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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]

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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
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Re: A five-digit number divisible by 3 is to be formed using [#permalink]

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New post 06 Aug 2012, 09:30
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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]

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Re: A five-digit number divisible by 3 is to be formed using [#permalink]

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Re: A five-digit number divisible by 3 is to be formed using [#permalink]

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New post 29 Feb 2016, 19:59
we can eliminate the incorrect answer choices by applying logic.

we definitely know that for number to be divisible, we must have the sum of the digits divisible by 3.
so we can use all the digits except for 0.
we can arrange these in 5! ways, or 120 ways.
now...
5+4+1+2+0 = 12, divisible by 3.
we can see that we can arrange these numbers in more than 2 ways. thus, A is eliminated.
B = 210-120 = 90. hold
C = 216-120 = 96 - hold
D - 217-120 = 97 - looks more like a prime number..not divisible by 2, by 3, by 5, by 7.
E - 225-120=105. 105=3*35=3*3*7 - we can't have 7, as we have max 6 digits in total, so definitely out.

between B and C:
90=3*3*2*5
96=4*4*3*2
well, we definitely can't use in the combination for 0,2,1,4,5 - 0 for tens of thousands, thus, we'll have max 4 possible ways to arrange for this digit.
B is out, and C remains.

this all takes less than 1 min to figure out.
Re: A five-digit number divisible by 3 is to be formed using   [#permalink] 29 Feb 2016, 19:59
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