Tough and tricky 4: addition problem

Question

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

AKProdigy87 · Answer

The answer is  D: 9 .

Logical deduction.

1) Two 2-digit numbers sum to a 3 digit number in the form AAA. Since AB < 100 and CD < 100, AB + CD must be less than 200. So AAA = 111, and A = 1.

2) Now we know, since A is equal to 1, that AB < 20. The only way that AB + CD = 111, if AB < 20, is that if CD > 90. Therefore, C = 9.

hgp2k · Answer

Answer is D = 9. 
Same logic as explained by AKProdigy87.

Bunuel · Answer

Set of Tough & tricky questions is now combined in one thread: tough-tricky-set-of-problms-85211.html You can continue discussions and see the solutions there.

khosru · Answer

sorry I disagree.
as 99 +12 =111 0r 12+99=111 C can be 9 or 1, again 76+25=111 or 25+76=111, C can be 2 or 7. so my answer is E.

Bunuel · Answer

That's not correct. AB and CD are two digit integers, their sum can give us only one 3-digit integer of a kind of AAA: 111. So, A=1 --> 1B +CD=111 . C can not be less than 9, because no 2-digit integer with first digit 1 (mean that it's <20) can be added to 2-digit integer less than 90 to have the sum 111 (if CD<90 meaning C<9 CD+1B<111) --> C=9. Answer: D.

Transcendentalist · Answer

Different approach...

A + C + x (Carryover) = AA.. Since all integers are positive C cant be zero...

=> C + x (Carryover) = 10

Now the only carryover that can come from B + D is 1

Hence C = 9

SrinathVangala · Answer

A + C gives A . So c must be either 0 or 10. I the answer choices there is no option with zero. so c must be 10.for c to be 10 we must get a carry over and c MUSt be 9 as the maximum carry over we can get is 1 . so D

pavanpaone · Answer

AB<100 and CD<100.So , 20<=AB+CD<200.From 20 to 199 which is in the form of AAA is only one number i.e 111.So A=1,and AB=1B<20 because the tens place has 1. 1B+CD=111,let us consider 1.) B=2 => CD = 99 AB+CD=12+99=111, C=9 2)B=3 => AB=13 =>CD = 98 . C=9 take B=4,5,6,7,8,9 ,we get C= 9

jlgdr · Answer

Correct, also C cannot be zero because the question says they are all positive integers

Tough and tricky 4: addition problem

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