1. **State the problem:** We need to find a 3-digit number with digits $a$, $b$, and $c$ such that: - The first digit $a$ is twice the third digit $c$. - The second digit $b$ is one less than the third digit $c$. - The sum of all digits $a + b + c = 11$. 2. **Write the equations:** $$a = 2c$$ $$b = c - 1$$ $$a + b + c = 11$$ 3. **Substitute $a$ and $b$ in the sum equation:** $$2c + (c - 1) + c = 11$$ 4. **Simplify the equation:** $$2c + c - 1 + c = 11$$ $$4c - 1 = 11$$ 5. **Add 1 to both sides:** $$4c - 1 + 1 = 11 + 1$$ $$4c = 12$$ 6. **Divide both sides by 4:** $$\cancel{4}c = \cancel{4}3$$ $$c = 3$$ 7. **Find $a$ and $b$ using $c=3$:** $$a = 2 \times 3 = 6$$ $$b = 3 - 1 = 2$$ 8. **Check the sum:** $$6 + 2 + 3 = 11$$ (correct) 9. **Final answer:** The 3-digit number is **623**.