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

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06 Aug 2012, 09:23

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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 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

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

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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.

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

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20 Sep 2017, 18:54

Can anyone please explain to me how do you calculate the total number of ways if 0 is in the first place? I was able to get to 120 for the first step but I am confused with the 0 part.

Can anyone please explain to me how do you calculate the total number of ways if 0 is in the first place? I was able to get to 120 for the first step but I am confused with the 0 part.

{0} - {1, 2, 4, 5}

0 is fixed at the first place. Four remaining numbers, {1, 2, 4, 5} can be arranged in 4! ways.
_________________

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

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21 Sep 2017, 08:01

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Bunuel wrote:

csaluja wrote:

Can anyone please explain to me how do you calculate the total number of ways if 0 is in the first place? I was able to get to 120 for the first step but I am confused with the 0 part.

{0} - {1, 2, 4, 5}

0 is fixed at the first place. Four remaining numbers, {1, 2, 4, 5} can be arranged in 4! ways.

This makes a lot of sense! 4!= 24, I was wondering why did you multiply 24 by 4 to get 96? Could you please explain this part as well? Would greatly appreciate it!

Can anyone please explain to me how do you calculate the total number of ways if 0 is in the first place? I was able to get to 120 for the first step but I am confused with the 0 part.

{0} - {1, 2, 4, 5}

0 is fixed at the first place. Four remaining numbers, {1, 2, 4, 5} can be arranged in 4! ways.

This makes a lot of sense! 4!= 24, I was wondering why did you multiply 24 by 4 to get 96? Could you please explain this part as well? Would greatly appreciate it!

Thanks a lot!

We don't multiply this by 4.

We have: {the total number of combinations} - {the number of combinations we don't want} = {the number of combinations we want}

5! - 4! = 4!(5 - 1) = 4!*4. So, we just factor out 4! from 5! - 4! to get 4!(5 - 1).