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Unknown Digits

Expert replies
by damilolaamele » Sun Mar 03, 2013 10:58 pm
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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Source: — Problem Solving |

by Anurag@Gurome » Mon Mar 04, 2013 12:02 am
damilolaamele 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?
Maximum possible value of the sum of 2 two-digit number is (99 + 99) = 198
In this case the sum is of the form AAA, i.e. a three-digit number whose all the three digits are same. Only possible sum is 111. This means A = 1

Now we can approach for finding C in two ways,
Method #1
  • As A = 1, AB < 20
    Hence, CD > (111 - 20) ---> CD > 91
    Hence, C = 9
Method #1
  • (1B + CD) = 111
    As B and D are distinct positive integers less than 9 and their sum has unit's digit 1, we must have a carry of 1 if we add them.

    Hence, (1 + 1 + C) = 11 --> C = 9
The correct answer is D.
Last edited by Anurag@Gurome on Mon Mar 04, 2013 12:16 am, edited 2 times in total.
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by Anurag@Gurome » Mon Mar 04, 2013 12:11 am
Algebraic Solution:
AB = (10A + B)
CD = (10C + D)
AAA = (100A + 10A + A)

So, (10A + B) + (10C + D) = (100A + 10A + A)
--> 10C + B + D = 100A + A
--> C = [101A - (B + D)]/10

Form my above post, only possible value of A is 1.
Now, maximum value of (B + D) is (9 + 8) = 17
Hence, C ≥ [101 - 17]/10 = 84/10 = 8.4

As C must be an integer, only possible value of C is 9.

The correct answer is D.
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by GMATGuruNY » Mon Mar 04, 2013 4:13 am
damilolaamele 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
Since 99 + 99 < 200, AB + CD < 200.
Thus, AAA = 111.

Since AB = 1B, AB≤19.
Since 111 = 19 + 92, CD≥92.

Thus, C=9.

The correct answer is D.
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by damilolaamele » Mon Mar 04, 2013 2:39 pm
Thank you soo much. This sure helped.
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