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AB + CD = AAA, where AB and CD are 2 digit numbers and AAA

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AB + CD = AAA, where AB and CD are 2 digit numbers and AAA [#permalink] New post 08 Jun 2005, 10:32
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B
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D
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71% (02:31) correct 29% (01:11) wrong based on 84 sessions
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

OPEN DISCUSSION OF THIS QUESTION IS HERE: ab-cd-aaa-where-ab-and-cd-are-two-digit-numbers-and-aa-136903.html
[Reveal] Spoiler: OA
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 [#permalink] New post 08 Jun 2005, 12:08
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Think I got it one way.

From AB + CD = AAA, we get 10A + B + 10C + D = 111A

Meaning, 101A = 10C + (B + D) ------------- (A)

Now, there are 2 possibilities,

1) B + D = A. Putting B+D=A in A, we get C = 10A. But that would violate the data given. Even if A=1 (minimum), C=10. So this condition is false.

2) B + D = 10 + A. Putting B+D=10+A in A, we get C = 10A - 1. Now, the only way C < 10, is when A=1.

Thus, we get A=1, C=9.

Any other ways to approach this??
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Re: Number properties - PS [#permalink] New post 09 Jun 2005, 07:57
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smcgrath12 wrote:
AB + CD = AAA, where AB and CD are 2 digit numbers and AAA is a 3 digit number. A,B,C and D are distinct positive numbers. In the above addition problem, what is the value of C?


The sum of two 2 digit numbers cannot be more than 200. Hence A had to be 1.

Since AB < 20 ( a=1) , CD > 90 , hence C = 9.

HMTG.
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 [#permalink] New post 08 Jun 2005, 11:57
Yep, thats right. Can you please explain how you got that? These kinda questions always stumps me.
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 [#permalink] New post 08 Jun 2005, 12:04
smcgrath12 wrote:
Yep, thats right. Can you please explain how you got that? These kinda questions always stumps me.


AB +
CD =
AAA

We know that A must be 1 since A + C (+1 maybe)= AA
(+1 maybe) means that an extra unit can be carried over from summing B and D.
The largest number AA can be cannot exceed 19.

Now since A=1 1+C = 11, C cannot be 10, so we need one unit to be carried over from B+D so C equals 9.

p.s. there are multiple combinations for B and D as long as B+D = 11
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 [#permalink] New post 08 Jun 2005, 16:37
9 for me...

I just picked AAA=111

then worked my way back, knowing A was 1!
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Re: Number properties - PS [#permalink] New post 09 Jun 2005, 02:28
smcgrath12 wrote:
AB + CD = AAA, where AB and CD are 2 digit numbers and AAA is a 3 digit number. A,B,C and D are distinct positive numbers. In the above addition problem, what is the value of C?


from AB+CD=AAA:
B+D=10+A
A+C+1=10A+A -> C+1=10A
A has to be 1 as if the sum of two 2-digit numbers equals a 3-digit number, the max that the 3-digit number can be is 198...
if A=1, C=9
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Re: AB + CD = AAA, where AB and CD are 2 digit numbers and AAA [#permalink] New post 24 Feb 2014, 08:49
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Re: AB + CD = AAA, where AB and CD are 2 digit numbers and AAA [#permalink] New post 24 Feb 2014, 10:48
Expert's post
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

Since AB and CD are two-digit integers, their sum can give us only one three digit integer of a kind of AAA: 111.

So, A=1 and we have 1B+CD=111

Now, C can not be less than 9, because no two-digit integer with first digit 1 (1B<20) can be added to two-digit integer less than 90, so that to have the sum 111 (if CD<90, so if C<9, CD+1B<111).

Hence C=9.

Answer: D.

OPEN DISCUSSION OF THIS QUESTION IS HERE: ab-cd-aaa-where-ab-and-cd-are-two-digit-numbers-and-aa-136903.html
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Re: AB + CD = AAA, where AB and CD are 2 digit numbers and AAA   [#permalink] 24 Feb 2014, 10:48
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