GMAT Question of the Day - Daily to your Mailbox; hard ones only

 It is currently 19 Jan 2019, 15:07

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

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

Events & Promotions

Events & Promotions in January
PrevNext
SuMoTuWeThFrSa
303112345
6789101112
13141516171819
20212223242526
272829303112
Open Detailed Calendar
• FREE Quant Workshop by e-GMAT!

January 20, 2019

January 20, 2019

07:00 AM PST

07:00 AM PST

Get personalized insights on how to achieve your Target Quant Score.
• Free GMAT Strategy Webinar

January 19, 2019

January 19, 2019

07:00 AM PST

09:00 AM PST

Aiming to score 760+? Attend this FREE session to learn how to Define your GMAT Strategy, Create your Study Plan and Master the Core Skills to excel on the GMAT.

M04-32

Author Message
TAGS:

Hide Tags

Math Expert
Joined: 02 Sep 2009
Posts: 52294

Show Tags

15 Sep 2014, 23:23
1
16
00:00

Difficulty:

95% (hard)

Question Stats:

44% (01:39) correct 56% (01:42) wrong based on 164 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

_________________
Math Expert
Joined: 02 Sep 2009
Posts: 52294

Show Tags

15 Sep 2014, 23:23
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 $$120+96=216$$

_________________
Intern
Joined: 03 Apr 2015
Posts: 10

Show Tags

05 Aug 2015, 05:31
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
_________________

Daniela

From 330 - Doing my best to beat the gmat!!

Manager
Joined: 19 Nov 2014
Posts: 68
Location: India
Concentration: Technology, General Management
Schools: ISB '18
WE: Information Technology (Computer Software)

Show Tags

05 Aug 2015, 06:21
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
Joined: 16 Aug 2013
Posts: 11

Show Tags

09 Aug 2015, 21:40
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
Joined: 21 Jan 2015
Posts: 21

Show Tags

17 Aug 2015, 07:11
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)

Show Tags

20 Nov 2015, 10:27
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

Math Expert
Joined: 02 Sep 2009
Posts: 52294

Show Tags

21 Nov 2015, 01:49
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

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
Joined: 02 Nov 2015
Posts: 21

Show Tags

24 Jul 2016, 02:26
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
Joined: 06 Jul 2013
Posts: 53
Location: United States

Show Tags

18 Sep 2018, 18:30
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.
Re: M04-32 &nbs [#permalink] 18 Sep 2018, 18:30
Display posts from previous: Sort by

M04-32

Moderators: chetan2u, Bunuel

 Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®.