Problem Solving

OA coming when a few people have attempted...

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

OK
A=1
allright>?

reason: no two digit no.s add to get more than 198 so
the sum is 111 ie AAA

then>
B cannot be 0
B>2
since CD cannot be 99
at extreme if B=9
CD = 92

therefor range of CD is from 92 to 98

My pick C=9

answer is D

AB
+ CD since B+D=AA and AA has to be 11 (can't be 22).
-------
AAA


then A+C+A= 2+C= AA (11) so C=9.

aim-wsc is right! i think. what is the OA?

OA:

Ans: 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

Therefore the correct answer is D (C = 9)

andre.heggli wrote:

aim-wsc wrote:OK
A=1
allright>?

reason: no two digit no.s add to get more than 198 so
the sum is 111 ie AAA

then>
B cannot be 0
B>2
since CD cannot be 99
at extreme if B=9
CD = 92

therefor range of CD is from 92 to 98

My pick C=9

answer is D

I don't fully understand how you exclude certain numbers, like why B>2 or A=1. Would anyone care to explain this in further detail or maybe reference to a source where I could read more about this kind of logical processing?

Thanks
André

The biggest 2 digit number you can form by adding two 2 digit numbers is 99+99 = 198. We know that AAA is certainly less than 198, since the digits in the 2 digit numbers are all different.

The only 3 digit number less than 198 where all 3 digits are the same is 111, Hence A = 1
Now since ABCD are distinct, you can try B =0, in which case AB = 10.
Now, CD could be anything, it would never give you 111 when added to AB, hence B cannot be 0 (Same logic goes for B=2, the only value you add to 12 to make 111 is 99, but C and D have to be distinct, hence B=2 is also wrong) and since it cant be 1 either, it has to be >2

This is a difficult question - would anything like this ever appear on the GMAT?

800guy wrote:OA coming when a few people have attempted...

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 A, B, C and D represent digits:
AB = 10A+B.
CD = 10C+D.
AAA = 100A+10A+A = 111A.

Substituting in the equation AB + CD = AAA, we get:
(10A+B) + (10C+D) = 111A.
10C + B + D = 101A.

The equation above shows that C=9.
If C=8, then we are left with the impossible equation B+D = 21. The sum of two digits cannot be 21.
Thus, C cannot be less than 9.

The correct answer is D.