It is currently 17 Nov 2017, 20:02

### 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

# The addition problem above shows four of the 24 different

Author Message
Manager
Joined: 10 Oct 2005
Posts: 51

Kudos [?]: 4 [0], given: 0

### Show Tags

03 Dec 2005, 17:30
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

The addition problem above shows four of the 24 different integers that can be formed by using each of the digits 1,2,3 & 4 exactly once in each integer. what is the sum of these 24 integers ?
A. 24000
B. 26664
C. 40440
D. 60000
E. 66660

How come there are only 24 possible combinations ? when I think there could be only 16 ??

Is it a combination questions ???

Kudos [?]: 4 [0], given: 0

Director
Joined: 10 Oct 2005
Posts: 713

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

### Show Tags

08 Dec 2005, 04:44
briozeal wrote:
The addition problem above shows four of the 24 different integers that can be formed by using each of the digits 1,2,3 & 4 exactly once in each integer. what is the sum of these 24 integers ?
A. 24000
B. 26664
C. 40440
D. 60000
E. 66660

How come there are only 24 possible combinations ? when I think there could be only 16 ??

Is it a combination questions ???

I don't how to call this type of question
But some combo definitely using here
There is 4! ways to place 4 integers 4!=24. Hence there is 24 different intgers consisting of 4 digits!
1234 2134
1342 2341
1423 2431
1243 2143
1324
1243
And so on!But you don't need to draw every possible combination of digits
You need only to remember that 4 digits may be arranged in 4!ways which is equal to 4*3*2*1=24 .For Example 5 digits may be arranged 5! ways 5!=5*4*3*2*1=120 Hope his helps!

By the way the answer Is E!I think
Later will explain if you need

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

Director
Joined: 14 Sep 2005
Posts: 984

Kudos [?]: 220 [0], given: 0

Location: South Korea

### Show Tags

08 Dec 2005, 08:22
First digit = 1 or 2 or 3 or 4
Second digit = 1 or 2 or 3 or 4 (= 10 or 20 or 30 or 40)
Third digit = 1 or 2 or 3 or 4 (= 100 or 200 or 300 or 400)

There are 24 ways of making a three digit number, and this means that each number is used 6 times in each digit.

100 * 6 = 600
200 * 6 = 1200
300 * 6 = 1800
400 * 6 = 2400
-----------------
Sum = 6000

10 * 6 = 60
20 * 6 = 120
30 * 6 = 180
40 * 6 = 240
-----------------
Sum = 600

1 * 6 = 6
2 * 6 = 12
3 * 6 = 18
4 * 6 = 24
----------------
Sum = 60

Total sum = 6660

What am I missing here?
_________________

Auge um Auge, Zahn um Zahn !

Kudos [?]: 220 [0], given: 0

Director
Joined: 14 Sep 2005
Posts: 984

Kudos [?]: 220 [0], given: 0

Location: South Korea

### Show Tags

08 Dec 2005, 08:28
gamjatang wrote:
First digit = 1 or 2 or 3 or 4
Second digit = 1 or 2 or 3 or 4 (= 10 or 20 or 30 or 40)
Third digit = 1 or 2 or 3 or 4 (= 100 or 200 or 300 or 400)

There are 24 ways of making a three digit number, and this means that each number is used 6 times in each digit.

100 * 6 = 600
200 * 6 = 1200
300 * 6 = 1800
400 * 6 = 2400
-----------------
Sum = 6000

10 * 6 = 60
20 * 6 = 120
30 * 6 = 180
40 * 6 = 240
-----------------
Sum = 600

1 * 6 = 6
2 * 6 = 12
3 * 6 = 18
4 * 6 = 24
----------------
Sum = 60

Total sum = 6660

What am I missing here?

I forgot to consider the fourth digit...

(1111*6)+(2222*6)+(3333*6)+(4444*6) = 66660
_________________

Auge um Auge, Zahn um Zahn !

Kudos [?]: 220 [0], given: 0

08 Dec 2005, 08:28
Display posts from previous: Sort by

# The addition problem above shows four of the 24 different

Moderator: chetan2u

 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®.