2023 CSAT — Q9

Add column by column and track carries — the digit sum falls out of the carry structure without ever pinning the individual digits.

The sum $1111$ has units $1$, tens $1$, hundreds $1$, thousands $1$.

• Units: $C + F$ ends in $1$. Distinct non-zero digits give $C + F \ge 1 + 2 = 3$, so $C + F = 11$ (carry $1$). It cannot be $1$ (too small for two non-zero digits) or $21$ (too large).
• Tens: $B + E + 1 = \dots 1$, so $B + E + 1 = 11 \Rightarrow B + E = 10$ (carry $1$).
• Hundreds: $A + D + 1 = \dots 1$, so $A + D + 1 = 11 \Rightarrow A + D = 10$ (carry $1$, which produces the leading $1$ of $1111$). ✓

Therefore

Answer: (d) 31.

You never need the actual digits — only the column sums. The leading $1$ of $1111$ is a pure carry, so the hundreds, tens, and units columns must read $A+D=10$, $B+E=10$, $C+F=11$. Carry propagation forces the units to $11$ (not $1$) because two distinct non-zero digits cannot sum to $1$. Adding the three column sums gives the answer; chasing specific assignments is the time trap.