clciotola wrote:

AB

+BA

---------

CDC

In the correctly worked computation above A, B, C and D represent distinct nonzero digits. What is the value of A+B+C+D

A. 14

B. 16

C. 18

D. 20

E. It cannot be determined from given information

From tens' digit D which does not equal C we know that A + B > 9 (i.e., we have to carry).

We know that AB and BA have inverted digits.

We know that CDC has to be 1_1 because

--In order to make D different from C, we have to carry.

--But you can't add two single digits in the ones' column and get anything greater than 17 (8+9), so the most you can carry is 10.

--If the most you can carry is 10, then in the tens' place, the most you can get is 180 (8_ + 9_ + 1_ = 18_) --

-- which means that C, in the hundreds' place, HAS to be 1.

We also know, where O = Odd and E = Even, that the equation must be

OE

EOOEO --->

C = 1 is odd. Only E + O = O

Similarly, for tens' column, we know we have to carry 1 (10). O + 1 = E. Then E + E = E

So C = 1, AB and BA must be EO and OE (or vice-versa), and A and B must sum to 11 .

First I tried

92

29121 - no, D must be different

BUT

83

38121 - bingo - and

74

47121 - and

65

56121

No matter which set you choose A + B + C + D = 14

Answer ASomething in this problem seems to be connected to 9: all the numbers that work are separated by 9 (38, 47, 56...). And all the nonzero digits show up as addend digits or sum digits - except 9.

_________________

In the depths of winter, I finally learned

that within me there lay an invincible summer.

-- Albert Camus, "Return to Tipasa"