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Math Expert V
Joined: 02 Sep 2009
Posts: 56244

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2
17 00:00

Difficulty:   95% (hard)

Question Stats: 42% (02:35) correct 58% (02:16) wrong based on 107 sessions

### HideShow timer Statistics 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

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Math Expert V
Joined: 02 Sep 2009
Posts: 56244

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4
7
Official Solution:

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

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 is 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} This set gives $$5!$$ numbers, as any combination of these digits would give us 5 digit number divisible by 3. $$5!=120$$.

{0, 1, 2, 4, 5}. Now, here we cannot 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) $$4!$$: $$5!-4!=4!(5-1)=4!*4=96$$.
Total $$120+96=216$$

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Intern  B
Joined: 03 Apr 2015
Posts: 10

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Hello,

Second step
{0, 1, 2, 4, 5}. Now, here we cannot 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.
Total 120+96=216
Why not :
5*4*2*1 = 40 (3 is not included in this combination)

Thanks a lot,

Daniela
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Daniela

From 330 - Doing my best to beat the gmat!!
Manager  Joined: 19 Nov 2014
Posts: 63
Location: India
Concentration: Technology, General Management
Schools: ISB '18
WE: Information Technology (Computer Software)

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trick is to know which numbers are divisible by 3 ,
whose sum of the digits are divisible by 3.

with 1 , 2 ,3,4, and 5 we can make following numbers :-

5*4*3*2*1 = 120 numbers ;

here the sum is 15 for the digits of the numbers formed formed ;

next are the numbers whose sum of the digits is 12 .

taking 0 , 1 , 2 ,4 ,5 :-
we can make 96 numbers:- 4*4*3*2*1 = 96 ( 4 IN STARTING AS A NUMBER CANT START WITH ZERO )

TOTAL NUMBERS FORMED = 120+96 = 216

OPTION E THEREBY .
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KUDOS pls if you like My Post
Intern  Joined: 16 Aug 2013
Posts: 11

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hi Bunuel

I agree with you about the solution. but what if we calculate all 5-digit numbers and then calculate 4-digit number. subtract them, then divide the result by 3(to find multiples of 3). I think this approach is correct but the result is quite different.
let's see :
1- 5-digit numbers: 5*5*4*3*2=600
2- 4-digit numbers: 5*5*4*3=300
3- (600-300)/3=100
?
Intern  Joined: 21 Jan 2015
Posts: 21

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I think this is a high-quality question and I agree with explanation. I think that this question is great.
Verbal Forum Moderator Joined: 15 Apr 2013
Posts: 181
Location: India
Concentration: General Management, Marketing
GMAT Date: 11-23-2015
GPA: 3.6
WE: Science (Other)

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Hello Bunuel,

I think question is flawed.

First Digit possibility 5 (except zero)
Second digit possibility 5 (Zero + remaining four digits)
Third digit possibility 4
Forth digit possibility 3
Fifth digit possibility 2

Total 5 X 5 X 4 X 3 X 2 = 600

Math Expert V
Joined: 02 Sep 2009
Posts: 56244

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WillGetIt wrote:
Hello Bunuel,

I think question is flawed.

First Digit possibility 5 (except zero)
Second digit possibility 5 (Zero + remaining four digits)
Third digit possibility 4
Forth digit possibility 3
Fifth digit possibility 2

Total 5 X 5 X 4 X 3 X 2 = 600

Have you read the highlighted part below?

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?
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Intern  Joined: 02 Nov 2015
Posts: 21

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I think this is a high-quality question and I agree with explanation. It is a super quality question and probably one of the best questions to understand number arrangements probability. It may not be possible for even smartest of minds to solve it correctly if it comes in 30-37 range of question number.

But even if we are hitting q50 or 51 and compelling GMAT to throw harder questions- will GMAT throw 19 nos. of- 700 level questions?

If somebody could throw some light.
Manager  B
Joined: 06 Jul 2013
Posts: 61
Location: United States
GMAT 1: 720 Q49 V38 ### Show Tags

Hi Bunuel,

Thanks for the elegant solution. Is this solvable via the slot method?

Since only 5 numbers can be used and because 0 cannot be the first digit: 4 x 4 x 3 x 2 x 1 gives me 96.
Intern  B
Joined: 28 Jul 2018
Posts: 11

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TippingPoint93 wrote:
Hi,

Thanks for the elegant solution. Is this solvable via the slot method?

Since only 5 numbers can be used and because 0 cannot be the first digit: 4 x 4 x 3 x 2 x 1 gives me 96.

Hi

I too solved the problem using the slot method. Should be fine.
Intern  B
Joined: 22 Dec 2018
Posts: 3

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I think this is a high-quality question and I agree with explanation. Re M04-32   [#permalink] 03 Jun 2019, 20:30
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# M04-32

Moderators: chetan2u, Bunuel   