It is currently 20 Oct 2017, 04:06

### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

# Events & Promotions

###### Events & Promotions in June
Open Detailed Calendar

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

Author Message
TAGS:

### Hide Tags

Manager
Status: struggling with GMAT
Joined: 06 Dec 2012
Posts: 205

Kudos [?]: 435 [2], given: 46

Concentration: Accounting
GMAT Date: 04-06-2013
GPA: 3.65
If the three unique positive digits A,B, and C are arranged [#permalink]

### Show Tags

09 Dec 2012, 09:59
2
KUDOS
6
This post was
BOOKMARKED
00:00

Difficulty:

75% (hard)

Question Stats:

51% (01:32) correct 49% (01:15) wrong based on 241 sessions

### HideShow timer Statistics

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

Kudos [?]: 435 [2], given: 46

VP
Status: Been a long time guys...
Joined: 03 Feb 2011
Posts: 1377

Kudos [?]: 1675 [1], given: 62

Location: United States (NY)
Concentration: Finance, Marketing
GPA: 3.75
Re: If the three unique positive digits A,B, and C are arranged [#permalink]

### Show Tags

09 Dec 2012, 20:16
1
KUDOS
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.
_________________

Kudos [?]: 1675 [1], given: 62

Manager
Joined: 04 Nov 2012
Posts: 90

Kudos [?]: 34 [0], given: 82

Location: United States
GPA: 3.7
WE: Science (Pharmaceuticals and Biotech)
Re: If the three unique positive digits A,B, and C are arranged [#permalink]

### Show Tags

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).
_________________

kudos are appreciated

Kudos [?]: 34 [0], given: 82

Manager
Joined: 26 Jul 2011
Posts: 120

Kudos [?]: 134 [0], given: 16

Location: India
WE: Marketing (Manufacturing)
Re: If the three unique positive digits A,B, and C are arranged [#permalink]

### Show Tags

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

Kudos [?]: 134 [0], given: 16

VP
Status: Been a long time guys...
Joined: 03 Feb 2011
Posts: 1377

Kudos [?]: 1675 [0], given: 62

Location: United States (NY)
Concentration: Finance, Marketing
GPA: 3.75
Re: If the three unique positive digits A,B, and C are arranged [#permalink]

### Show Tags

09 Dec 2012, 22:49
_________________

Kudos [?]: 1675 [0], given: 62

Math Expert
Joined: 02 Sep 2009
Posts: 41892

Kudos [?]: 128949 [1], given: 12185

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

### Show Tags

10 Dec 2012, 01:50
1
KUDOS
Expert's post
7
This post was
BOOKMARKED
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.
_________________

Kudos [?]: 128949 [1], given: 12185

Manager
Joined: 26 Jul 2011
Posts: 120

Kudos [?]: 134 [0], given: 16

Location: India
WE: Marketing (Manufacturing)
Re: If the three unique positive digits A,B, and C are arranged [#permalink]

### Show Tags

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..

Kudos [?]: 134 [0], given: 16

Intern
Joined: 24 Apr 2012
Posts: 48

Kudos [?]: 24 [0], given: 1

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

### Show Tags

10 Dec 2012, 04:38
1
This post was
BOOKMARKED
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).
_________________

www.mnemoniceducation.com

Kudos [?]: 24 [0], given: 1

Math Expert
Joined: 02 Sep 2009
Posts: 41892

Kudos [?]: 128949 [0], given: 12185

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

### Show Tags

10 Dec 2012, 04:41
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
_________________

Kudos [?]: 128949 [0], given: 12185

GMAT Club Legend
Joined: 09 Sep 2013
Posts: 16640

Kudos [?]: 273 [0], given: 0

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

### Show Tags

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).

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.
_________________

Kudos [?]: 273 [0], given: 0

SVP
Status: The Best Or Nothing
Joined: 27 Dec 2012
Posts: 1853

Kudos [?]: 2622 [0], given: 193

Location: India
Concentration: General Management, Technology
WE: Information Technology (Computer Software)
Re: If the three unique positive digits A,B, and C are arranged [#permalink]

### Show Tags

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

_________________

Kindly press "+1 Kudos" to appreciate

Kudos [?]: 2622 [0], given: 193

GMAT Club Legend
Joined: 09 Sep 2013
Posts: 16640

Kudos [?]: 273 [0], given: 0

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

### Show Tags

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.
_________________

Kudos [?]: 273 [0], given: 0

BSchool Forum Moderator
Joined: 12 Aug 2015
Posts: 2212

Kudos [?]: 837 [0], given: 595

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

### Show Tags

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
_________________

Give me a hell yeah ...!!!!!

Kudos [?]: 837 [0], given: 595

Director
Joined: 26 Oct 2016
Posts: 694

Kudos [?]: 179 [0], given: 855

Location: United States
Schools: HBS '19
GMAT 1: 770 Q51 V44
GPA: 4
WE: Education (Education)
Re: If the three unique positive digits A,B, and C are arranged [#permalink]

### Show Tags

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 three-digit 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 three-digit 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.

_________________

Thanks & Regards,
Anaira Mitch

Kudos [?]: 179 [0], given: 855

Re: If the three unique positive digits A,B, and C are arranged   [#permalink] 13 Mar 2017, 02:28
Display posts from previous: Sort by