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The addition problem above shows four of the 24 different in

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The addition problem above shows four of the 24 different in  [#permalink]

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New post 14 May 2019, 06:04
The solutions above are great. However, I believe this can be answered using simple logic and leveraging your answer choices.

The question tells us that there are 24 integers. Given there are 4 unique integers, we know there will be 6 of each (given the symmetry).

Here's where common sense comes handy: we know that 6 unique numbers will begin with the digit 4; therefore, they will have a minimum sum of 4*6000, or 24,000. Similarly, 6 unique numbers will begin with 3 and will have a minimum sum of 3000*6, or 18,000. Therefore, we have a minimum total sum of:

6*4,000 + 6*3,000 + 6*2,000 + 6*1,000 = 24,000 + 18,000 + 12,000 + 6,000 = 60,000. Logically, our answer will be greater than 60,000. Only one answer choice remains --> E.
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The addition problem above shows four of the 24 different in  [#permalink]

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New post 14 May 2019, 07:39
student26 wrote:
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 1,2,3,4 exact;y once in each integer. What is the sum of these 24 integers?

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


add least and greatest integers:
1234+4321=5555
5555/2=2777.5
2777.5*24=66,660
E
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Re: The addition problem above shows four of the 24 different in  [#permalink]

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New post 06 Aug 2019, 16:52
student26 wrote:
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 1,2,3,4 exact;y once in each integer. What is the sum of these 24 integers?

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


24 total/4 numbers = 6
The sum of each number will be the number's digit placement * 6
6*1(1000 + 100 + 10 + 1) = 6,666
6*2(1000 + 100 + 10 + 1) = 12 * 1111 = 13,332 [...remember, 11*12 = 132, the same idea here]
6*3(1000 + 100 + 10 + 1) = 18 * 1111 = 19,998
6*4(1000 + 100 + 10 + 1) = 24 * 1111 = 26,664

Start adding, Units digit = 6+2+8+4 = 20, so 0 + 2 carryover ... 0
Tens digit = 6+3+9+6 + 2 carryover = 26, so 6 + 2 carryover ... 60
We can see only E matches
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Re: The addition problem above shows four of the 24 different in   [#permalink] 06 Aug 2019, 16:52

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