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# Find the sum of all the four digit numbers formed using the

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Manager
Joined: 08 Sep 2010
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Find the sum of all the four digit numbers formed using the [#permalink]

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24 Oct 2010, 10:03
1
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Question Stats:

86% (00:49) correct 14% (00:28) wrong based on 27 sessions

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Find the sum of all the four digit numbers formed using the digits 2,3,4 and 5 without repetition.

I will post OA and OE tomorrow.

Kudos [?]: 331 [1], given: 21

Manager
Joined: 08 Sep 2010
Posts: 223

Kudos [?]: 331 [12], given: 21

Location: India
WE 1: 6 Year, Telecom(GSM)
Re: Nice question and a good way to solve.... [#permalink]

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24 Oct 2010, 10:22
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We can form a total of 4! or 24 numbers.When we add all these numbers ,let us look at the contribution of of the digit 2 to the sum.

When 2 occurs in the thousand place in a particular number,its contribution in the total will be 2000.the number of numbers that can be formed
with 2 in the thousand place is 3! i.e 6 numbers.Hence when 2 is in the thousands place its contribution to the sum is 3! * 2000

Similarly when 2 occurs in the hundreds place, its contribution to the sum is 3! * 200

Similarly when 2 occurs in the tenth place, its contribution to the sum is 3! * 20

Similarly when 2 occurs in the unit place, its contribution to the sum is 3! * 10

The total contribution of 2 to the sum is 3! *(2000+200+20+1)=3!*2222
In a similar manner ,the contribution of 3,4, and 5 to sum will respectively be 3!*3333,3!*4444 and 3!*5555

Hence total sum using the aove four digits 3!*(2222+3333+4444+5555) i.e 3! *(2+3+4+5) * 1111

Now we can generalize the above

If all the possible n digit numbers using n distinct digits are formed ,the sum of all the numbers so formed is equal to
(n-1)! * ( sum of the n digits ) *( 1111...n times)

Consider giving me KUDOS if this is useful.

Kudos [?]: 331 [12], given: 21

CEO
Status: Nothing comes easy: neither do I want.
Joined: 12 Oct 2009
Posts: 2754

Kudos [?]: 1925 [1], given: 235

Location: Malaysia
Concentration: Technology, Entrepreneurship
Schools: ISB '15 (M)
GMAT 1: 670 Q49 V31
GMAT 2: 710 Q50 V35
Re: Nice question and a good way to solve.... [#permalink]

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24 Oct 2010, 10:24
1
KUDOS
ankitranjan wrote:
Find the sum of all the four digit numbers formed using the digits 2,3,4 and 5 without repetition.

I will post OA and OE tomorrow.

If u find this post useful ,consider giving me KUDOS.

Along with the question, do post the options.

if you keep the digit 2 at the one's digit. The total numbers that can be formed is 3! = 6
=> 2 occurs at one's digit 6 times. Similarly all the 4 numbers occurs at one's digit 6 times.

Similar all the 4 numbers occurs at all the 4 digits - one's,ten's, hundred's, thousand's- 6 times.

=> total sum = ( 1000+100+10+1 ) *( 2+3+4+5)*6

= 1111*14*6 = 1111*84 = 93324
_________________

Fight for your dreams :For all those who fear from Verbal- lets give it a fight

Money Saved is the Money Earned

Jo Bole So Nihaal , Sat Shri Akaal

GMAT Club Premium Membership - big benefits and savings

Gmat test review :
http://gmatclub.com/forum/670-to-710-a-long-journey-without-destination-still-happy-141642.html

Kudos [?]: 1925 [1], given: 235

Math Expert
Joined: 02 Sep 2009
Posts: 42578

Kudos [?]: 135425 [2], given: 12692

Re: Nice question and a good way to solve.... [#permalink]

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24 Oct 2010, 10:33
2
KUDOS
Expert's post
5
This post was
BOOKMARKED
ankitranjan wrote:
Find the sum of all the four digit numbers formed using the digits 2,3,4 and 5 without repetition.

I will post OA and OE tomorrow.

If u find this post useful ,consider giving me KUDOS.

Similar questions:

permutation-88357.html?hilit=digit%20using
manhattan-arithimetic-chapter-1-need-help-88869.html?hilit=digit%20using#p670812
sum-of-all-3-digit-nos-with-88864.html?hilit=digit%20using
m04-70602.html?hilit=digit%20using
testing-number-properties-91007.html?hilit=digit%20using
can-someone-help-94836.html?hilit=repetition

Actually there is the direct formula for this kind of problems. Of course it's better to understand the concept, then to memorize the formula but in case someone is interested here it is:

1. Sum of all the numbers which can be formed by using the $$n$$ digits without repetition is: (n-1)!*(sum of the digits)*(111…..n times).

2. Sum of all the numbers which can be formed by using the $$n$$ digits (repetition being allowed) is: $$n^{n-1}$$*(sum of the digits)*(111…..n times).

Hope it helps.
_________________

Kudos [?]: 135425 [2], given: 12692

CEO
Status: Nothing comes easy: neither do I want.
Joined: 12 Oct 2009
Posts: 2754

Kudos [?]: 1925 [0], given: 235

Location: Malaysia
Concentration: Technology, Entrepreneurship
Schools: ISB '15 (M)
GMAT 1: 670 Q49 V31
GMAT 2: 710 Q50 V35
Re: Nice question and a good way to solve.... [#permalink]

### Show Tags

24 Oct 2010, 10:41
Yes understanding is very imp as pointed by bunuel.

The question becomes interesting if total sum of all the 3 digits number is asked.
Then the give formula is modified.

the 111...n becomes 111....m where m = 3 for 3 digit number.
_________________

Fight for your dreams :For all those who fear from Verbal- lets give it a fight

Money Saved is the Money Earned

Jo Bole So Nihaal , Sat Shri Akaal

GMAT Club Premium Membership - big benefits and savings

Gmat test review :
http://gmatclub.com/forum/670-to-710-a-long-journey-without-destination-still-happy-141642.html

Kudos [?]: 1925 [0], given: 235

Manager
Joined: 27 Jul 2010
Posts: 190

Kudos [?]: 47 [0], given: 15

Location: Prague
Schools: University of Economics Prague
Re: Nice question and a good way to solve.... [#permalink]

### Show Tags

02 Feb 2011, 02:52
gurpreetsingh wrote:
Yes understanding is very imp as pointed by bunuel.

The question becomes interesting if total sum of all the 3 digits number is asked.
Then the give formula is modified.

the 111...n becomes 111....m where m = 3 for 3 digit number.

Do I get it?
When you have to sum all the 3 digit numbers out of 2,3,4,5 than the answer is:

without repetition:
(3-1)!*14*111= 3108

with repetition:
(3^2)*14*111=13986
_________________

You want somethin', go get it. Period!

Kudos [?]: 47 [0], given: 15

Senior Manager
Joined: 13 Aug 2012
Posts: 457

Kudos [?]: 569 [0], given: 11

Concentration: Marketing, Finance
GPA: 3.23
Re: Find the sum of all the four digit numbers formed using the [#permalink]

### Show Tags

20 Dec 2012, 04:29
How many options do I have? 4*3*2*1
How many digits will be repeated equally in the options? 4

$$=\frac{4*3*2*1}{4}*14*1111=84*1111=64*1111=93,324$$
_________________

Impossible is nothing to God.

Kudos [?]: 569 [0], given: 11

Intern
Joined: 24 Apr 2013
Posts: 1

Kudos [?]: 5 [0], given: 23

Location: United States
GMAT Date: 01-01-2014
GPA: 3.5
WE: Project Management (Consulting)
Re: Find the sum of all the four digit numbers formed using the [#permalink]

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27 Apr 2013, 07:21
Hi Bunuel,
So in the below question if we apply the formula shouldn't we consider 4 as n?
Find the sum of all the four digit numbers formed using the digits 2,3,4 and 5 without repetition.
(4-1)*(2+3+4+5)*1111
Aybige

Kudos [?]: 5 [0], given: 23

Math Expert
Joined: 02 Sep 2009
Posts: 42578

Kudos [?]: 135425 [0], given: 12692

Re: Find the sum of all the four digit numbers formed using the [#permalink]

### Show Tags

28 Apr 2013, 02:28
aybige wrote:
Hi Bunuel,
So in the below question if we apply the formula shouldn't we consider 4 as n?
Find the sum of all the four digit numbers formed using the digits 2,3,4 and 5 without repetition.
(4-1)!*(2+3+4+5)*1111
Aybige

Correct, but you've missed factorial (!). It should be: (4-1)!*(2+3+4+5)*(1111)=93324.
_________________

Kudos [?]: 135425 [0], given: 12692

Re: Find the sum of all the four digit numbers formed using the   [#permalink] 28 Apr 2013, 02:28
Display posts from previous: Sort by
First, the content structure of the provided text is as follows: It starts with a timestamp, followed by a section header and a brief description of a service, and then a question is presented.

Now, the analysis based on the provided format:
single_question
- the text contains a single mathematical question: "find the sum of all the four digit numbers formed using the digits 2,3,4 and 5 without repetition."
Given a text passage, determine:
    1. If it contains a mathematical question (including word problems or calculations)
    2. If an answer to that mathematical question is provided in the text

First, describe the content structure you found in 1-2 sentences.

Then provide your analysis in the following format:
TYPE: <one of the following values>
- NO_QUESTION
- SINGLE_QUESTION
- SINGLE_QUESTION_WITH_ANSWER
- MULTIPLE_QUESTIONS
- MULTIPLE_QUESTIONS_WITH_ANSWERS
system
It is currently 12 Dec 2017, 23:29

### GMAT Club Daily Prep

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# Events & Promotions

###### Events & Promotions in June
Open Detailed Calendar

# Find the sum of all the four digit numbers formed using the

Author Message
TAGS:

### Hide Tags

Manager
Joined: 08 Sep 2010
Posts: 223

Kudos [?]: 331 [1], given: 21

Location: India
WE 1: 6 Year, Telecom(GSM)
Find the sum of all the four digit numbers formed using the [#permalink]

### Show Tags

24 Oct 2010, 10:03
1
KUDOS
5
This post was
BOOKMARKED
00:00

Difficulty:

(N/A)

Question Stats:

86% (00:49) correct 14% (00:28) wrong based on 27 sessions

### HideShow timer Statistics

Find the sum of all the four digit numbers formed using the digits 2,3,4 and 5 without repetition.

I will post OA and OE tomorrow.

Kudos [?]: 331 [1], given: 21

Manager
Joined: 08 Sep 2010
Posts: 223

Kudos [?]: 331 [12], given: 21

Location: India
WE 1: 6 Year, Telecom(GSM)
Re: Nice question and a good way to solve.... [#permalink]

### Show Tags

24 Oct 2010, 10:22
12
KUDOS
3
This post was
BOOKMARKED
We can form a total of 4! or 24 numbers.When we add all these numbers ,let us look at the contribution of of the digit 2 to the sum.

When 2 occurs in the thousand place in a particular number,its contribution in the total will be 2000.the number of numbers that can be formed
with 2 in the thousand place is 3! i.e 6 numbers.Hence when 2 is in the thousands place its contribution to the sum is 3! * 2000

Similarly when 2 occurs in the hundreds place, its contribution to the sum is 3! * 200

Similarly when 2 occurs in the tenth place, its contribution to the sum is 3! * 20

Similarly when 2 occurs in the unit place, its contribution to the sum is 3! * 10

The total contribution of 2 to the sum is 3! *(2000+200+20+1)=3!*2222
In a similar manner ,the contribution of 3,4, and 5 to sum will respectively be 3!*3333,3!*4444 and 3!*5555

Hence total sum using the aove four digits 3!*(2222+3333+4444+5555) i.e 3! *(2+3+4+5) * 1111

Now we can generalize the above

If all the possible n digit numbers using n distinct digits are formed ,the sum of all the numbers so formed is equal to
(n-1)! * ( sum of the n digits ) *( 1111...n times)

Consider giving me KUDOS if this is useful.

Kudos [?]: 331 [12], given: 21

CEO
Status: Nothing comes easy: neither do I want.
Joined: 12 Oct 2009
Posts: 2754

Kudos [?]: 1925 [1], given: 235

Location: Malaysia
Concentration: Technology, Entrepreneurship
Schools: ISB '15 (M)
GMAT 1: 670 Q49 V31
GMAT 2: 710 Q50 V35
Re: Nice question and a good way to solve.... [#permalink]

### Show Tags

24 Oct 2010, 10:24
1
KUDOS
ankitranjan wrote:
Find the sum of all the four digit numbers formed using the digits 2,3,4 and 5 without repetition.

I will post OA and OE tomorrow.

If u find this post useful ,consider giving me KUDOS.

Along with the question, do post the options.

if you keep the digit 2 at the one's digit. The total numbers that can be formed is 3! = 6
=> 2 occurs at one's digit 6 times. Similarly all the 4 numbers occurs at one's digit 6 times.

Similar all the 4 numbers occurs at all the 4 digits - one's,ten's, hundred's, thousand's- 6 times.

=> total sum = ( 1000+100+10+1 ) *( 2+3+4+5)*6

= 1111*14*6 = 1111*84 = 93324
_________________

Fight for your dreams :For all those who fear from Verbal- lets give it a fight

Money Saved is the Money Earned

Jo Bole So Nihaal , Sat Shri Akaal

GMAT Club Premium Membership - big benefits and savings

Gmat test review :
http://gmatclub.com/forum/670-to-710-a-long-journey-without-destination-still-happy-141642.html

Kudos [?]: 1925 [1], given: 235

Math Expert
Joined: 02 Sep 2009
Posts: 42578

Kudos [?]: 135425 [2], given: 12692

Re: Nice question and a good way to solve.... [#permalink]

### Show Tags

24 Oct 2010, 10:33
2
KUDOS
Expert's post
5
This post was
BOOKMARKED
ankitranjan wrote:
Find the sum of all the four digit numbers formed using the digits 2,3,4 and 5 without repetition.

I will post OA and OE tomorrow.

If u find this post useful ,consider giving me KUDOS.

Similar questions:

permutation-88357.html?hilit=digit%20using
manhattan-arithimetic-chapter-1-need-help-88869.html?hilit=digit%20using#p670812
sum-of-all-3-digit-nos-with-88864.html?hilit=digit%20using
m04-70602.html?hilit=digit%20using
testing-number-properties-91007.html?hilit=digit%20using
can-someone-help-94836.html?hilit=repetition

Actually there is the direct formula for this kind of problems. Of course it's better to understand the concept, then to memorize the formula but in case someone is interested here it is:

1. Sum of all the numbers which can be formed by using the $$n$$ digits without repetition is: (n-1)!*(sum of the digits)*(111…..n times).

2. Sum of all the numbers which can be formed by using the $$n$$ digits (repetition being allowed) is: $$n^{n-1}$$*(sum of the digits)*(111…..n times).

Hope it helps.
_________________

Kudos [?]: 135425 [2], given: 12692

CEO
Status: Nothing comes easy: neither do I want.
Joined: 12 Oct 2009
Posts: 2754

Kudos [?]: 1925 [0], given: 235

Location: Malaysia
Concentration: Technology, Entrepreneurship
Schools: ISB '15 (M)
GMAT 1: 670 Q49 V31
GMAT 2: 710 Q50 V35
Re: Nice question and a good way to solve.... [#permalink]

### Show Tags

24 Oct 2010, 10:41
Yes understanding is very imp as pointed by bunuel.

The question becomes interesting if total sum of all the 3 digits number is asked.
Then the give formula is modified.

the 111...n becomes 111....m where m = 3 for 3 digit number.
_________________

Fight for your dreams :For all those who fear from Verbal- lets give it a fight

Money Saved is the Money Earned

Jo Bole So Nihaal , Sat Shri Akaal

GMAT Club Premium Membership - big benefits and savings

Gmat test review :
http://gmatclub.com/forum/670-to-710-a-long-journey-without-destination-still-happy-141642.html

Kudos [?]: 1925 [0], given: 235

Manager
Joined: 27 Jul 2010
Posts: 190

Kudos [?]: 47 [0], given: 15

Location: Prague
Schools: University of Economics Prague
Re: Nice question and a good way to solve.... [#permalink]

### Show Tags

02 Feb 2011, 02:52
gurpreetsingh wrote:
Yes understanding is very imp as pointed by bunuel.

The question becomes interesting if total sum of all the 3 digits number is asked.
Then the give formula is modified.

the 111...n becomes 111....m where m = 3 for 3 digit number.

Do I get it?
When you have to sum all the 3 digit numbers out of 2,3,4,5 than the answer is:

without repetition:
(3-1)!*14*111= 3108

with repetition:
(3^2)*14*111=13986
_________________

You want somethin', go get it. Period!

Kudos [?]: 47 [0], given: 15

Senior Manager
Joined: 13 Aug 2012
Posts: 457

Kudos [?]: 569 [0], given: 11

Concentration: Marketing, Finance
GPA: 3.23
Re: Find the sum of all the four digit numbers formed using the [#permalink]

### Show Tags

20 Dec 2012, 04:29
How many options do I have? 4*3*2*1
How many digits will be repeated equally in the options? 4

$$=\frac{4*3*2*1}{4}*14*1111=84*1111=64*1111=93,324$$
_________________

Impossible is nothing to God.

Kudos [?]: 569 [0], given: 11

Intern
Joined: 24 Apr 2013
Posts: 1

Kudos [?]: 5 [0], given: 23

Location: United States
GMAT Date: 01-01-2014
GPA: 3.5
WE: Project Management (Consulting)
Re: Find the sum of all the four digit numbers formed using the [#permalink]

### Show Tags

27 Apr 2013, 07:21
Hi Bunuel,
So in the below question if we apply the formula shouldn't we consider 4 as n?
Find the sum of all the four digit numbers formed using the digits 2,3,4 and 5 without repetition.
(4-1)*(2+3+4+5)*1111
Aybige

Kudos [?]: 5 [0], given: 23

Math Expert
Joined: 02 Sep 2009
Posts: 42578

Kudos [?]: 135425 [0], given: 12692

Re: Find the sum of all the four digit numbers formed using the [#permalink]

### Show Tags

28 Apr 2013, 02:28
aybige wrote:
Hi Bunuel,
So in the below question if we apply the formula shouldn't we consider 4 as n?
Find the sum of all the four digit numbers formed using the digits 2,3,4 and 5 without repetition.
(4-1)!*(2+3+4+5)*1111
Aybige

Correct, but you've missed factorial (!). It should be: (4-1)!*(2+3+4+5)*(1111)=93324.
_________________

Kudos [?]: 135425 [0], given: 12692

Re: Find the sum of all the four digit numbers formed using the   [#permalink] 28 Apr 2013, 02:28
Display posts from previous: Sort by
user
First, the content structure of the provided text is as follows: It starts with a timestamp, followed by a section header and a brief description of a service, and then a question is presented.

Now, the analysis based on the provided format:

TYPE: SINGLE_QUESTION
- The text contains a single mathematical question: "Find the sum of all the four digit numbers formed using the digits 2,3,4 and 5 without repetition."