|
Author |
Message |
|
TAGS:
|
|
|
Senior Manager
Status: struggling with GMAT
Joined: 06 Dec 2012
Posts: 322
Location: Bangladesh
Concentration: Accounting
GMAT Date: 04-06-2013
GPA: 3.65
Followers: 5
Kudos [?]:
29
[0], given: 46
|
If the three unique positive digits A,B, and C are arranged [#permalink]
 09 Dec 2012, 09:59
Question Stats:
42% (02:45) correct
57% (00:36) wrong based on 3 sessions
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
|
|
|
|
|
|
|
Verbal GMAT Forum Moderator
Status: Preparing for the another shot...!
Joined: 03 Feb 2011
Posts: 1326
Location: India
Concentration: Finance, Marketing
GMAT 1: 720 Q V
GPA: 3.75
Followers: 64
Kudos [?]:
315
[1] , given: 52
|
Re: If the three unique positive digits A,B, and C are arranged [#permalink]
09 Dec 2012, 20:16
1
This post received
KUDOS
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.
_________________
Prepositional Phrases Clarified|Elimination of BEING| Absolute Phrases Clarified
Rules For Posting
|
|
|
|
|
|
Manager
Joined: 04 Nov 2012
Posts: 65
Location: United States
Concentration: Finance, International Business
GMAT Date: 02-23-2013
GPA: 3.7
WE: Science (Pharmaceuticals and Biotech)
Followers: 0
Kudos [?]:
21
[0], given: 71
|
Re: If the three unique positive digits A,B, and C are arranged [#permalink]
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).
_________________
kudos are appreciated
|
|
|
|
|
|
Manager
Joined: 26 Jul 2011
Posts: 121
Location: India
WE: Marketing (Manufacturing)
Followers: 1
Kudos [?]:
14
[0], given: 15
|
Re: If the three unique positive digits A,B, and C are arranged [#permalink]
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
|
|
|
|
|
|
Verbal GMAT Forum Moderator
Status: Preparing for the another shot...!
Joined: 03 Feb 2011
Posts: 1326
Location: India
Concentration: Finance, Marketing
GMAT 1: 720 Q V
GPA: 3.75
Followers: 64
Kudos [?]:
315
[0], given: 52
|
Re: If the three unique positive digits A,B, and C are arranged [#permalink]
09 Dec 2012, 22:49
|
|
|
|
|
|
GMAT Club team member
Joined: 02 Sep 2009
Posts: 11632
Followers: 1802
Kudos [?]:
9607
[0], given: 829
|
Re: If the three unique positive digits A,B, and C are arranged [#permalink]
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: (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.htmlwhat-is-the-sum-of-all-3-digit-positive-integers-that-can-be-78143.htmlfind-the-sum-of-all-the-four-digit-numbers-formed-using-the-103523.htmlwhat-is-the-sum-of-all-4-digit-numbers-that-can-be-formed-94836.htmlwhat-is-the-sum-of-all-3-digit-positive-integers-that-can-be-78143.htmlfind-the-sum-of-all-the-four-digit-numbers-which-are-formed-88357.htmlfind-the-sum-of-all-3-digit-nos-that-can-be-formed-by-88864.htmlthere-are-24-different-four-digit-integers-than-can-be-141891.htmlHope it helps.
_________________
PLEASE READ AND FOLLOW: 11 Rules for Posting!!!
RESOURCES: [GMAT MATH BOOK]; 1. Triangles; 2. Polygons; 3. Coordinate Geometry; 4. Factorials; 5. Circles; 6. Number Theory
COLLECTION OF QUESTIONS:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. NEW!!!
DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS ; 9 Devil's Dozen!!!; 10 Number Properties set. NEW!!!
What are GMAT Club Tests?
25 extra-hard Quant Tests
Find out what's new at GMAT Club - latest features and updates
|
|
|
|
|
|
Manager
Joined: 26 Jul 2011
Posts: 121
Location: India
WE: Marketing (Manufacturing)
Followers: 1
Kudos [?]:
14
[0], given: 15
|
Re: If the three unique positive digits A,B, and C are arranged [#permalink]
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..
|
|
|
|
|
|
Intern
Joined: 24 Apr 2012
Posts: 44
Followers: 0
Kudos [?]:
7
[0], given: 1
|
Re: If the three unique positive digits A,B, and C are arranged [#permalink]
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).
_________________
www.mnemoniceducation.com
TURN ON YOUR MINDS!!!
|
|
|
|
|
|
GMAT Club team member
Joined: 02 Sep 2009
Posts: 11632
Followers: 1802
Kudos [?]:
9607
[0], given: 829
|
Re: If the three unique positive digits A,B, and C are arranged [#permalink]
10 Dec 2012, 04:41
|
|
|
|
|
|
|
Re: If the three unique positive digits A,B, and C are arranged
[#permalink]
10 Dec 2012, 04:41
|
|
|
|
|
|
|
|
|
|
|
close
