1. **State the problem:** We need to find the greatest number $n$ of six-volt batteries connected to a circuit with resistance $R=500$ ohms such that the current $I$ does not exceed 0.25 amperes. 2. **Recall the formula:** Ohm's law states that current $I$ is given by $$I = \frac{V}{R}$$ where $V$ is the potential difference and $R$ is the resistance. 3. **Express the potential difference:** The total potential difference from $n$ batteries each of 6 volts is $$V = 6n$$ 4. **Substitute into Ohm's law:** $$I = \frac{6n}{500}$$ 5. **Set the current limit:** We want $$I \leq 0.25$$ Substitute $I$: $$\frac{6n}{500} \leq 0.25$$ 6. **Solve for $n$:** Multiply both sides by 500: $$6n \leq 0.25 \times 500$$ $$6n \leq 125$$ Divide both sides by 6: $$n \leq \frac{125}{6} \approx 20.8333$$ 7. **Interpret the result:** Since $n$ must be an integer number of batteries, the greatest integer $n$ satisfying the inequality is $$n = 20$$ **Final answer:** The greatest number of six-volt batteries that can be used without exceeding 0.25 amperes is **20**.