={1, 2, 3, 4, 5} and 15-3={0, 1, 2, 4, 5}. Meaning that no other 5 from given six will total the number divisible by 3.

Second step:

We have two set of numbers:
1, 2, 3, 4, 5 and 0, 1, 2, 4, 5. How many 5 digit numbers can be formed using these two sets:

1, 2, 3, 4, 5 --> 5! as any combination of these digits would give us 5 digit number divisible by 3. 5!=120.

0, 1, 2, 4, 5 --> here we can not use 0 as the first digit, otherwise number won't be any more 5 digit and become 4 digit. So, desired # would be total combinations 5!, minus combinations with 0 as the first digit (combination of 4) 4! --> 5!-4!=4!(5-1)=4!*4=96

120+96=216

Answer: C.
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Re: A five-digit number divisible by 3 is to be formed using [#permalink] New post 26 Aug 2014, 03:33
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Re: A five-digit number divisible by 3 is to be formed using   [#permalink] 26 Aug 2014, 03:33
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