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

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Manager
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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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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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Kudos [?]: 1067 [1] , given: 62

Re: If the three unique positive digits A,B, and C are arranged [#permalink]

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09 Dec 2012, 20:16
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Expert's post
The 6 numbers are:
ABC, ACB, BCA, BAC, CAB, CBA.
or 222(A+B+C).
Among the answer choices , the sum is not divisible by 37.
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Kudos [?]: 34 [0], given: 82

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.
or 222(A+B+C).
Among the answer choices , the sum is not divisible by 37.

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]

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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
Expert's post
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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
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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.
or 222(A+B+C).
Among the answer choices , the sum is not divisible by 37.

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:
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]

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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
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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]

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10 Dec 2012, 04:41
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]

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18 Jun 2014, 08:13
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]

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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

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Re: If the three unique positive digits A,B, and C are arranged [#permalink]

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23 Aug 2015, 20:18
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]

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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.
or 222(A+B+C).
Among the answer choices , the sum is not divisible by 37.

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:
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.

does gmat really ask such kind of questions?
i got it right and its excellent for practise but really?
thanks
Re: If the three unique positive digits A,B, and C are arranged   [#permalink] 14 Mar 2016, 08:47
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