GMAT Question of the Day - Daily to your Mailbox; hard ones only

It is currently 16 Oct 2018, 05:14

Close

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
Your Progress

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

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

Close

Request Expert Reply

Confirm Cancel

In list L above, there are 3 positive integers, where each

  new topic post reply Question banks Downloads My Bookmarks Reviews Important topics  
Author Message
TAGS:

Hide Tags

Intern
Intern
avatar
Joined: 01 Dec 2012
Posts: 33
Concentration: Finance, Operations
GPA: 2.9
Reviews Badge
In list L above, there are 3 positive integers, where each  [#permalink]

Show Tags

New post Updated on: 18 Dec 2012, 02:50
5
18
00:00
A
B
C
D
E

Difficulty:

  85% (hard)

Question Stats:

48% (02:30) correct 52% (03:02) wrong based on 203 sessions

HideShow timer Statistics

List L: ABC, BCA, CAB

In list L above, there are 3 positive integers, where each of A, B, and C is a different nonzero digit. Which of the following is the sum of all the positive integers that MUST be factors of the sum of the integers in list L?

A. 47
B. 114
C. 152
D. 161
E. 488

Originally posted by MOKSH on 17 Dec 2012, 16:41.
Last edited by Bunuel on 18 Dec 2012, 02:50, edited 1 time in total.
Renamed the topic and edited the question.
Most Helpful Expert Reply
Math Expert
User avatar
V
Joined: 02 Sep 2009
Posts: 49914
Re: In list L above, there are 3 positive integers, where each  [#permalink]

Show Tags

New post 18 Dec 2012, 03:02
7
4
MOKSH wrote:
List L: ABC, BCA, CAB

In list L above, there are 3 positive integers, where each of A, B, and C is a different nonzero digit. Which of the following is the sum of all the positive integers that MUST be factors of the sum of the integers in list L?

A. 47
B. 114
C. 152
D. 161
E. 488


Sum = (100A + 10B + C) + (100B + 10C + A) + (100C + 10A + B) = 111*(A + B + C).

So, we have that the sum will be a multiple of 111=3*37.

As for A + B + C: if A=1, B=2 and C=4, then A + B + C = 7, so the sum will also be divisible by 7 BUT if A=1, B=2 and C=8, then A + B + C = 11, so the sum will also be divisible by 11. This implies that A + B + C will not produce the same factors for all possible values of A, B and C.

Therefore, we can say that 111*(A + B + C) MUST be divisible only by 111 --> so, by 1, 3, 37, and 111 --> 1 + 3 + 37 + 111 = 152.

Answer: C.

Hope it's clear.

P.S. Please read carefully and follow: rules-for-posting-please-read-this-before-posting-133935.html Pay attention to the rule #3.
_________________

New to the Math Forum?
Please read this: Ultimate GMAT Quantitative Megathread | All You Need for Quant | PLEASE READ AND FOLLOW: 12 Rules for Posting!!!

Resources:
GMAT Math Book | Triangles | Polygons | Coordinate Geometry | Factorials | Circles | Number Theory | Remainders; 8. Overlapping Sets | PDF of Math Book; 10. Remainders | GMAT Prep Software Analysis | SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS) | Tricky questions from previous years.

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. ,11 Mixed Questions, 12 Fresh Meat

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., 11 New DS set.


What are GMAT Club Tests?
Extra-hard Quant Tests with Brilliant Analytics

General Discussion
VP
VP
User avatar
Status: Been a long time guys...
Joined: 03 Feb 2011
Posts: 1170
Location: United States (NY)
Concentration: Finance, Marketing
GPA: 3.75
GMAT ToolKit User Premium Member
Re: In list L above, there are 3 positive integers, where each  [#permalink]

Show Tags

New post 18 Dec 2012, 03:21
3
1
The list L consists of :
ABC, BCA, CAB.

Sum of these numbers will be \(111(A+B+C)\).
Since A, B, C are distinct non-zero digits, hence minimum value of A+B+C=6. Hence the sum will be 111*6 or 111*7 or 111*8.........

So atleast \(111\) will be a factor of the sum of the integers.

On prime factorization, we will get 37 and 3 as the prime factors.

Sum of the factors:
{a^ (p+1) - 1}{b^ (q+1) - 1}{c^ (r+1) - 1} / (a-1)*(b-1)*(c-1)

Here a=37, b=3, p=1, q=1

On applying the formula:
we get \((1368*8)/(36*2)\)
or \(152\).
Hope that helps.
_________________

Prepositional Phrases Clarified|Elimination of BEING| Absolute Phrases Clarified
Rules For Posting
www.Univ-Scholarships.com

Manager
Manager
avatar
Joined: 02 Sep 2012
Posts: 211
Location: United States
Concentration: Entrepreneurship, Finance
GMAT Date: 07-25-2013
GPA: 3.83
WE: Architecture (Computer Hardware)
Re: In list L above, there are 3 positive integers, where each  [#permalink]

Show Tags

New post 20 Dec 2012, 11:16
very well explained bunnel
Manager
Manager
avatar
Joined: 02 Sep 2012
Posts: 211
Location: United States
Concentration: Entrepreneurship, Finance
GMAT Date: 07-25-2013
GPA: 3.83
WE: Architecture (Computer Hardware)
Re: In list L above, there are 3 positive integers, where each  [#permalink]

Show Tags

New post 20 Dec 2012, 11:24
1
To explain still elaborately
take any number of the format given:123+231+312=666==>111(1+2+3)
So now the question asks for which of the following will be the factors of all the three postive integers.
Only 111 can be the factor of all 3 +ve integers
So the factors of 111 are 1,3,37,111..
Making a sum ofall the factors gives the answer E
Intern
Intern
User avatar
Joined: 24 Apr 2012
Posts: 48
Re: In list L above, there are 3 positive integers, where each  [#permalink]

Show Tags

New post 21 Dec 2012, 04:12
Ans:

the sum of three numbers would be 100(A+B+C) +10(A+B+C) (A+B+C)= 111(A+B+C) ,

so we need to find the sum of factors of 111 which would be 1+3+37+111=152,

Therefore the answer is (C).
_________________

www.mnemoniceducation.com

TURN ON YOUR MINDS!!!

Math Expert
User avatar
V
Joined: 02 Sep 2009
Posts: 49914
Re: In list L above, there are 3 positive integers, where each  [#permalink]

Show Tags

New post 13 Jul 2013, 06:47
Senior Manager
Senior Manager
User avatar
P
Joined: 08 Jun 2013
Posts: 442
Location: India
Schools: INSEAD Jan '19
GMAT 1: 200 Q1 V1
GPA: 3.82
WE: Engineering (Other)
GMAT ToolKit User Premium Member
Re: In list L above, there are 3 positive integers, where each  [#permalink]

Show Tags

New post 22 Sep 2018, 19:50
KAPLAN OE :

The question is very involved and somewhat convoluted sounding! It asks us about distinct digits that are arranged in different orders to form 3 different numbers, which tells us that the digits each variable represents will remain constant, but each number’s value will change depending on the position each digit occupies (units, tens, hundreds). We need to determine the sum of all the factors of the sum of the three numbers. Since the question asks about those integers that must be factors, we should be wary of finding integers that could be factors. This question is testing both our calculation skills and knowledge of the position of digits in numbers.

Identify the Task:

We need to determine the integers in the list, their sum, and the factors of that sum.

Approach Strategically:

The calculations involved will make Backsolving a challenge. Since the question asks which integers must be factors, if we decide to Pick Numbers, we should be aware that we could wind up finding integers that are factors only with the values we pick, and as such we might need to pick multiple sets of numbers. This strategy is certainly a possibility, but doing the Straightforward Math will be our fastest option.

The first digit to the left of the decimal represents the ones or units’ digit. The digit to the immediate left of the units’ digit is the tens’ digit (and its value is 10 times whatever digit is in that place), and the digit to the immediate left of the tens’ digit is the hundreds’ digit (and its value is 100 times whatever digit is in that place). Therefore the value of the integer ABC is equal to 100A + 10B + C; the value of the integer BCA is equal to 100B + 10C + A; and the value of the integer CAB is equal to 100C + 10A + B.

The sum of the integers ABC, BCA, and CAB can therefore be expressed as:

, which equals 111(A + B + C)

The sum of the integers in list L must be a multiple of 111, but depending on the value of (A + B + C) it could be a multiple of several other numbers. Any positive integer that is a factor of 111 also must be a factor of the sum of the integers in list L. 111 = 3 × 37, and both 3 and 37 are prime numbers. So the positive integers that are factors of 111 are 1, 3, 37, and 111. When we sum these, we get 1 + 3 + 37 + 111 = 152. Answer Choice (C) is correct.

Confirm your Answer:

We can also Pick Numbers to verify this. The sum of the numbers in list L is 111(A + B + C). If A = 1, B = 2, and C = 4, then A + B + C =1 + 2 + 4 = 7. With these values, the sum 111(A + B + C) of the numbers in list L would have the factors: 1, 3, 37, 111, and 7.

However, If A = 1, B = 2, and C = 5, then A + B + C =1 + 2 + 5 = 8. With these values, the sum 111(A + B + C) of the numbers in list L has the factors 1, 3, 37, 111, and 8. Neither 7 nor 8 MUST be a factor of the sum, but 1, 3, 37, and 111 always must be.

If we picked numbers from the beginning, we would see something similar. If A = 1, B = 2 and C = 4, our list would be comprised of 124, 241 and 412. The sum would be 777. The prime factors would therefore be 3, 7 and 37, and all of the factors would be 1, 3, 7, 21, 37, 111, 259, 777. Without picking another set of numbers we could not easily determine which of these must be factors of the sum.

If A = 1, B = 2 and C = 5, our list would be comprised of 125, 251, and 512. The sum would be 888. The prime factors would be 2, 2, 2, 3 and 37 and all of the factors would be 1, 2, 3, 4, 6, 8, 12, 24, 37, 74, 111, 148, 222, 296, 444, 888. Only 1, 3, 37 and 111 are common to both lists.
_________________

It seems Kudos button not working correctly with all my posts...

Please check if it is working with this post......

is it?....

Anyways...Thanks for trying :cool:

GMAT Club Bot
Re: In list L above, there are 3 positive integers, where each &nbs [#permalink] 22 Sep 2018, 19:50
Display posts from previous: Sort by

In list L above, there are 3 positive integers, where each

  new topic post reply Question banks Downloads My Bookmarks Reviews Important topics  


Copyright

GMAT Club MBA Forum Home| About| Terms and Conditions and Privacy Policy| GMAT Club Rules| Contact| Sitemap

Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne

Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®.