Author 
Message 
TAGS:

Hide Tags

Director
Status: Finally Done. Admitted in Kellogg for 2015 intake
Joined: 25 Jun 2011
Posts: 537
Location: United Kingdom
Concentration: International Business, Strategy
GPA: 2.9
WE: Information Technology (Consulting)

Given that a, b, c, and d are different nonzero digits and [#permalink]
Show Tags
31 Jan 2012, 17:51
2
This post received KUDOS
15
This post was BOOKMARKED
Question Stats:
40% (02:36) correct
60% (02:12) wrong based on 184 sessions
HideShow timer Statistics
Given that a, b, c, and d are different nonzero digits and that 10d + 11c < 100 – a, which of the following could not be a solution to the addition problem below? abdc + dbca (A) 3689 (B) 6887 (C) 8581 (D) 9459 (E) 16091 This is the original question and its beyond my head to solve this. can someone please help and try to explain the concept of this? Also , can you please tell how to solve if the question says COULD BE a solution to the addition problem?
Official Answer and Stats are available only to registered users. Register/ Login.
_________________
Best Regards, E.
MGMAT 1 > 530 MGMAT 2> 640 MGMAT 3 > 610 GMAT ==> 730



Math Expert
Joined: 02 Sep 2009
Posts: 39589

Re: Digits [#permalink]
Show Tags
31 Jan 2012, 18:14
14
This post received KUDOS
Expert's post
6
This post was BOOKMARKED
enigma123 wrote: Given that a, b, c, and d are different nonzero digits and that 10d + 11c < 100 – a, which of the following could not be a solution to the addition problem below? abdc + dbca (A) 3689 (B) 6887 (C) 8581 (D) 9459 (E) 16091
This is the original question and its beyond my head to solve this. can someone please help and try to explain the concept of this?
Also , can you please tell how to solve if the question says COULD BE a solution to the addition problem? 10d + 11c < 100 – a > 10d+11c+a<100 > (10d+c)+(10c+a)<100. (10d+c) is the way of writing twodigit integer dc and (10c+a) is the way of writing twodigit integer ca. Look at the sum: ab dc+db ca Now, as twodigit integer dc + twodigit integer ca is less than 100, then there won't be carried over 1 to the hundreds place and as b+b=2b=even then the hundreds digit of the given sum must be even too. Thus 8 581 could not be the sum of abdc+dbca (for any valid digits of a, b, c, and d). Answer: C. As for "could be true" questions: try our new "Must or Could be True Questions" tag  search.php?search_id=tag&tag_id=193 to learn more about this type of questions. Hope it helps.
_________________
New to the Math Forum? Please read this: 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? Extrahard Quant Tests with Brilliant Analytics



Director
Status: Finally Done. Admitted in Kellogg for 2015 intake
Joined: 25 Jun 2011
Posts: 537
Location: United Kingdom
Concentration: International Business, Strategy
GPA: 2.9
WE: Information Technology (Consulting)

Re: Digits [#permalink]
Show Tags
31 Jan 2012, 18:21
1
This post received KUDOS
Bunuel  thanks for this buddy. Its almost clear apart from how did you get Now, as twodigit integer dc + twodigit integer ca is less than 100 then there won't be carried over 1 to the hundreds place and as b+b must be ecen then the hundreds digit of the given sum must be even too
_________________
Best Regards, E.
MGMAT 1 > 530 MGMAT 2> 640 MGMAT 3 > 610 GMAT ==> 730



Math Expert
Joined: 02 Sep 2009
Posts: 39589

Re: Digits [#permalink]
Show Tags
31 Jan 2012, 18:27



Director
Status: Finally Done. Admitted in Kellogg for 2015 intake
Joined: 25 Jun 2011
Posts: 537
Location: United Kingdom
Concentration: International Business, Strategy
GPA: 2.9
WE: Information Technology (Consulting)

Re: Given that a, b, c, and d are different nonzero digits and [#permalink]
Show Tags
31 Jan 2012, 18:29
Perfect!!!! What an explanation. Thanks a ton.
_________________
Best Regards, E.
MGMAT 1 > 530 MGMAT 2> 640 MGMAT 3 > 610 GMAT ==> 730



Manager
Status: MBA Aspirant
Joined: 12 Jun 2010
Posts: 176
Location: India
Concentration: Finance, International Business
WE: Information Technology (Investment Banking)

Re: Given that a, b, c, and d are different nonzero digits and [#permalink]
Show Tags
31 Jan 2012, 21:08
enigma123 wrote: Given that a, b, c, and d are different nonzero digits and that 10d + 11c < 100 – a, which of the following could not be a solution to the addition problem below? abdc + dbca (A) 3689 (B) 6887 (C) 8581 (D) 9459 (E) 16091
This is the original question and its beyond my head to solve this. can someone please help and try to explain the concept of this?
Also , can you please tell how to solve if the question says COULD BE a solution to the addition problem? Given 10d+11c+a<100 now we have to find the sum of > 1000a+100b+10d+c+1000d+100b+10c+a we simplify it further to get > 1000a+1000d+200b+(11c+10d+a) we know that the value in bracket has to be less than 100 now take the options  let us take C which is 8581 we can break it to 8000+500+81 > since each of a,b,c and d is an integer so we cannot find a value of b to get the sum of 500 Thus C is the answer



Intern
Joined: 06 Nov 2011
Posts: 37
Location: Germany
Concentration: Entrepreneurship, General Management
GMAT Date: 03102012
GPA: 3

Re: Given that a, b, c, and d are different nonzero digits and [#permalink]
Show Tags
19 Feb 2012, 15:55
Isn't it be easier just to look for b? The hundreds digit of the sum must be even. There is only one answer choice where the hundreds digit is odd. Posted from GMAT ToolKit
Last edited by M3tm4n on 20 Feb 2012, 04:16, edited 1 time in total.



Math Expert
Joined: 02 Sep 2009
Posts: 39589

Re: Given that a, b, c, and d are different nonzero digits and [#permalink]
Show Tags
19 Feb 2012, 22:36



Intern
Joined: 06 Nov 2011
Posts: 37
Location: Germany
Concentration: Entrepreneurship, General Management
GMAT Date: 03102012
GPA: 3

Re: Given that a, b, c, and d are different nonzero digits and [#permalink]
Show Tags
20 Feb 2012, 04:22
Bunuel wrote: M3tm4n wrote: Isn't it be easier just to look for b? The hundreds digit of the sum must be even. There is online one answer choice where the hundreds digit is odd. Posted from GMAT ToolKitNo that's not correct. The hundreds digit will be even if there is no carried over 1 from the sum of the tens digits. So you should check this first. Please refer to the complete solution above. Hope it helps. Okay, got it. Thank you.



Intern
Joined: 02 May 2013
Posts: 24

Re: Given that a, b, c, and d are different nonzero digits and [#permalink]
Show Tags
03 Jun 2013, 08:16
enigma123 wrote: Given that a, b, c, and d are different nonzero digits and that 10d + 11c < 100 – a, which of the following could not be a solution to the addition problem below?
abdc + dbca
(A) 3689 (B) 6887 (C) 8581 (D) 9459 (E) 16091
This is the original question and its beyond my head to solve this. can someone please help and try to explain the concept of this?
Also , can you please tell how to solve if the question says COULD BE a solution to the addition problem? Bunuel's solution is always the perfect one, but I just wanted to share what I did. Pretty much the same as above, though... 10d + 11c + a = 10d + 10c + c + a < 100 this means, 10d + 10c <= 90 (since it cannot be equal to hundred) then, 10(d+c) <= 90 thus, d + c < 10 and also, c + a < 10 This concludes that hundreds digit needs to be even because there is no carry over.



Senior Manager
Joined: 03 Dec 2012
Posts: 339

Re: Given that a, b, c, and d are different nonzero digits and [#permalink]
Show Tags
15 Nov 2013, 21:25
this is certainly a 750+ level question. it wud take up a lot of time in the xam.



Intern
Joined: 19 Aug 2013
Posts: 18
Location: Germany
Concentration: International Business, Technology
GMAT 1: 580 Q35 V35 GMAT 2: 690 Q44 V40
GPA: 3.85
WE: Information Technology (Consulting)

Re: Given that a, b, c, and d are different nonzero digits and [#permalink]
Show Tags
21 Nov 2013, 07:42
Hey Bunuel, most is clear, except for the starting point: Quote: 10d + 11c < 100 – a > 10d+11c+a<100 > (10d+c)+(10c+a)<100 Why is 10d + 11c < 100 – a the same as (10d+c)+(10c+a)<100 ??? That's not clear for me... :/



Math Expert
Joined: 02 Sep 2009
Posts: 39589

Re: Given that a, b, c, and d are different nonzero digits and [#permalink]
Show Tags
21 Nov 2013, 08:11



Intern
Joined: 19 Aug 2013
Posts: 18
Location: Germany
Concentration: International Business, Technology
GMAT 1: 580 Q35 V35 GMAT 2: 690 Q44 V40
GPA: 3.85
WE: Information Technology (Consulting)

Re: Given that a, b, c, and d are different nonzero digits and [#permalink]
Show Tags
21 Nov 2013, 08:13
Ok, got it now...I think I worked on too many problems today, so I oversaw that... Anyway, thanks!



GMAT Club Legend
Joined: 09 Sep 2013
Posts: 15916

Re: Given that a, b, c, and d are different nonzero digits and [#permalink]
Show Tags
27 Jun 2015, 23:50
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.
_________________
GMAT Books  GMAT Club Tests  Best Prices on GMAT Courses  GMAT Mobile App  Math Resources  Verbal Resources



GMAT Club Legend
Joined: 09 Sep 2013
Posts: 15916

Re: Given that a, b, c, and d are different nonzero digits and [#permalink]
Show Tags
01 Jan 2017, 14:44
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.
_________________
GMAT Books  GMAT Club Tests  Best Prices on GMAT Courses  GMAT Mobile App  Math Resources  Verbal Resources



Intern
Joined: 24 Feb 2017
Posts: 3

Re: Given that a, b, c, and d are different nonzero digits and [#permalink]
Show Tags
03 Mar 2017, 10:08
Bunuel wrote: enigma123 wrote: Given that a, b, c, and d are different nonzero digits and that 10d + 11c < 100 – a, which of the following could not be a solution to the addition problem below? abdc + dbca (A) 3689 (B) 6887 (C) 8581 (D) 9459 (E) 16091
This is the original question and its beyond my head to solve this. can someone please help and try to explain the concept of this?
Also , can you please tell how to solve if the question says COULD BE a solution to the addition problem? 10d + 11c < 100 – a > 10d+11c+a<100 > (10d+c)+(10c+a)<100. (10d+c) is the way of writing twodigit integer dc and (10c+a) is the way of writing twodigit integer ca. Look at the sum: ab dc+db ca Now, as twodigit integer dc + twodigit integer ca is less than 100, then there won't be carried over 1 to the hundreds place and as b+b=2b=even then the hundreds digit of the given sum must be even too. Thus 8 581 could not be the sum of abdc+dbca (for any valid digits of a, b, c, and d). Answer: C. As for "could be true" questions: [b]try our new "Must or Could be True Questions" to learn more about this type of questions. Hope it helps. I adopted a different approach and was able to eliminate a different option. The numeric form of abdc + dbca, lets call it sum S, is (1001a + 1010d + 200b + 11c). From the the other given condition 10d+11c<100a implies 11c <100a10d. When we substitute for 11c in S, we get 1001a + 1010d + 200b + 11c < 1001a + 1010d + 200b + (100a10d) => S < 1000a + 1000d + 200b + 100 The minumum value of S (given that a,b,c,d are digits i.e. positive DIFFERENT integers) will be 3700. So, option A (3689) can never be a value of S.



Manager
Joined: 03 Jan 2017
Posts: 201

Re: Given that a, b, c, and d are different nonzero digits and [#permalink]
Show Tags
18 Mar 2017, 13:29
if we take b digit for a quick analysis, 2b in the sum is supposed to b even. C is the only answer that stands out




Re: Given that a, b, c, and d are different nonzero digits and
[#permalink]
18 Mar 2017, 13:29








Similar topics 
Author 
Replies 
Last post 
Similar Topics:


85


Let a,b,c, and d be nonzero real numbers

amitpaul527 
9 
17 Mar 2017, 09:23 

3


If a, b, c, d, and x are nonzero numbers, which of the following condi

Bunuel 
2 
06 Nov 2016, 20:57 

16


If a, b, and c are different nonnegative digits, which of

monsoon1 
7 
23 Mar 2017, 19:12 

40


How many different arrangements of A, B, C, D, and E are pos

rxs0005 
16 
18 Nov 2016, 04:33 

7


a, b, c, d and e are all nonzero distinct single digit integers. What

ykaiim 
5 
27 Jun 2016, 21:51 



