Summer is Coming! Join the Game of Timers Competition to Win Epic Prizes. Registration is Open. Game starts Mon July 1st.

It is currently 16 Jul 2019, 13:52

Close

GMAT Club Daily Prep

Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track
Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

Close

Request Expert Reply

Confirm Cancel

M04-32

  new topic post reply Question banks Downloads My Bookmarks Reviews Important topics  
Author Message
TAGS:

Hide Tags

Find Similar Topics 
Math Expert
User avatar
V
Joined: 02 Sep 2009
Posts: 56244
M04-32  [#permalink]

Show Tags

New post 16 Sep 2014, 00:23
2
17
00:00
A
B
C
D
E

Difficulty:

  95% (hard)

Question Stats:

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

HideShow timer Statistics


Math Expert
User avatar
V
Joined: 02 Sep 2009
Posts: 56244
Re M04-32  [#permalink]

Show Tags

New post 16 Sep 2014, 00:23
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 \(120+96=216\)

Answer: E
_________________
Intern
Intern
User avatar
B
Joined: 03 Apr 2015
Posts: 10
Concentration: General Management, International Business
Re: M04-32  [#permalink]

Show Tags

New post 05 Aug 2015, 06:31
Hello,

Could you please explain me:

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
_________________
Daniela

From 330 - Doing my best to beat the gmat!!
Manager
Manager
avatar
Joined: 19 Nov 2014
Posts: 63
Location: India
Concentration: Technology, General Management
Schools: ISB '18
WE: Information Technology (Computer Software)
GMAT ToolKit User Reviews Badge
Re: M04-32  [#permalink]

Show Tags

New post 05 Aug 2015, 07:21
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 .
_________________
KUDOS pls if you like My Post
Intern
Intern
avatar
Joined: 16 Aug 2013
Posts: 11
GMAT ToolKit User
Re: M04-32  [#permalink]

Show Tags

New post 09 Aug 2015, 22:40
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
Intern
avatar
Joined: 21 Jan 2015
Posts: 21
GMAT ToolKit User
Re M04-32  [#permalink]

Show Tags

New post 17 Aug 2015, 08:11
I think this is a high-quality question and I agree with explanation. I think that this question is great.
Verbal Forum Moderator
avatar
Joined: 15 Apr 2013
Posts: 181
Location: India
Concentration: General Management, Marketing
GMAT Date: 11-23-2015
GPA: 3.6
WE: Science (Other)
GMAT ToolKit User Reviews Badge
Re: M04-32  [#permalink]

Show Tags

New post 20 Nov 2015, 11:27
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

Please assist.
Math Expert
User avatar
V
Joined: 02 Sep 2009
Posts: 56244
Re: M04-32  [#permalink]

Show Tags

New post 21 Nov 2015, 02:49
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

Please assist.


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?
_________________
Intern
Intern
avatar
Joined: 02 Nov 2015
Posts: 21
Re M04-32  [#permalink]

Show Tags

New post 24 Jul 2016, 03:26
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
Manager
avatar
B
Joined: 06 Jul 2013
Posts: 61
Location: United States
GMAT 1: 720 Q49 V38
Re: M04-32  [#permalink]

Show Tags

New post 18 Sep 2018, 19:30
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
Intern
avatar
B
Joined: 28 Jul 2018
Posts: 11
Re: M04-32  [#permalink]

Show Tags

New post 24 Feb 2019, 08:54
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
Intern
User avatar
B
Joined: 22 Dec 2018
Posts: 3
GMAT ToolKit User
Re M04-32  [#permalink]

Show Tags

New post 03 Jun 2019, 20:30
I think this is a high-quality question and I agree with explanation.
GMAT Club Bot
Re M04-32   [#permalink] 03 Jun 2019, 20:30
Display posts from previous: Sort by

M04-32

  new topic post reply Question banks Downloads My Bookmarks Reviews Important topics  

Moderators: chetan2u, Bunuel






cron

Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne