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The addition problem above shows four of the 24 different in [#permalink]
 03 Nov 2010, 00:34
Question Stats:
57% (02:02) correct
42% (00:56) wrong based on 9 sessions
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
Last edited by Bunuel on 06 Nov 2012, 03:17, edited 1 time in total.
Renamed the topic and edited the question.
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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 Answer : e
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Thanks for the great explaination shrouded1!
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The addition problem [#permalink]
08 Feb 2011, 05:09
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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Re: The addition problem [#permalink]
08 Feb 2011, 06:26
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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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Re: The addition problem [#permalink]
08 Feb 2011, 06:48
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The addition problem above shows four of the 24 different [#permalink]
22 Jan 2013, 05:49
1234
1243
1324
.......
.......
+4321
Q. The addition problem above shows four of the 24 different integers that can be formed using each of the digits 1,2,3,4 exactly once in each integer. What is the sum of these 24 integers?
A) 24000
B) 26664
c) 40440
D) 60000
E) 66660
I'm not too sure what concept is tested here so not sure what to tag this particular question type.
Last edited by fozzzy on 22 Jan 2013, 08:23, edited 1 time in total.
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Re: The addition problem above shows four of the 24 different [#permalink]
22 Jan 2013, 05:58
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Re: The addition problem [#permalink]
22 Jan 2013, 10:44
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.htmlcan-someone-help-94836.htmlsum-of-all-3-digit-nos-with-88864.htmlpermutation-88357.htmlsum-of-3-digit-s-78143.htmlCould 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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Re: The addition problem [#permalink]
22 Jan 2013, 21:07
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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The addition problem above shows four of the 24 different intege [#permalink]
21 May 2013, 07:35
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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The addition problem above shows four of the 24 different intege
[#permalink]
21 May 2013, 07:35
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