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Difficult Math Question #5

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Difficult Math Question #5

by 800guy » Thu Aug 31, 2006 6:17 pm
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
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by aim-wsc » Thu Aug 31, 2006 7:46 pm
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
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Yep, i think it is D too.

by bensanderson » Fri Sep 01, 2006 6:45 am
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?
good luck!
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by 800guy » Tue Sep 05, 2006 7:43 pm
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)
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by anhe123 » Wed Aug 25, 2010 8:48 am
Last edited by anhe123 on Fri Jul 29, 2016 1:07 am, edited 1 time in total.
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by kmittal82 » Wed Aug 25, 2010 9:33 am
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
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by havok » Thu Apr 21, 2011 10:10 pm
This is a difficult question - would anything like this ever appear on the GMAT?
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by GMATGuruNY » Fri Apr 22, 2011 3:20 am
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.
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by Ian Stewart » Fri Apr 22, 2011 10:30 am
havok wrote:This is a difficult question - would anything like this ever appear on the GMAT?
Yes, I've seen similar problems. It's quite fast to solve if you notice where to start:

AB
+CD
AAA

We are adding two numbers less than 100; their sum must be less than 200, so A=1. Thus we have

1B
+CD
111

Now 1B is less than 20 so CD must be greater than 90 to get a sum of 111, and C=9.
For online GMAT math tutoring, or to buy my higher-level Quant books and problem sets, contact me at ianstewartgmat at gmail.com

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