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

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If the three unique positive digits A,B, and C are arranged [#permalink]  09 Dec 2012, 08:59
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Question Stats:

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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
[Reveal] Spoiler: OA
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Re: If the three unique positive digits A,B, and C are arranged [#permalink]  09 Dec 2012, 19:16
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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]  09 Dec 2012, 20:06
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 non-divisor is 11 (answer D).
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Re: If the three unique positive digits A,B, and C are arranged [#permalink]  09 Dec 2012, 21:42
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]  09 Dec 2012, 21:49
Expert's post
Can you please let me know your approach?
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Re: If the three unique positive digits A,B, and C are arranged [#permalink]  10 Dec 2012, 00:50
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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: $$(n-1)!*(sum \ of \ the \ digits)*(111... \ n \ times)$$.

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 to prctice:
the-addition-problem-above-shows-four-of-the-24-different-in-104166.html
what-is-the-sum-of-all-3-digit-positive-integers-that-can-be-78143.html
find-the-sum-of-all-the-four-digit-numbers-formed-using-the-103523.html
what-is-the-sum-of-all-4-digit-numbers-that-can-be-formed-94836.html
what-is-the-sum-of-all-3-digit-positive-integers-that-can-be-78143.html
find-the-sum-of-all-the-four-digit-numbers-which-are-formed-88357.html
find-the-sum-of-all-3-digit-nos-that-can-be-formed-by-88864.html
there-are-24-different-four-digit-integers-than-can-be-141891.html

Hope it helps.
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Re: If the three unique positive digits A,B, and C are arranged [#permalink]  10 Dec 2012, 02:10
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]  10 Dec 2012, 03:38
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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]  10 Dec 2012, 03:41
Expert's post
priyamne wrote:
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).

222 IS divisible by 37: 37*6=222. Check here: if-the-three-unique-positive-digits-a-b-and-c-are-arranged-143836.html#p1152825
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Re: If the three unique positive digits A,B, and C are arranged [#permalink]  18 Jun 2014, 07:13
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Re: If the three unique positive digits A,B, and C are arranged [#permalink]  19 Jun 2014, 01:45
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]  23 Aug 2015, 19:18
Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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Re: If the three unique positive digits A,B, and C are arranged   [#permalink] 23 Aug 2015, 19:18
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# If the three unique positive digits A,B, and C are arranged

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