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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, 05:04
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, 06:39
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, 15:52
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, 18:22
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]

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New post 30 Dec 2019, 06:56
Top Contributor
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


Since we're adding 24 numbers, we know that:
Six numbers will be in the form 1 _ _ _
Six numbers will be in the form 2 _ _ _
Six numbers will be in the form 3 _ _ _
Six numbers will be in the form 4 _ _ _

Let's first see what the sum is when we say all 24 numbers are 1000, 2000, 3000 or 4000
The sum = (6)(1000) + (6)(2000) + (6)(3000) + (6)(4000)
= 6(1000 + 2000 + 3000 + 4000)
= 6(10,000)
= 60,000

Since the 24 numbers are actually greater than 1000, 2000, etc, we know that the actual sum must be greater than 60,000

Answer: E

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

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New post 17 Jan 2020, 19:40
For every digit when repition is not allowed, in each unit or tens or hundreds or thousands place we will fixed combinations.
When we have four digits to choose in each place, we have 3*2*1combinations for each digit in each place to fix. Hope that part is clear.

So basically we have 6 combinations of each of 1,2,3,4 in each of unit or tens or hundreds or thousands place.

Now sum the combinations . ( 6*1+6*2+6*3+6*4) = 60 .

It means on adding each place I get 60 ..
Now it's simple addtion alogorithm that we do in primary classes.

Unit place sum is 60 and hence place 0 and carry 6 to tens place..
Tens place sum is also 60 and add the carried 6 making it 66. Place 6 and carry 6 to hundred place .
Hundreds place sum is 60 and add carried 6 making it 66 again. Place 6 and carry 6 to thousand place.

Now thousand place sum is also 60 and add carried 6 making 66 at the thousand place.

So we have 66660. HENCE THE ANSWER.

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Re: The addition problem above shows four of the 24 different integers tha   [#permalink] 17 Jan 2020, 19:40

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