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# 1234 1243 1324 ... ... 4321 The addition problem above shows

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Senior Manager
Joined: 11 May 2004
Posts: 454
Location: New York
1234 1243 1324 ... ... 4321 The addition problem above shows [#permalink]

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10 Dec 2005, 20:34
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

1234
1243
1324
...
...
4321

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

a. 24000
b. 26664
c. 40440
d. 60000
e. 66660
SVP
Joined: 24 Sep 2005
Posts: 1885

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10 Dec 2005, 21:04
desiguy wrote:
1234
1243
1324
...
...
4321

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

a. 24000
b. 26664
c. 40440
d. 60000
e. 66660

24 differents integers -----> 1,2,3,4 are in thousands digit six times each AND 1,2,3,4 are in hundreds digit six times each AND 1,2,3,4 are in tens digit six times each AND 1,2,3,4 are in unit digit six times each.

That is to say:
for digit 1: 6*( 1000+100+10+1)
for digit 2: 6* (2000+200+20+2)= 12 * (1000+100+10+1)
for digit 3: 6* ( 3000+300+30+3)= 18 * (1000+100+10+1)
for digit 4: 6*( 4000+400+40+4)= 24*(1000+100+10+1)

-----> the sum of these 24 numbers= (1000+100+10+1)* ( 6+12+18+24)
= 1111 * 60 = 66660
Senior Manager
Joined: 11 May 2004
Posts: 454
Location: New York

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11 Dec 2005, 09:43
Good job. E is the OA.
Director
Joined: 10 Oct 2005
Posts: 718

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12 Dec 2005, 03:01
Intern
Joined: 09 Sep 2005
Posts: 18

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12 Dec 2005, 03:08
desiguy wrote:
1234
1243
1324
...
...
4321

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

a. 24000
b. 26664
c. 40440
d. 60000
e. 66660

A=6*(1+2+3+4)*1000+6*(1+2+3+4)*100+6*(1+2+3+4)*10+6*(1+2+3+4)
Re: Number Properties 2   [#permalink] 12 Dec 2005, 03:08
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