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The addition problem above shows four of the 24 different integers tha

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Re: The addition problem above shows four of the 24 different integers tha  [#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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Re: The addition problem above shows four of the 24 different integers tha  [#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 integers tha  [#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 integers tha  [#permalink]

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New post 15 Nov 2019, 19:22
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


PS78602.01




Method 1: Sum Each Column of Digits



We need positive integers having 4 digits.

S = __ __ __ __

Since no repetition allowed, we can make 4*3*2*1 = 24 such positive integers.


Now imagine writing these 24 numbers one below the other to add.

1234
1243
...
...
x 24 combinations

When we add them, noticing the symmetry we know that there will be 6 1's in units digits, 6 2's, 6 3's and 6 4's. So units digits will add up to (1+2+3+4)*6.

Similarly, tens digits will add up (1+2+3+4)*6*10
Similarly, hundreds digits will add up (1+2+3+4)*6*100
Similarly, thousands digits will add up (1+2+3+4)*6*100)

Adding all of them:
(1+2+3+4)*6 + (1+2+3+4)*6*10 + (1+2+3+4)*6*100 + (1+2+3+4)*6*1000 = (1+2+3+4)*6 * (1 + 10 + 100 + 1000) = 10 * 6 * 1111 = 66,660

ANSWER: E


Method 2: Direct Formula



Sum of all n digit numbers formed by n non-zero digits without repetition is:

(n−1)!∗(sum of the digits)∗(111... n times)

= (4-3)! * (1 + 2 + 3 + 4) * (1111)
= 6 * 10 * 1111
= 66,660

Answer: E
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Re: The addition problem above shows four of the 24 different integers tha   [#permalink] 15 Nov 2019, 19:22

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