1. **State the problem:** Mark's car gets 30 miles per gallon and Jodi's car gets 21 miles per gallon. Both used a whole number of gallons to travel from their home to the theater. We need to find the shortest distance that satisfies this condition. 2. **Understand the problem:** If the distance is $d$ miles, then for Mark, the gallons used is $\frac{d}{30}$, and for Jodi, it is $\frac{d}{21}$. Both must be whole numbers. 3. **Set up the conditions:** $$\frac{d}{30} = m, \quad m \in \mathbb{Z}^+$$ $$\frac{d}{21} = j, \quad j \in \mathbb{Z}^+$$ 4. **Find the shortest distance $d$:** Since $d$ must be divisible by both 30 and 21, $d$ must be a common multiple of 30 and 21. 5. **Calculate the least common multiple (LCM):** Prime factorization: $$30 = 2 \times 3 \times 5$$ $$21 = 3 \times 7$$ LCM takes the highest powers of all primes: $$\text{LCM}(30,21) = 2 \times 3 \times 5 \times 7 = 210$$ 6. **Conclusion:** The shortest distance is $\boxed{210}$ miles. 7. **Check options:** - a. 420 (multiple of 210) - b. 300 (not multiple of 21) - c. 630 (multiple of 210) - d. 210 (LCM and shortest) **Answer: d. 210 miles**