1. **State the problem:** Find six numbers whose mean is 3 and range is 6. 2. **Recall the formulas:** - Mean of $n$ numbers $x_1, x_2, \ldots, x_n$ is given by $$\text{Mean} = \frac{x_1 + x_2 + \cdots + x_n}{n}$$ - Range is the difference between the maximum and minimum numbers: $$\text{Range} = \max(x_i) - \min(x_i)$$ 3. **Apply the mean condition:** Given mean = 3 and $n=6$, so $$\frac{x_1 + x_2 + x_3 + x_4 + x_5 + x_6}{6} = 3$$ Multiply both sides by 6: $$x_1 + x_2 + x_3 + x_4 + x_5 + x_6 = 18$$ 4. **Apply the range condition:** Range = 6 means $$\max(x_i) - \min(x_i) = 6$$ 5. **Choose numbers to satisfy both conditions:** Let the minimum number be $a$, then the maximum number is $a + 6$. 6. **Find the other four numbers:** We want the sum of all six numbers to be 18: $$a + (a+6) + x_3 + x_4 + x_5 + x_6 = 18$$ Simplify: $$2a + 6 + x_3 + x_4 + x_5 + x_6 = 18$$ $$x_3 + x_4 + x_5 + x_6 = 18 - 2a - 6 = 12 - 2a$$ 7. **Example choice:** Choose $a=1$ (minimum number), then maximum is $1+6=7$. Sum of other four numbers: $$12 - 2(1) = 12 - 2 = 10$$ 8. **Distribute 10 among four numbers:** For simplicity, choose all four numbers equal: $$x_3 = x_4 = x_5 = x_6 = \frac{10}{4} = 2.5$$ 9. **Final six numbers:** $$1, 2.5, 2.5, 2.5, 2.5, 7$$ 10. **Verify:** - Mean: $$\frac{1 + 2.5 + 2.5 + 2.5 + 2.5 + 7}{6} = \frac{18}{6} = 3$$ - Range: $$7 - 1 = 6$$ **Answer:** The six numbers can be $1, 2.5, 2.5, 2.5, 2.5, 7$.