If ABC and DEF are both 3-digit numbers such that A, B, C, D, E and F are distinct non-zero digits such that ABC + DEF = 1111, then what is the value of  A + B + C + D + E + F ?

(UPSC CSAT – 2023)

1. 28
2. 29
3. 30
4. 31

Solution: Ans.(4)

ABC + DEF = 1111

(A×100 + B×10 + C) + (D×100 + E×10 + F) = 1000 + 100 + 11

(A + D)×100 + (B + E)×10 + (C + F) = 10×100 + 10×10 + 11

So (A + D) = 10, (B + E) = 10, (C + F) = 11

A + B + C + D + E + F = 10+10+11 = 31