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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?

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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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.
_________________

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 non-divisor is 11 (answer D).
_________________

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

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: \(n^{n-1}*(sum \ of \ the \ digits)*(111... \ n \ times)\).

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..

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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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).

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..

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).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________

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: \((n-1)!*(sum \ of \ the \ digits)*(111... \ n \ times)\).

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

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
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14 Mar 2016, 08:47

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