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# 1,234 1,243 1,324 ... ... +4,321 _______

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VP
Joined: 06 Jun 2004
Posts: 1020
Location: CA
1,234 1,243 1,324 ... ... +4,321 _______   [#permalink]

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20 Dec 2005, 00:44
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1,234
1,243
1,324
...
...
+4,321
_______

The addition problem above shows four of the 24 different integers that can be formed by using each of the digits 1, 2 3, and 4 exactly once in each integer. What is the sume of these 24 integers?

(A) 24,000
(B) 26,664
(C) 40,440
(D) 60,000
(E) 66,660

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GMAT Club Legend
Joined: 07 Jul 2004
Posts: 4906
Location: Singapore

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20 Dec 2005, 01:05
Keeping 1 in the thousands place, we can have 6 arrangements for 2,3 and 4:

1234
1243
1324
1342
1432
1423

Keeping 2 in the thousands place, we can have 6 arrangements as well for 1,3,4
This time round, in the ones, tens and hundreds and thousands position, we will have 1,3,4 occuring twice in their respective positions.

Therefore, if we add the 24 numbers, we will have 1,2,3,4 occuring 6 times in their respective places.

For ones position, we will have 1,2,3,4 occuring 6 times. This gives us 6+12+18+24 = 60. So we keep 0 and carry 6 to the tens position.

For tens position, we will have 1,2,3,4 occuring 6 times. This gives us 60+6 = 66. Keep 6 and carry 6 to the hundreds positions.

For hundreds position, the sum is now 60+6 = 66. Keep 6 and carry 6 to thousands positions.

For thousands position, the sum is 66 with no carry over.

The sum of the 24 numbers is therefore 66,660 (E)
Director
Joined: 09 Jul 2005
Posts: 566

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20 Dec 2005, 02:59
Each column sums 60. the answer should be E.
Senior Manager
Joined: 15 Apr 2005
Posts: 410
Location: India, Chennai

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20 Dec 2005, 03:24
There are 6 different numbers with 1 in the thousands place, and similarly 6 different numbers for 2,3 and 4.

So approximately we will have
6 * 1000 = 6000 +
6 * 2000 = 12000 +
6 * 3000 = 18000 +
6 * 4000 = 24000
= 60000

We have just counted for the thousand's digit. So if we have to count the hundreds digit, tens and unit digit, it should be obvious that the number is > 60000. Hence E.
Manager
Joined: 12 Nov 2005
Posts: 77

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20 Dec 2005, 10:31
1

If one can remember, here's the formula for that.

The sum of all the numbers formed from n digits is

(n-1)! *(Sum of Digits)*(111...n times)

Here in this case, n =(1,2,3,4)
Therefore Sum = (4-1)!* (1+2+3+4)*(1111) .....
= 6*10*1111
=66660
Intern
Joined: 26 Sep 2005
Posts: 3
Location: Rochester Hills,MI

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20 Dec 2005, 13:05
1
This is when you don't remember formulae.
My approach is from units place... when you look at the given answers all the number have different digits in tens place.
So attack from units end
There are 24 numbers with these numbers arranged ==> there are 6
numbers with each digit in units place.
1 * 6 = 06
2 * 6 = 12
3 * 6 = 18
4 * 6 = 24
=======
60

the last 2 digits should be 60 ====> E
I don't know if this is the right approach... Please correct me, if I am deviating
Senior Manager
Joined: 03 Nov 2005
Posts: 346
Location: Chicago, IL

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20 Dec 2005, 13:34
1234
1324
1432

2134
2314
2431

.......
4231

There are 12 possible combinations with 1,2,3,4 where each digit is used 3 times in each place. The sum =1+2+3+4=10

So, sum=3*10*1000+3*10*100+3*10+10+ 3*10=33330
To find the sum of 24 numbers, multiply by 2=66660
_________________

Hard work is the main determinant of success

Senior Manager
Joined: 11 Nov 2005
Posts: 313
Location: London

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20 Dec 2005, 15:35
I liked the method of 'krisrini' simple and easy to follow!

good work
Director
Joined: 14 Sep 2005
Posts: 955
Location: South Korea

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21 Dec 2005, 02:11
This is interesting.

The sum of the following

1234
1324
1432
2134
2314
2431
...
4231

is same as the sum of the following;

1111
1111
1111
1111
1111
1111
2222
2222
2222
2222
2222
2222
3333
3333
3333
3333
3333
3333
4444
4444
4444
4444
4444
4444

Obviously, it's 66,660.

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This discussion does not meet community quality standards. It has been retired.

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Joined: 09 Sep 2013
Posts: 8179
Re: 1,234 1,243 1,324 ... ... +4,321 _______   [#permalink]

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04 Sep 2018, 08:39
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Re: 1,234 1,243 1,324 ... ... +4,321 _______ &nbs [#permalink] 04 Sep 2018, 08:39
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