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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?

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.

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 _________________

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.

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

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.

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.

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.

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).

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.

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..

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