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Tough and tricky 4: addition problem

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Bunuel
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Tough and tricky 4: addition problem [#permalink] New post   11 Oct 2009, 17:42
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Tough and tricky 4: addition problem

AB + CD = AAA, where AB and CD are two-digit numbers and AAA is a three digit number; A, B, C, and D are distinct positive integers. In the addition problem above, what is the value of C?

(A) 1

(B) 3

(C) 7

(D) 9

(E) Cannot be determined

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Re: Tough and tricky 4: addition problem [#permalink] New post   11 Oct 2009, 17:55
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The answer is D: 9.

Logical deduction.

1) Two 2-digit numbers sum to a 3 digit number in the form AAA. Since AB < 100 and CD < 100, AB + CD must be less than 200. So AAA = 111, and A = 1.

2) Now we know, since A is equal to 1, that AB < 20. The only way that AB + CD = 111, if AB < 20, is that if CD > 90. Therefore, C = 9.
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Re: Tough and tricky 4: addition problem [#permalink] New post   11 Oct 2009, 19:17
Answer is D = 9.
Same logic as explained by AKProdigy87.
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Re: Tough and tricky 4: addition problem [#permalink] New post   12 Oct 2009, 09:35
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Set of Tough & tricky questions is now combined in one thread: tough-tricky-set-of-problms-85211.html

You can continue discussions and see the solutions there.
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Re: Tough and tricky 4: addition problem [#permalink] New post   12 May 2013, 02:52
AKProdigy87 wrote:
The answer is D: 9.

Logical deduction.

1) Two 2-digit numbers sum to a 3 digit number in the form AAA. Since AB < 100 and CD < 100, AB + CD must be less than 200. So AAA = 111, and A = 1.

2) Now we know, since A is equal to 1, that AB < 20. The only way that AB + CD = 111, if AB < 20, is that if CD > 90. Therefore, C = 9.



sorry I disagree.
as 99 +12 =111 0r 12+99=111 C can be 9 or 1, again 76+25=111 or 25+76=111, C can be 2 or 7. so my answer is E.
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Re: Tough and tricky 4: addition problem [#permalink] New post   12 May 2013, 03:04
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khosru wrote:
AKProdigy87 wrote:
The answer is D: 9.

Logical deduction.

1) Two 2-digit numbers sum to a 3 digit number in the form AAA. Since AB < 100 and CD < 100, AB + CD must be less than 200. So AAA = 111, and A = 1.

2) Now we know, since A is equal to 1, that AB < 20. The only way that AB + CD = 111, if AB < 20, is that if CD > 90. Therefore, C = 9.



sorry I disagree.
as 99 +12 =111 0r 12+99=111 C can be 9 or 1, again 76+25=111 or 25+76=111, C can be 2 or 7. so my answer is E.


That's not correct.

AB and CD are two digit integers, their sum can give us only one 3-digit integer of a kind of AAA: 111. So, A=1 --> 1B+CD=111.

C can not be less than 9, because no 2-digit integer with first digit 1 (mean that it's <20) can be added to 2-digit integer less than 90 to have the sum 111 (if CD<90 meaning C<9 CD+1B<111) --> C=9.

Answer: D.
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Re: Tough and tricky 4: addition problem [#permalink] New post   12 May 2013, 11:11
Different approach...

A + C + x (Carryover) = AA.. Since all integers are positive C cant be zero...

=> C + x (Carryover) = 10

Now the only carryover that can come from B + D is 1

Hence C = 9
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Re: Tough and tricky 4: addition problem [#permalink] New post   13 May 2013, 01:46
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Bunuel wrote:
Tough and tricky 4: addition problem

AB + CD = AAA, where AB and CD are two-digit numbers and AAA is a three digit number; A, B, C, and D are distinct positive integers. In the addition problem above, what is the value of C?

(A) 1

(B) 3

(C) 7

(D) 9

(E) Cannot be determined



A + C gives A . So c must be either 0 or 10. I the answer choices there is no option with zero. so c must be 10.for c to be 10 we must get a carry over and c MUSt be 9 as the maximum carry over we can get is 1 . so D
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Re: Tough and tricky 4: addition problem [#permalink] New post   13 May 2013, 04:11
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Bunuel wrote:
Tough and tricky 4: addition problem

AB + CD = AAA, where AB and CD are two-digit numbers and AAA is a three digit number; A, B, C, and D are distinct positive integers. In the addition problem above, what is the value of C?

(A) 1

(B) 3

(C) 7

(D) 9

(E) Cannot be determined


AB<100 and CD<100.So , 20<=AB+CD<200.From 20 to 199 which is in the form of AAA is only one number i.e 111.So A=1,and AB=1B<20 because the tens place has 1.

1B+CD=111,let us consider
1.) B=2 => CD = 99 AB+CD=12+99=111, C=9

2)B=3 => AB=13 =>CD = 98 . C=9

take B=4,5,6,7,8,9 ,we get C= 9
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Re: Tough and tricky 4: addition problem [#permalink] New post   26 Sep 2013, 06:17
SrinathVangala wrote:
Bunuel wrote:
Tough and tricky 4: addition problem

AB + CD = AAA, where AB and CD are two-digit numbers and AAA is a three digit number; A, B, C, and D are distinct positive integers. In the addition problem above, what is the value of C?

(A) 1

(B) 3

(C) 7

(D) 9

(E) Cannot be determined



A + C gives A . So c must be either 0 or 10. I the answer choices there is no option with zero. so c must be 10.for c to be 10 we must get a carry over and c MUSt be 9 as the maximum carry over we can get is 1 . so D


Correct, also C cannot be zero because the question says they are all positive integers

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This topic has been closed and archived due to inactivity or violation of community quality standards. No more replies are possible here.
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Re: Tough and tricky 4: addition problem [#permalink]
26 Sep 2013, 06:17
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