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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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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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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.

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

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New post 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.

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

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New post 20 Sep 2017, 22:47
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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.
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Re: A five-digit number divisible by 3 is to be formed using [#permalink]

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New post 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!

Thanks a lot!

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

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New post 21 Sep 2017, 08:05
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csaluja wrote:
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!

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).

Hope it's clear.
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Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


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

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New post 21 Sep 2017, 08:24
Understood! Thanks a lot Bunuel! Kudos given!

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Re: A five-digit number divisible by 3 is to be formed using   [#permalink] 21 Sep 2017, 08:24
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