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Intern  Joined: 10 Sep 2008
Posts: 31

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3
10
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

Spoiler: :: OA
E

Source: GMAT Club Tests - hardest GMAT questions

For an integer to be divisible by 3 sum of its digits must be divisible by 3. Then you can have only 15= 1+2+3+4+5 or
12= 0+1+2+4+5 as the sum. In the first case there are 5! permutations, and in the second there are 4*4! because 0 can not be the first digit. Therefore, the answer is 5! + (4*4!) - 216.

My only question here is: How do we come up with 4*4! for the second sum permutations? The 0 that can't be in first place has only four options so I think the logic is..how many times can we arrange one number into four spots? 4 differnt ways. Then 4! comes from how do we arrange 4 numbers into five spots? Help with the 4*4! explanation. I really wish these answers were more explanatory.

Thank you!
Intern  Joined: 17 Sep 2008
Posts: 4
Schools: Chicago Booth, Wharton, MIT, Haas

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3
1
Start from the leftmost digit :

1st digit - can be filled in 4 ways (1,2,4,5).
2nd digit- can then be filled in 4 ways again - 0 , and 3 of the remaining digits from(1,2,4,5).
3rd digit - 3 ways
4th digit - 2 ways
5th digit - 1 way

So the total number of possibilities is 4*4*3*2*1, which is 4*4!
Manager  Joined: 12 Apr 2009
Posts: 128

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1
mmond4 wrote:
Start from the leftmost digit :

1st digit - can be filled in 4 ways (1,2,4,5).
2nd digit- can then be filled in 4 ways again - 0 , and 3 of the remaining digits from(1,2,4,5).
3rd digit - 3 ways
4th digit - 2 ways
5th digit - 1 way

So the total number of possibilities is 4*4*3*2*1, which is 4*4!

I believe the explanation above is wrong.

For an integer to be divisible by 3 the sum of its digits must be divisible by 3. Then you can have only $$15 = 1+2+3+4+5$$ or $$12 = 0+1+2+4+5$$ as the sum. In the first case there are $$5!$$ permutations, and in the second - $$4*4!$$ since 0 can not be the first digit. Therefore, the answer is $$5!+4*4!=216$$ .

che,
dg
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Intern  Joined: 15 Feb 2010
Posts: 8

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the questionn "...divisible by 3, without repeating the digits." <- does this mean that 3 shouldn't be repeated more than once for each digit?
and why do multiply by 4
Intern  Joined: 23 Dec 2009
Posts: 35
Schools: HBS 2+2
WE 1: Consulting
WE 2: Investment Management

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1
This question would've taken me way too long to figure out on the real thing; would've started looking at answer choices... 1) Need the digits to sum to a multiple of 3 (to satisfy "divisible by 3" rule):

12345 does the job
01425 does the job

2) Need to make a 5 digit number:

5! counts the number of ways of arranging 1, 2, 3, 4, 5
5!-4! counts the number of ways of arranging 0, 1, 2, 4, 5 taking out numbers that start with 0 (otherwise we would be counting 4 digit numbers as well!)

3) 5! + (5!-4!) = 120 + 96 = 216
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dczuchta wrote:
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?

For an integer to be divisible by 3 sum of its digits must be divisible by 3. Then you can have only 15= 1+2+3+4+5 or
12= 0+1+2+4+5 as the sum. In the first case there are 5! permutations, and in the second there are 4*4! because 0 can not be the first digit. Therefore, the answer is 5! + (4*4!) - 216.

My only question here is: How do we come up with 4*4! for the second sum permutations? The 0 that can't be in first place has only four options so I think the logic is..how many times can we arrange one number into four spots? 4 differnt ways. Then 4! comes from how do we arrange 4 numbers into five spots? Help with the 4*4! explanation. I really wish these answers were more explanatory.

Thank you!

This question was posted in PS subforum. Below is my solution from there:

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 this 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, 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

Hope it helps.
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Manager  Joined: 24 Aug 2010
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Location: Finland
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Answer is 216. Thanks for the explanation
Intern  Joined: 17 Apr 2010
Posts: 44

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5! + (5! - 4!) - 216 . its E
Manager  Joined: 07 Aug 2010
Posts: 50

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thanks for all the explanations.

started with the sum of digits should be divisible by 3. got stuck in figuring out the factorials _________________
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Manager  Joined: 27 Jul 2010
Posts: 132
Location: Prague
Schools: University of Economics Prague

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tiruraju wrote:
5! + (5! - 4!) - 216 . its E

How you can come to the this formula? I just calculate with: 5! +4*4!
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Intern  Joined: 01 Oct 2010
Posts: 4

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1
Here is the way I solved it hope it helps.....

for numbers 0,1,2,3,4,5 we have to find how many five digit numbers can be formed that's divisible by 3

For a number to be divisible by 3 sum of the digits have to be divisible by 3

let take 1,2,3,4,5 we leave the zero for now.

we see the sum of digits are 1+2+3+4+5=15 all numbers that have these no will be divisible by 3

So no of ways these numbers can be arranged is 5!=120 ...i)

Now again we take zero we have to see the sum of digits must be divisible by 3

for zero only be possible for 0,1,2,4,5 => 0+1+2+4+5=12 rest does not add up to mutiple of 3

No of ways =5!=120 however we have to see the 0 is not the first digit.(then its a 4 digit no.)

[so when 0 is first digit no of permutations are 4!=24
numbers formed without 0 being first digit = 120 -24= 96 ...ii)

I solved using this method and trust me it's really quick .....
Manager  Joined: 21 Nov 2010
Posts: 82

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mmond4 wrote:
Start from the leftmost digit :

1st digit - can be filled in 4 ways (1,2,4,5).
2nd digit- can then be filled in 4 ways again - 0 , and 3 of the remaining digits from(1,2,4,5).
3rd digit - 3 ways
4th digit - 2 ways
5th digit - 1 way

So the total number of possibilities is 4*4*3*2*1, which is 4*4!

Kudos! I had the same question and that was a great explanation.
Manager  Joined: 20 Nov 2010
Posts: 120

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Nice question
The answer is 5! + 4*4! = 216
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CR notes
http://gmatclub.com/forum/massive-collection-of-verbal-questions-sc-rc-and-cr-106195.html#p832142
http://gmatclub.com/forum/1001-ds-questions-file-106193.html#p832133
http://gmatclub.com/forum/gmat-prep-critical-reasoning-collection-106783.html
http://gmatclub.com/forum/how-to-get-6-0-awa-my-guide-64327.html
http://gmatclub.com/forum/how-to-get-6-0-awa-my-guide-64327.html?hilit=chineseburned
Manager  Joined: 01 Nov 2010
Posts: 208
Location: India
Concentration: Technology, Marketing
GMAT Date: 08-27-2012
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WE: Marketing (Manufacturing)

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good question.
ans : 5! + 4.4! = 216
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Director  Joined: 28 Jun 2011
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Bunuel

Excellent explanation as always~
Intern  Status: Preparing for GMAT
Joined: 19 Sep 2012
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Location: India
GMAT Date: 01-31-2013
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when 1 2 3 4 5
5! = 120

when 0 1 2 3 5 excluding zero starting = 120 - 4! =96.

So total will be

120+96=216 . 'E'.

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Chennai, India. Re: M04 # 32   [#permalink] 12 Oct 2012, 05:47
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# M04 # 32

Moderator: Bunuel  