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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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02 Nov 2010, 23:34
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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
[Reveal] Spoiler: OA

Last edited by Bunuel on 06 Nov 2012, 02:17, edited 1 time in total.
Renamed the topic and edited the question.
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02 Nov 2010, 23:45
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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

Using the symmetry in the numbers involved (All formed using all possible combinations of 1,2,3,4), and we know there are 24 of them. We know there will be 6 each with the units digits as 1, as 2, as 3 and as 4. And the same holds true of the tens, hundreds and thousands digit.

The sum is therefore = (1 + 10 + 100 + 1000) * (1*6 +2*6 +3*6 +4*6) = 1111 * 6 * 10 = 66660

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08 Feb 2011, 05:26
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1,2,3,4 can be arranged in 4! = 24 ways

The units place of all the integers will have six 1's, six 2's, six 3's and six 4's
Likewise,
The tens place of all the integers will have six 1's, six 2's, six 3's and six 4's
The hundreds place of all the integers will have six 1's, six 2's, six 3's and six 4's
The thousands place of all the integers will have six 1's, six 2's, six 3's and six 4's

Addition always start from right(UNITS) to left(THOUSANDS);

Units place addition; 6(1+2+3+4) = 60.
Unit place of the result: 0
carried over to tens place: 6

Tens place addition; 6(1+2+3+4) = 60 + 6(Carried over from Units place) = 66
Tens place of the result: 6
carried over to hunderes place: 6

Hundreds place addition; 6(1+2+3+4) = 60 + 6(Carried over from tens place) = 66
Hundreds place of the result: 6
carried over to thousands place: 6

Thousands place addition; 6(1+2+3+4) = 60 + 6(Carried over from hundreds place) = 66
Thousands place of the result: 6
carried over to ten thousands place: 6

Ten thousands place of the result: 0+6(Carried over from thousands place) = 6

Result: 66660

Ans: "E"
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08 Feb 2011, 05:48
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Merging similar topics.

Formulas for such kind of problems (just in case):

1. Sum of all the numbers which can be formed by using the $$n$$ digits without repetition is: $$(n-1)!*(sum \ of \ the \ digits)*(111... \ n \ times)$$.

2. Sum of all the numbers which can be formed by using the $$n$$ digits (repetition being allowed) is: $$n^{n-1}*(sum \ of \ the \ digits)*(111... \ n \ times)$$.

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22 Jan 2013, 20:07
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hellscream wrote:

Could you tell me the way to calculate the sum which the repetition is allowed? For example: from 1,2,3,4. how can we calculate the sum of four digit number that formed from 1,2,3,4 and repetition is allowed?

The logic is no different from 'no repetition allowed' question. The only thing different is the number of numbers you can make.

How many numbers can you make using the four digits 1, 2, 3 and 4 if repetition is allowed?
You can make 4*4*4*4 = 256 numbers (there are 4 options for each digit)

1111
1112
1121
... and so on till 4444

By symmetry, each digit will appear equally in each place i.e. in unit's place, of the 256 numbers, 64 will have 1, 64 will have 2, 64 will have 3 and 64 will have 4.
Same for 10s, 100s and 1000s place.

Sum = 1000*(64*1 + 64*2 + 64*3 + 64*4) + 100*(64*1 + 64*2 + 64*3 + 64*4) + 10*(64*1 + 64*2 + 64*3 + 64*4) + 1*(64*1 + 64*2 + 64*3 + 64*4)
= (1000 + 100 + 10 + 1)(64*1 + 64*2 + 64*3 + 64*4)
= 1111*64*10 = 711040

or use the formula given by Bunuel above:
Sum of all the numbers which can be formed by using the digits (repetition being allowed) is:$$n^{n-1}$$*Sum of digits*(111...n times)
=$$4^3*(1+2+3+4)*(1111) = 711040$$ (Same calculation as above)
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09 Nov 2010, 05:24
Thanks for the great explaination shrouded1!
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08 Feb 2011, 04:09
the addition problem below shows four of the 24 different integers that can be formed by using each of the digits 1,2,3, and 4 exactly once in each integer.what is the sum of these 24 integers?

1,234
1,243
1,324
.......
.......
+4,321

a) 24,000
b) 26,664
c) 40,440
d) 60,000
e) 66,660
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22 Jan 2013, 09:44
Bunuel wrote:
Merging similar topics.

Formulas for such kind of problems (just in case):

1. Sum of all the numbers which can be formed by using the $$n$$ digits without repetition is: $$(n-1)!*(sum \ of \ the \ digits)*(111... \ n \ times)$$.

2. Sum of all the numbers which can be formed by using the $$n$$ digits (repetition being allowed) is: $$n^{n-1}*(sum \ of \ the \ digits)*(111... \ n \ times)$$.

Similar questions:
nice-question-and-a-good-way-to-solve-103523.html
can-someone-help-94836.html
sum-of-all-3-digit-nos-with-88864.html
permutation-88357.html
sum-of-3-digit-s-78143.html

Could you tell me the way to calculate the sum which the repetition is allowed? For example: from 1,2,3,4. how can we calculate the sum of four digit number that formed from 1,2,3,4 and repetition is allowed?
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The addition problem above shows four of the 24 different intege [#permalink]

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21 May 2013, 06:35
This can be solved much easier by realizing that, since the number of four term permutations is 4!, and that summing the a sequence to its reverse gives

1234 +4321 = 5555

1243 +3421 = 5555

we may see that there are 4!/2 pairings we can make, giving us

5555(12) = 66660
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Re: The addition problem above shows four of the 24 different in [#permalink]

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16 Jun 2014, 03:04
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Re: The addition problem above shows four of the 24 different in [#permalink]

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28 Sep 2014, 09:41
For those who could not memorize the formular, you can guess the answer in 30 secs:
Since we have 24 numbers, we will have 6 of 1 thousand something, 6 of 2 thousand something, 6 of 3 thousand something, and 6 of 4 thousand something
So,
6x1(thousand something) = 6 (thousand something)
6x2(thousand something) = 12 (thousand something)
6x3(thousand something) = 18 (thousand something)
6x4(thousand something) = 24 (thousand something)
Add them all 6+12 +18 + 24 = 60 (thousand something)
----> E
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Re: The addition problem above shows four of the 24 different in [#permalink]

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11 Nov 2015, 23:06
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Re: The addition problem above shows four of the 24 different in [#permalink]

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15 Aug 2016, 13:38
each term is repeating 6 times..(total 24/4=4)
now at unit...each term will repeat 6 times... (1x6+ 2x6 + 3x6 + 4x6 = 60) , so unit digit is "0" and 6 remaining.

repeating same.....total of 24 ten digit will be 60 + 6 from total of unit ,so ten digit will be 6.

only E has 60 as last two digits.
Re: The addition problem above shows four of the 24 different in   [#permalink] 15 Aug 2016, 13:38
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