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

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

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31 Jan 2012, 17:51
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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

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

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Posts: 39589
Re: Digits [#permalink]

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31 Jan 2012, 18:14
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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).

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.
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Director
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Re: Digits [#permalink]

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31 Jan 2012, 18:21
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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
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E.

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

Math Expert
Joined: 02 Sep 2009
Posts: 39589
Re: Digits [#permalink]

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31 Jan 2012, 18:27
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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.
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Re: Given that a, b, c, and d are different nonzero digits and [#permalink]

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31 Jan 2012, 18:29
Perfect!!!! What an explanation. Thanks a ton.
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E.

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

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

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

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

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

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

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

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

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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... :-/
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Posts: 39589
Re: Given that a, b, c, and d are different nonzero digits and [#permalink]

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

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

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27 Jun 2015, 23:50
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Re: Given that a, b, c, and d are different nonzero digits and [#permalink]

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01 Jan 2017, 14:44
Hello from the GMAT Club BumpBot!

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

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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).

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

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

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