A 3-digit number ABC, on multiplication with D gives 37DD where A, B, C and D are different non-zero digits. What is the value of A + B + C ?

• The equation is $ABC \times D = 37DD$, which can be written as $ABC \times D = 3700 + 11D$.

• Since $ABC$ is a 3-digit number, $100 \le ABC \le 999$. Thus, $100D \le 3700 + 11D \le 999D$.

• From $100D \le 3700 + 11D$, we get $89D \le 3700$, so $D \le 41.5$. Since $D$ is a single digit, $D \le 9$.

• From $3700 + 11D \le 999D$, we get $3700 \le 988D$, so $D \ge 3.74$. Thus, $D$ can be 4, 5, 6, 7, 8, or 9.

• We test these values for $D$ to see if $ABC = (3700 + 11D) / D$ is an integer with distinct non-zero digits $A, B, C$ different from $D$.

• If $D=4$, $ABC = (3700 + 44)/4 = 3744/4 = 936$. Digits are 9, 3, 6, 4. All are distinct and non-zero. $A=9, B=3, C=6$. Sum $A+B+C = 9+3+6 = 18$.

• If $D=5$, $ABC = (3700 + 55)/5 = 3755/5 = 751$. Digits are 7, 5, 1, 5. Here $B=D=5$, so digits are not all different. Invalid.

• If $D=6$, $ABC = (3700 + 66)/6 = 3766/6 = 627.66...$ Not an integer. Invalid.

• If $D=7$, $ABC = (3700 + 77)/7 = 3777/7 = 539.57...$ Not an integer. Invalid.

• If $D=8$, $ABC = (3700 + 88)/8 = 3788/8 = 473.5$. Not an integer. Invalid.

• If $D=9$, $ABC = (3700 + 99)/9 = 3799/9 = 422.11...$ Not an integer. Invalid.

• Only $D=4$ yields a valid solution where $ABC=936$ and all digits $A, B, C, D$ are distinct non-zero digits.

• Therefore, $A=9, B=3, C=6$, and $A+B+C = 18$.