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 ?

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Accepted Answer

The correct answer is option A. Given the equation: ABC × D = 37DD.
We can express 37DD as 3700 + 11D. So, ABC = (3700 + 11D) / D = 3700/D + 11.
Since ABC is a 3-digit number (100 ≤ ABC ≤ 999) and A, B, C, D are different non-zero digits (1-9), we can test values for D:
- If D=1, ABC = 3700/1 + 11 = 3711 (Not a 3-digit number).
- If D=2, ABC = 3700/2 + 11 = 1850 + 11 = 1861 (Not a 3-digit number).
- If D=3, ABC = 3700/3 + 11 = 1233.33 + 11 (Not an integer).
- If D=4, ABC = 3700/4 + 11 = 925 + 11 = 936. Here, A=9, B=3, C=6. All digits (9, 3, 6, 4) are different and non-zero. This is a valid solution.
  A+B+C = 9+3+6 = 18.
- If D=5, ABC = 3700/5 + 11 = 740 + 11 = 751. Here, A=7, B=5, C=1. But B=D=5, which violates the condition that digits must be different.
- Other values of D (6, 7, 8, 9) will not yield an integer for 3700/D or will result in ABC being less than 100 or having repeated digits with D.
Thus, the only valid solution is A=9, B=3, C=6, and D=4. The sum A+B+C = 18.

Q26. 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 ?

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