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If the three unique positive digits A,B, and C are arranged [#permalink]
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09 Dec 2012, 09:59
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If the three unique positive digits A,B, and C are arranged in all possible sequences (ABC,ACB,BAC,ETC....),then the sum of all the resulting three digit integers must be divisible by each of the following EXCEPT? (A) 2 (B) 3 (C) 6 (D) 11 (E) 37
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Re: If the three unique positive digits A,B, and C are arranged [#permalink]
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09 Dec 2012, 20:16
The 6 numbers are: ABC, ACB, BCA, BAC, CAB, CBA. Adding these would give 200A+200B+200C+20A+20B+20C+2(A+B+C). or 222(A+B+C). Among the answer choices , the sum is not divisible by 37. Hence the answer.
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Re: If the three unique positive digits A,B, and C are arranged [#permalink]
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09 Dec 2012, 21:06
Marcab wrote: The 6 numbers are: ABC, ACB, BCA, BAC, CAB, CBA. Adding these would give 200A+200B+200C+20A+20B+20C+2(A+B+C). or 222(A+B+C). Among the answer choices , the sum is not divisible by 37. Hence the answer. Clever approach. Apparently the post is wrong then as it says the nondivisor is 11 (answer D).
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Re: If the three unique positive digits A,B, and C are arranged [#permalink]
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09 Dec 2012, 22:42
The answer indeed is 11. However I reached there by actually doing the permutations of digits 1, 2, 3 and adding them up and checking for the divisibility furthur. I am sure there should be some other approach



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Re: If the three unique positive digits A,B, and C are arranged [#permalink]
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09 Dec 2012, 22:49



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Re: If the three unique positive digits A,B, and C are arranged [#permalink]
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10 Dec 2012, 01:50
Marcab wrote: If the three unique positive digits A,B, and C are arranged in all possible sequences (ABC,ACB,BAC,ETC....),then the sum of all the resulting three digit integers must be divisible by each of the following EXCEPT?
(A) 2 (B) 3 (C) 6 (D) 11 (E) 37
The 6 numbers are: ABC, ACB, BCA, BAC, CAB, CBA. Adding these would give 200A+200B+200C+20A+20B+20C+2(A+B+C). or 222(A+B+C). Among the answer choices , the sum is not divisible by 37. Hence the answer. 222 is not divisible by 11 but it is divisible by 37, thus the answer is D, not E. Generally:1. Sum of all the numbers which can be formed by using the \(n\) digits without repetition is: \((n1)!*(sum \ of \ the \ digits)*(111... \ n \ times)\). 2. Sum of all the numbers which can be formed by using the \(n\) digits ( repetition being allowed) is: \(n^{n1}*(sum \ of \ the \ digits)*(111... \ n \ times)\). Similar questions to prctice: theadditionproblemaboveshowsfourofthe24differentin104166.htmlwhatisthesumofall3digitpositiveintegersthatcanbe78143.htmlfindthesumofallthefourdigitnumbersformedusingthe103523.htmlwhatisthesumofall4digitnumbersthatcanbeformed94836.htmlwhatisthesumofall3digitpositiveintegersthatcanbe78143.htmlfindthesumofallthefourdigitnumberswhichareformed88357.htmlfindthesumofall3digitnosthatcanbeformedby88864.htmlthereare24differentfourdigitintegersthancanbe141891.htmlHope it helps.
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Re: If the three unique positive digits A,B, and C are arranged [#permalink]
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10 Dec 2012, 03:10
Mine is a random approach...the question here is "Except" kind off so one can pick any number. I picked 123 and listed all the possible permutations
123 132 231 213 312 321
Which sums up to 1332. one can see that the number "1332" is clearly divisible by 3 and 2 and hence 6. also it is not divisible by 11. for 37 I actually divided the number by 37 and found it as divisible. and hence answer is D....Bunuels' formula is an excellent way to do with..



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Re: If the three unique positive digits A,B, and C are arranged [#permalink]
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10 Dec 2012, 04:38
Ans: There would be six combinations of numbers, let A=x, B=y, C=z. Now ABC= 100x+10y+z and if we add all the six numbers we get 222(x+y+z). We check that 222 is divisible by all of the options except 37. The answer is (E).
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Re: If the three unique positive digits A,B, and C are arranged [#permalink]
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10 Dec 2012, 04:41



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Re: If the three unique positive digits A,B, and C are arranged [#permalink]
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19 Jun 2014, 02:45
ratinarace wrote: Mine is a random approach...the question here is "Except" kind off so one can pick any number. I picked 123 and listed all the possible permutations
123 132 231 213 312 321
Which sums up to 1332. one can see that the number "1332" is clearly divisible by 3 and 2 and hence 6. also it is not divisible by 11. for 37 I actually divided the number by 37 and found it as divisible. and hence answer is D....Bunuels' formula is an excellent way to do with.. Did in the same way; shortened the addition part Picked up first two & so on as below.. 255 + 466 + 933 = 1332 Fails divisibility check of 11 Answer = D
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Re: If the three unique positive digits A,B, and C are arranged [#permalink]
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14 Mar 2016, 08:47
Bunuel wrote: Marcab wrote: If the three unique positive digits A,B, and C are arranged in all possible sequences (ABC,ACB,BAC,ETC....),then the sum of all the resulting three digit integers must be divisible by each of the following EXCEPT?
(A) 2 (B) 3 (C) 6 (D) 11 (E) 37
The 6 numbers are: ABC, ACB, BCA, BAC, CAB, CBA. Adding these would give 200A+200B+200C+20A+20B+20C+2(A+B+C). or 222(A+B+C). Among the answer choices , the sum is not divisible by 37. Hence the answer. 222 is not divisible by 11 but it is divisible by 37, thus the answer is D, not E. Generally:1. Sum of all the numbers which can be formed by using the \(n\) digits without repetition is: \((n1)!*(sum \ of \ the \ digits)*(111... \ n \ times)\). 2. Sum of all the numbers which can be formed by using the \(n\) digits ( repetition being allowed) is: \(n^{n1}*(sum \ of \ the \ digits)*(111... \ n \ times)\). Similar questions to prctice: theadditionproblemaboveshowsfourofthe24differentin104166.htmlwhatisthesumofall3digitpositiveintegersthatcanbe78143.htmlfindthesumofallthefourdigitnumbersformedusingthe103523.htmlwhatisthesumofall4digitnumbersthatcanbeformed94836.htmlwhatisthesumofall3digitpositiveintegersthatcanbe78143.htmlfindthesumofallthefourdigitnumberswhichareformed88357.htmlfindthesumofall3digitnosthatcanbeformedby88864.htmlthereare24differentfourdigitintegersthancanbe141891.htmlHope it helps. does gmat really ask such kind of questions? i got it right and its excellent for practise but really? Honest answer expected as always thanks
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Re: If the three unique positive digits A,B, and C are arranged [#permalink]
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13 Mar 2017, 02:28
First, consider a random three digit number as an example: 375 = 100(3) + 10(7) + 1(5), because 3 is in the hundreds place, 7 is in the tens place, and 5 is in the ones place. More generally, ABC = 100(A) + 10(B) + 1(C). For digit problems, particularly those that involve “shuffling” or permutations of digits, we must think about place value. Since we are dealing with three unique digits, the number of possible sequences will be 3! or 6. If we write them out, we can see a pattern: ABC ACB BAC BCA CAB CBA Notice that each unique digit appears exactly twice in each column, so each column individually sums to 2(A + B + C). Sum of the hundreds column: 100 × 2(A + B + C) = 200(A + B + C) Sum of the tens column: 10 × 2(A + B + C) = 20(A + B + C) Sum of the ones column: 1 × 2(A + B + C) = 2(A + B + C) Altogether, the sum of the threedigit numbers is (200 + 20 + 2)(A + B + C) = 222(A + B + C). Regardless of the values of A, B, and C, the sum of the threedigit numbers must be divisible by 222 and all of its factors: 1, 2, 3, 6, 37, 74, 111, and 222. The only answer choice that is not among these is 11. The correct answer is D.
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Re: If the three unique positive digits A,B, and C are arranged [#permalink]
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04 May 2018, 14:54
mun23 wrote: If the three unique positive digits A,B, and C are arranged in all possible sequences (ABC,ACB,BAC,ETC....),then the sum of all the resulting three digit integers must be divisible by each of the following EXCEPT?
(A) 2 (B) 3 (C) 6 (D) 11 (E) 37 let digits be 2,3,4 possible sequences=6 [(234+432)/2]*6=1998=sum of six 3digit integers 1998/37=54 1998/11=181.6 D




Re: If the three unique positive digits A,B, and C are arranged
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