It is currently 18 Nov 2017, 20:29

### 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

# Given that a, b, c, and d are different nonzero digits and

Author Message
TAGS:

### Hide Tags

Director
Status: Finally Done. Admitted in Kellogg for 2015 intake
Joined: 25 Jun 2011
Posts: 533

Kudos [?]: 4211 [2], given: 217

Location: United Kingdom
GMAT 1: 730 Q49 V45
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
KUDOS
23
This post was
BOOKMARKED
00:00

Difficulty:

95% (hard)

Question Stats:

40% (01:46) correct 60% (02:26) wrong based on 238 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?
[Reveal] Spoiler: OA

_________________

Best Regards,
E.

MGMAT 1 --> 530
MGMAT 2--> 640
MGMAT 3 ---> 610
GMAT ==> 730

Kudos [?]: 4211 [2], given: 217

Math Expert
Joined: 02 Sep 2009
Posts: 42249

Kudos [?]: 132625 [17], given: 12326

### Show Tags

31 Jan 2012, 18:14
17
KUDOS
Expert's post
10
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 two-digit integer dc and (10c+a) is the way of writing two-digit integer ca. Look at the sum:

abdc
+dbca
-----

Now, as two-digit integer dc + two-digit 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 8581 could not be the sum of abdc+dbca (for any valid digits of a, b, c, and d).

Hope it helps.
_________________

Kudos [?]: 132625 [17], given: 12326

Director
Status: Finally Done. Admitted in Kellogg for 2015 intake
Joined: 25 Jun 2011
Posts: 533

Kudos [?]: 4211 [1], given: 217

Location: United Kingdom
GMAT 1: 730 Q49 V45
GPA: 2.9
WE: Information Technology (Consulting)

### Show Tags

31 Jan 2012, 18:21
1
KUDOS
Bunuel - thanks for this buddy. Its almost clear apart from how did you get

Now, as two-digit integer dc + two-digit 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

Kudos [?]: 4211 [1], given: 217

Math Expert
Joined: 02 Sep 2009
Posts: 42249

Kudos [?]: 132625 [1], given: 12326

### Show Tags

31 Jan 2012, 18:27
1
KUDOS
Expert's post
enigma123 wrote:
Bunuel - thanks for this buddy. Its almost clear apart from how did you get

Now, as two-digit integer dc + two-digit 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

Two-digit integer dc + two-digit integer ca is less than 100: suppose they are 12 and 23:
ab12
+db23
-----
XY35

So, there is no carried over 1 to hundreds place or to b+b. Next, b+b=2b=even so the hundreds digit of the given sum, Y in our example, must be even too.

Hope it's clear.
_________________

Kudos [?]: 132625 [1], given: 12326

Director
Status: Finally Done. Admitted in Kellogg for 2015 intake
Joined: 25 Jun 2011
Posts: 533

Kudos [?]: 4211 [0], given: 217

Location: United Kingdom
GMAT 1: 730 Q49 V45
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

Kudos [?]: 4211 [0], given: 217

Manager
Status: MBA Aspirant
Joined: 12 Jun 2010
Posts: 171

Kudos [?]: 102 [0], given: 1

Location: India
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

Kudos [?]: 102 [0], given: 1

Intern
Joined: 06 Nov 2011
Posts: 36

Kudos [?]: 8 [0], given: 50

Location: Germany
Concentration: Entrepreneurship, General Management
GMAT Date: 03-10-2012
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.

Kudos [?]: 8 [0], given: 50

Math Expert
Joined: 02 Sep 2009
Posts: 42249

Kudos [?]: 132625 [0], given: 12326

Re: Given that a, b, c, and d are different nonzero digits and [#permalink]

### Show Tags

19 Feb 2012, 22:36
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 ToolKit

No 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.
_________________

Kudos [?]: 132625 [0], given: 12326

Intern
Joined: 06 Nov 2011
Posts: 36

Kudos [?]: 8 [0], given: 50

Location: Germany
Concentration: Entrepreneurship, General Management
GMAT Date: 03-10-2012
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 ToolKit

No 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.

Kudos [?]: 8 [0], given: 50

Intern
Joined: 02 May 2013
Posts: 24

Kudos [?]: 20 [0], given: 76

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.

Kudos [?]: 20 [0], given: 76

Intern
Joined: 19 Aug 2013
Posts: 18

Kudos [?]: 11 [0], given: 0

Location: Germany
Schools: HKU '15 (A)
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... :-/

Kudos [?]: 11 [0], given: 0

Math Expert
Joined: 02 Sep 2009
Posts: 42249

Kudos [?]: 132625 [0], given: 12326

Re: Given that a, b, c, and d are different nonzero digits and [#permalink]

### Show Tags

21 Nov 2013, 08:11
wfmd wrote:
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... :-/

10d + 11c < 100 – a

Add a to both sides: 10d + 11c + a< 100;
10d + (10c + c) + a< 100;
(10d+c)+(10c+a)<100.

Hope it's clear.
_________________

Kudos [?]: 132625 [0], given: 12326

Intern
Joined: 19 Aug 2013
Posts: 18

Kudos [?]: 11 [0], given: 0

Location: Germany
Schools: HKU '15 (A)
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!

Kudos [?]: 11 [0], given: 0

Intern
Joined: 24 Feb 2017
Posts: 38

Kudos [?]: 11 [0], given: 38

Schools: CBS '20 (S)
GMAT 1: 760 Q50 V42
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 two-digit integer dc and (10c+a) is the way of writing two-digit integer ca. Look at the sum:

abdc
+dbca
-----

Now, as two-digit integer dc + two-digit 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 8581 could not be the sum of abdc+dbca (for any valid digits of a, b, c, and d).

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<100-a implies 11c <100-a-10d.

When we substitute for 11c in S, we get 1001a + 1010d + 200b + 11c < 1001a + 1010d + 200b + (100-a-10d)
=> 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.

Kudos [?]: 11 [0], given: 38

Manager
Joined: 03 Jan 2017
Posts: 197

Kudos [?]: 9 [0], given: 4

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

Kudos [?]: 9 [0], given: 4

Intern
Joined: 25 Jul 2011
Posts: 44

Kudos [?]: 5 [0], given: 1058

Location: India
Concentration: Strategy, Operations
GMAT 1: 740 Q49 V41
GPA: 3.5
WE: Engineering (Energy and Utilities)
Given that a, b, c, and d are different nonzero digits and [#permalink]

### Show Tags

19 Aug 2017, 07:11
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?

This is one of the easiest methods;

abdc
+ dbca

Clearly unit digit of sum will depend upon (c+a) and tenth digit will depend upon (d+c);
We have 10d + 11c < 100 – a
or we can say 10d+11c+a< 100
or 10d+10c+c+a< 100
or 10(d+c)+ (c+a)< 100
Now just go through the options;

1. 3689 so (c+a)=9 and we have 10(d+c)+ (c+a)< 100..so (d+c)< 9...here it is 8 (tenth digit) ..so possible
2. 6887 so (c+a)=7 or 17 and we have 10(d+c)+ (c+a)< 100..so (d+c)<9 or 8...here it is 8 (tenth digit) ..so possible
3. 8581 so (c+a)=11..it cant be 1 as a, b, c, and d are different nonzero digits and we have 10(d+c)+ (c+a)< 100..so (d+c)< 8...here it is 8 (tenth digit) ..which can not be possible....So option C..

Give kudos if you find this solution useful. :)
_________________

Kudos [?]: 5 [0], given: 1058

Given that a, b, c, and d are different nonzero digits and   [#permalink] 19 Aug 2017, 07:11
Display posts from previous: Sort by