It is currently 22 Sep 2017, 02:44

### 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 June
Open Detailed Calendar

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

Author Message
TAGS:

### Hide Tags

Intern
Joined: 09 Jun 2012
Posts: 1

Kudos [?]: 12 [5], given: 2

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

### Show Tags

06 Aug 2012, 09:23
5
KUDOS
12
This post was
BOOKMARKED
00:00

Difficulty:

55% (hard)

Question Stats:

70% (02:00) correct 30% (02:15) wrong based on 131 sessions

### HideShow timer Statistics

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
[Reveal] Spoiler: OA

Kudos [?]: 12 [5], given: 2

Math Expert
Joined: 02 Sep 2009
Posts: 41682

Kudos [?]: 124392 [5], given: 12078

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

### Show Tags

06 Aug 2012, 09:30
5
KUDOS
Expert's post
5
This post was
BOOKMARKED
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

_________________

Kudos [?]: 124392 [5], given: 12078

GMAT Club Legend
Joined: 09 Sep 2013
Posts: 17594

Kudos [?]: 270 [0], given: 0

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

### Show Tags

26 Aug 2014, 04:33
Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________

Kudos [?]: 270 [0], given: 0

GMAT Club Legend
Joined: 09 Sep 2013
Posts: 17594

Kudos [?]: 270 [0], given: 0

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

### Show Tags

19 Sep 2015, 23:11
Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________

Kudos [?]: 270 [0], given: 0

CEO
Joined: 17 Jul 2014
Posts: 2589

Kudos [?]: 379 [0], given: 177

Location: United States (IL)
Concentration: Finance, Economics
Schools: Stanford '20
GMAT 1: 650 Q49 V30
GPA: 3.92
WE: General Management (Transportation)
Re: A five-digit number divisible by 3 is to be formed using [#permalink]

### Show Tags

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.

Kudos [?]: 379 [0], given: 177

GMAT Club Legend
Joined: 09 Sep 2013
Posts: 17594

Kudos [?]: 270 [0], given: 0

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

### Show Tags

20 Apr 2017, 10:03
Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________

Kudos [?]: 270 [0], given: 0

Manager
Joined: 27 Dec 2016
Posts: 82

Kudos [?]: 26 [0], given: 488

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

### Show Tags

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.

Kudos [?]: 26 [0], given: 488

Math Expert
Joined: 02 Sep 2009
Posts: 41682

Kudos [?]: 124392 [1], given: 12078

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

### Show Tags

20 Sep 2017, 22:47
1
KUDOS
Expert's post
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.
_________________

Kudos [?]: 124392 [1], given: 12078

Manager
Joined: 27 Dec 2016
Posts: 82

Kudos [?]: 26 [1], given: 488

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

### Show Tags

21 Sep 2017, 08:01
1
KUDOS
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!

Kudos [?]: 26 [1], given: 488

Math Expert
Joined: 02 Sep 2009
Posts: 41682

Kudos [?]: 124392 [2], given: 12078

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

### Show Tags

21 Sep 2017, 08:05
2
KUDOS
Expert's post
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.
_________________

Kudos [?]: 124392 [2], given: 12078

Manager
Joined: 27 Dec 2016
Posts: 82

Kudos [?]: 26 [0], given: 488

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

### Show Tags

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

Kudos [?]: 26 [0], given: 488

Re: A five-digit number divisible by 3 is to be formed using   [#permalink] 21 Sep 2017, 08:24
Similar topics Replies Last post
Similar
Topics:
How many numbers with 3 different digits can be formed using only the 3 31 Mar 2015, 02:45
29 How many five-digit numbers can be formed from the digits 0, 17 10 Nov 2016, 07:45
4 How many positive five-digit numbers can be formed with 0,3 and 5 2 28 Mar 2017, 05:03
82 How many five-digit numbers can be formed using digits 27 22 Jan 2017, 07:02
8 A FIVE DIGIT NUMBER DIVISIBLE BY 3 IS TO BE FORMED USING THE 7 22 Mar 2017, 04:42
Display posts from previous: Sort by