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find the number

by daretodream » Fri Feb 19, 2010 3:28 am
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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by harsh.champ » Fri Feb 19, 2010 3:54 am
daretodream wrote: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
We can write it as 10A + B + 10C + D = 100A + 10A + 1A
=>B + 10C + D = 101A

I think we have 4 variables and 1 equation.
So, C cannot be determined.
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by ajith » Fri Feb 19, 2010 4:07 am
daretodream wrote: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 =1 (Because the 100s digit of of sum of two 2 digit numbers cannot be anything else than 1)

B+D = 11 ( Since the sum is 111)

C = 9 ( A is 1 , C has to be 9)
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by thephoenix » Fri Feb 19, 2010 4:10 am
daretodream wrote: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 + CD = AAA
Since AB and CD are two digit numbers, then AAA must be 111
Therefore 1B + CD = 111
B can assume any value between 3 and 9
If B = 3, then CD = 111-13 = 98 and C = 9
If B = 9, then CD = 111-19 = 92 and C = 9
So for all B between 3 & 9, C = 9

so D
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by harsh.champ » Fri Feb 19, 2010 4:14 am
ajith wrote:
daretodream wrote: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 =1 (Because the 100s digit of of sum of two 2 digit numbers cannot be anything else than 1)

B+D = 11 ( Since the sum is 111)

C = 9 ( A is 1 , C has to be 9)
Hey ajith,
Thanks for the solution.
I had made a mistake in the soln.
The most imp. point was that
A =1 (Because the 100s digit of of sum of two 2 digit numbers cannot be anything else than 1)

I have found that I have a weakness in such puzzle type questions.
Is there any guide for it??
It takes time and effort to explain, so if my comment helped you please press Thanks button :)



Just because something is hard doesn't mean you shouldn't try,it means you should just try harder.

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