1. Good post? |

## Digits

AB + CD = AAA, where AB and CD are two-digit numbers and AAA is a three digit number; A, B, C, and D is 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

2. Good post? |
Here is my take:
A can't be other than 1. (Because if A is 2 then AAA would be in 200s, which is not possible, because 2 two digit numbers of which one is in 20s can't never add upto 200s, similarly for any number higher than 2, it is not possible).

When A = 1 total of 2 numbers = 111

1B + CD = 111

D can't be 1 (as no 2 digits are same) so B + D has to be 11. For the total to reach to 3 digits number C has to be 9.

3. Good post? |
We can represent AB+CD=AAA as;

10A+B + 10C+D = 100A+10A+A
i.e., B+10C+D=101A

(A can only be 1 for AB+CD=AAA to be true)

i.e., 10C = 101-B-D

(and we know that A, B, C & D are all distinct integers > 0)

i.e., B & D must add up to 11 (since they can't add upto 1 or 21 or more!)
i.e., 10C = 90

=> C = 9 (Answer is D)

4. Good post? |
krovvidy... dont you think Skbala's approach is quicker...

i mean think about it AB<100 CD<100
AB+CD < 200

that implies AAA < 200 the only possible value of AAA is 111
When A = 1

10*1 +B + CD = 111

B+CD = 101

for CD + B to be 101 C has to be 9

D can't be 1 (as no 2 digits are same) so B + D has to be 11. For the total to reach to 3 digits number C has to be 9.

5. Good post? |
@Dynamo you're right ... if you look at it closely both the approaches are same! ... I explained the solution in more steps!!

Anyways, thanks skbala ... you've posted a good solution in your first post ... keep it going (rather coming!) ...

6. Good post? |
AB+CD=AAA
AB<100 and CD<100 so,AB+CD<200
so,AAA=111 is the only acceptable solution
so,A=1
at this point we have, 1B+CD=111
1B is less than or equal to 20 that means CD must be greater than or equal to 90 to reach AB+CD=111.
thats why c=9 is the ans.

7. Good post? |
Thanks Krovvidy!

8. Good post? |
When I attempted this problem I came up with:
79
+32
=111

So how come C cannot be 3?

9. Good post? |
Originally Posted by mickgreen58
When I attempted this problem I came up with:
79
+32
=111

So how come C cannot be 3?
because answer has to be in form AAA. Hence in your case AB = 79, i.e A = 7. and answer is 111, where A = 1. The two values of A should match

10. Good post? |
Originally Posted by Lhomme
because answer has to be in form AAA. Hence in your case AB = 79, i.e A = 7. and answer is 111, where A = 1. The two values of A should match
Duh!!

Thanks for pointing that out

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