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# How many five-digit numbers can be formed using digits

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How many five-digit numbers can be formed using digits [#permalink]

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22 Oct 2009, 13:20
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How many five digit numbers can be formed using digits 0,1,2,3,4,5, Which are divisible by 3, whithout any of the digits repeating?

A. 15
B. 96
C. 120
D. 181
E. 216
[Reveal] Spoiler: OA
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22 Oct 2009, 13:59
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TheRob wrote:
How many five digit numbers can be formed using digits 0,1,2,3,4,5, Which are divisible by 3, whithout any of the digits repeating?

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=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!=96

120+96=216

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06 Jan 2012, 06:48
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A number is divisible by 3 if sum of its digits is a multiple of 3.

With the given set of digits, there are two possible combinations of 5 digits each-

A. [1,2,3,4,5] No. of possible 5 digit numbers: 5!= 120
B. [0,1,2,4,5] No. of possible 5 digit numbers: 4*4!=96 [the number can't start with a 0]

A+B= 120+96= 216

E
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Last edited by blink005 on 06 Jan 2012, 08:27, edited 1 time in total.
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26 Oct 2009, 05:21
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Only 2 sets are possible

case (1) 1,2,3,4,5
case (2) 0,1,2,4,5.

case (1) : there will 5! ways to form the number = 120

case (2) ; there will 4*4*3*2*1 = 96 ways

So total no.of ways = 120+96 = 216 ways
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29 Apr 2010, 22:51
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By the property of divisibility by 3 i.e "a no: is divisible by 3, if the sum of the digits is divisible by 3"(e.g= 12-->1+2=3)

so from 0,1,2,3,4,5 the set of 5 digit no:s that can be formed which is divisible by 3 are 0,1,2,4,5(sum=12) & 1,2,3,4,5(sum=15)

from first set(0,1,2,4,5) no:s formed are 96 i.e first digit can be formed from any 4 no: except 0, second digit from 4 no: except digit used at first place,3rd from rest 3 , 4th from rest 2 no: and in fifth remaining digit since no repetition allowed.

from second set(1,2,3,4,5) no:s formed are 120 i.e first digit can be formed from any 5 digits, second digit from 4 no: except digit used at first place,3rd from rest 3 , 4th from rest 2 no: and in fifth remaining digit since no repetition allowed.

so total 120+96=216
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06 Feb 2011, 14:18
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ajit257 wrote:
so lets say we were asked a multiple of 5 so in that case we would have to withdraw the digit 5 ..is that correct ?

5 or 0, as 0 is also a multiple of 5.

AGAIN: we have (sum of 6 digits)=(multiple of 3). Question what digit should we withdraw so that the sum of the remaining 5 digits remain a multiple of 3? Answer: the digit which is itself a multiple of 3.

Below might help to understand this concept better.

If integers $$a$$ and $$b$$ are both multiples of some integer $$k>1$$ (divisible by $$k$$), then their sum and difference will also be a multiple of $$k$$ (divisible by $$k$$):
Example: $$a=6$$ and $$b=9$$, both divisible by 3 ---> $$a+b=15$$ and $$a-b=-3$$, again both divisible by 3.

If out of integers $$a$$ and $$b$$ one is a multiple of some integer $$k>1$$ and another is not, then their sum and difference will NOT be a multiple of $$k$$ (divisible by $$k$$):
Example: $$a=6$$, divisible by 3 and $$b=5$$, not divisible by 3 ---> $$a+b=11$$ and $$a-b=1$$, neither is divisible by 3.

If integers $$a$$ and $$b$$ both are NOT multiples of some integer $$k>1$$ (divisible by $$k$$), then their sum and difference may or may not be a multiple of $$k$$ (divisible by $$k$$):
Example: $$a=5$$ and $$b=4$$, neither is divisible by 3 ---> $$a+b=9$$, is divisible by 3 and $$a-b=1$$, is not divisible by 3;
OR: $$a=6$$ and $$b=3$$, neither is divisible by 5 ---> $$a+b=9$$ and $$a-b=3$$, neither is divisible by 5;
OR: $$a=2$$ and $$b=2$$, neither is divisible by 4 ---> $$a+b=4$$ and $$a-b=0$$, both are divisible by 4.

Hope it's clear.
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07 Feb 2011, 03:39
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0,1,2,3,4,5

One digit will have to remain out for all 5 digit numbers;

if 0 is out; Leftover digits will be 1,2,3,4,5 = Sum(1,2,3,4,5)=15.
5! = 120 numbers

if 1 is out; Leftover digits will be 0,2,3,4,5 = Sum(0,2,3,4,5)=14. Ignore(Not divisible by 3)

if 3 is out; Leftover digits will be 0,1,2,4,5 = Sum(0,1,2,4,5)=12.
4*4! = 4*24 = 96

if 4 is out; Leftover digits will be 0,1,2,3,5 = Sum(0,1,2,3,5)=11. Ignore
if 5 is out; Leftover digits will be 0,1,2,3,4 = Sum(0,1,2,3,4)=10. Ignore

Total count of numbers divisible by 3 = 120+96 = 216

Ans: "E"
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Re: How many five-digit numbers can be formed using digits [#permalink]

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28 Dec 2012, 05:54
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TheRob wrote:
How many five digit numbers can be formed using digits 0,1,2,3,4,5, Which are divisible by 3, whithout any of the digits repeating?

A. 15
B. 96
C. 120
D. 181
E. 216

0 + 1 + 2 + 3 + 4 + 5 = 15

To form 5-digit number, we can remove a digit and the sum should still be divisible by 3.

15 - 1 = 14
15 - 2 = 13
15 - 3 = 12 BINGO!
15 - 4 = 11
15 - 5 = 10

Possible = {5,4,3,2,1} and {5,4,0,2,1}

There are 5! = 120 ways to arrange {5,4,3,2,1}
There are 5! - 5!/5 = 96 ways to arrange {5,4,0,2,1} since 0 cannot start the five number digit.

120 + 96 = 216

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Re: How many five-digit numbers can be formed using digits [#permalink]

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13 Jan 2013, 23:40
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I did in 1 min 18 sec.
At first I wanted to choose a set of five digits, but started to worry about the complications with the leading zero.

Then I thought that the last digits could always be chosen in only two ways so as to ensure divisibility by three - however, I quickly realized that I would not get all different digits.

Then I realized that once I get a number I can keep permuting the digits while still getting valid numbers.

In an attempt to avoid the leading zero I tried 12345 and noticed that it was divisible by 3. Thus, I've got 5!=120 answers and immediately eliminated two answers, A and B.

Then I addressed the case of a leading zero. Since I wanted to preserve divisibility by 3, I quickly saw that I could only use 0 instead of 3. Thus, the only other possible set was {0, 1, 2, 4, 5}. I tried adding another 5! and got 240, so the answer was slightly less than that.

After that I knew I had to subtract 4!=24 to account for all the possibilities with a leading zero, which left me with 240-24=216. This is how I do such problems...
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16 Jun 2014, 22:11
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ronr34 wrote:
I tried to do as follows:
take all 5 digit numbers possible : 5 *5*4*3*2
divide by 3 to get all numbers divisible by 3.

What is wrong with this logic?

We cannot do this because we have the asymmetric 0 as one of the digits. The number of 5 digit numbers that can be formed with 0, 1, 2, 3 and 4 is different from the number of 5 digit numbers that can be formed with 1, 2, 3, 4 and 5 (because 0 cannot be the first digit).

Had the digits been 1, 2, 3, 4, 5 and 6, then your method would have been correct.

If 0 is included:
{0, 1, 2, 3, 4} --> 96 5-digit numbers possible with this set.
{0, 1, 2, 3, 5} --> 96 5-digit numbers possible with this set.
{0, 1, 2, 4, 5} --> 96 5-digit numbers possible with this set. - All these numbers are divisible by 3
{0, 1, 3, 4, 5} --> 96 5-digit numbers possible with this set.
{0, 2, 3, 4, 5} --> 96 5-digit numbers possible with this set.
{1, 2, 3, 4, 5} --> 120 5-digit numbers possible with this set. - All these numbers are divisible by 3
The number of 5 digit numbers in these sets is not the same - Sets with 0 have fewer numbers

If 0 is not included:
{1, 2, 3, 4, 5} --> 120 5-digit numbers possible with this set.
{1, 2, 3, 4, 6} --> 120 5-digit numbers possible with this set.
{1, 2, 3, 5, 6} --> 120 5-digit numbers possible with this set. - All these numbers are divisible by 3
{1, 2, 4, 5, 6} --> 120 5-digit numbers possible with this set.
{1, 3, 4, 5, 6} --> 120 5-digit numbers possible with this set.
{2, 3, 4, 5, 6} --> 120 5-digit numbers possible with this set. - All these numbers are divisible by 3
Here exactly 1/3rd of the numbers will be divisible by 3.
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22 Oct 2009, 14:14
Thanks Bunuel for the great explanation . +1 Kudos...

Why don't I borrow your math brain for my GMAT LOL
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23 Oct 2009, 23:06
srini123 wrote:
Thanks Bunuel for the great explanation . +1 Kudos...

Why don't I borrow your math brain for my GMAT LOL

I have been thinking the same for quite some days!
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25 Oct 2009, 13:42
YOu're awesome Bunuel !!
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20 Mar 2010, 02:02
aramjung 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?

(C) 2008 GMAT Club - m04#32

15
96
120
181
216

HOW DO YOU SOLVE THIS, I CAN'T UNDERSTAND IT

For a number to be divisible by 3 sum of its digits should be divisible by 3, so in 0,1,2,3,4,5 set of digits that can be together
1,2,3,4,5, and 0,1,2,4,5 (drop one digit at a time and sum rest of the others to find this )
in first set numbers that can be there = 5*4*3*2*1
in second set = 4*4*3*2*1 (since zero cannot be the last digit)

total = 5*4*3*2*1 + 4*4*3*2*1 = 4! * (5+4) = 24 * 9 = 216, hence 216 is the answer.
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20 Mar 2010, 07:06
thank you sooo much!!!!!!!!
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06 Feb 2011, 13:48
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.

i understood the first part but did not get the second part 15-3={0, 1, 2, 4, 5}. Meaning that no other 5 from given six will total the number divisible by 3. ..Could you please explain it in a little bit more detail. Thanks
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06 Feb 2011, 13:55
ajit257 wrote:
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.

i understood the first part but did not get the second part 15-3={0, 1, 2, 4, 5}. Meaning that no other 5 from given six will total the number divisible by 3. ..Could you please explain it in a little bit more detail. Thanks

The sum of the given digits is already a multiple of 3 (15), in order the sum of 5 digits to be a multiple of 3 you must withdraw a digit which is itself a multiple of 3, otherwise (multiple of 3) - (non-multiple of 3) = (non-multiple of 3).
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06 Feb 2011, 13:59
so lets say we were asked a multiple of 5 so in that case we would have to withdraw the digit 5 ..is that correct ?
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06 Feb 2011, 14:20
Bunuel...awesome ..thanks a ton!
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How many five digit numbers can be formed using digits 0,1,2,3,4 [#permalink]

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13 Nov 2012, 18:31
12345

and

01234

are the two set of numbers which are divisible by 3 when added.

12345->5.4.3.2.1 ways = 120
01234->4.4.3.2.1 ways = 96

Total 216
How many five digit numbers can be formed using digits 0,1,2,3,4   [#permalink] 13 Nov 2012, 18:31

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