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Senior Manager
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1234 1243 1324 ... ... 4321 The addition problem above shows [#permalink] New post 10 Dec 2005, 19:34
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
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Re: Number Properties 2 [#permalink] New post 10 Dec 2005, 20:04
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
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 [#permalink] New post 11 Dec 2005, 08:43
Good job. E is the OA.
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 [#permalink] New post 12 Dec 2005, 02:01
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Re: Number Properties 2 [#permalink] New post 12 Dec 2005, 02:08
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, 02:08
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