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Given that a, b, c, and d are different nonzero digits and that 10d +  [#permalink]

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Question Stats: 32% (02:34) correct 68% (02:42) wrong based on 257 sessions

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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?

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Re: Given that a, b, c, and d are different nonzero digits and that 10d +  [#permalink]

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28
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?

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).

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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Re: Given that a, b, c, and d are different nonzero digits and that 10d +  [#permalink]

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1
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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Re: Given that a, b, c, and d are different nonzero digits and that 10d +  [#permalink]

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1
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 that 10d +  [#permalink]

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Perfect!!!! What an explanation. Thanks a ton.
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Re: Given that a, b, c, and d are different nonzero digits and that 10d +  [#permalink]

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1
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 that 10d +  [#permalink]

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1
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

Originally posted by M3tm4n on 19 Feb 2012, 15:55.
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 that 10d +  [#permalink]

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1
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 that 10d +  [#permalink]

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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 that 10d +  [#permalink]

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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?

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 that 10d +  [#permalink]

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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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Re: Given that a, b, c, and d are different nonzero digits and that 10d +  [#permalink]

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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 that 10d +  [#permalink]

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Ok, got it now...I think I worked on too many problems today, so I oversaw that...
Anyway, thanks!
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GMAT 1: 760 Q50 V42 Re: Given that a, b, c, and d are different nonzero digits and that 10d +  [#permalink]

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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).

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 that 10d +  [#permalink]

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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
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Re: Given that a, b, c, and d are different nonzero digits and that 10d +  [#permalink]

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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?

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. :)
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Re: Given that a, b, c, and d are different nonzero digits and that 10d +  [#permalink]

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1
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).

As for "could be true" questions: try our new "Must or Could be True Questions" tag - http://gmatclub.com/forum/search.php?se ... tag_id=193 to learn more about this type of questions.

Hope it helps.

Best explanation ever. couldn't be more clear.
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Re: Given that a, b, c, and d are different nonzero digits and that 10d +  [#permalink]

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A few things here :
(1) If there a numerical term in an equation/inequality, a question can be asked why THIS NUMBER here (NOT OTHER NUMBER).
(2) In additional & subtraction questions with UNKNOWN variables, an IMPORTANT question: is there any CARRY FORWARD value?
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Re: Given that a, b, c, and d are different nonzero digits and that 10d +  [#permalink]

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In this ADDITION question:
(1) Why 100 is here, not 100000000 is here. (Perhaps anything related with LAST two digit is being TESTED because in addition or subtraction question, the unknown variables are given in 100x+10y+z format).
Lets Check:
(a) Is the given info “10d+11c+a” related with LAST two digit of first term (abdc)?.......NO
(b) Is the given info “10d+11c+a” related with LAST two digit of second term (dbca)?.......NO
(c) Is the given info “10d+11c+a” related with LAST two digit of SUM of those two terms (abdc+dbca)?.......YES. because last two digit of the sum is [(10d+c)+ (10c+a)]=10d+11c+a
GOT it: last two digit of the sum is LESS than 100.

(2) That means there is NO CARRY FORWARD value to the HUNDRED digit.
Since hundred digit in two terms is “b” , sum in hundred digit should be 2m . i.e. EVEN.
All options here have EVEN numbers except “c”, which has ODD (5) number .
So, C could not be the SUM of the two terms. Re: Given that a, b, c, and d are different nonzero digits and that 10d +   [#permalink] 05 Sep 2019, 21:27
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