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ATQ,
Let the three-digit number be 100a+10b+c. Given: c = b+2 and (100a+10b+c)−311 = 13. Simplify:100a+10b+c = 324. Using c=b+2:100a+11b+2 = 324 ⟹ 100a+11b=322. Try a=3:100×3+11b = 322 ⟹ b=2. c=b+2=4. The original number is 324. Sum of digits: 3+2+4=9.
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