1. **State the problem:** Miya bought the same number of burgers and rolls. Burgers come in boxes of 8 costing 3 each, rolls come in packets of 6 costing 1.5 each. We want to find the smallest total amount spent. 2. **Define variables:** Let $n$ be the number of burgers (and rolls) Miya bought. 3. **Constraints:** Burgers are sold in boxes of 8, so $n$ must be a multiple of 8. Rolls are sold in packets of 6, so $n$ must be a multiple of 6. 4. **Find the smallest $n$:** Since $n$ must be a multiple of both 8 and 6, $n$ must be a multiple of the least common multiple (LCM) of 8 and 6. Calculate LCM: $$\text{LCM}(8,6) = \frac{8 \times 6}{\text{GCD}(8,6)}$$ The greatest common divisor (GCD) of 8 and 6 is 2. So, $$\text{LCM}(8,6) = \frac{48}{2} = 24$$ 5. **Calculate the number of boxes and packets:** Number of burger boxes = $\frac{24}{8} = 3$ Number of roll packets = $\frac{24}{6} = 4$ 6. **Calculate total cost:** Cost of burgers = $3 \times 3 = 9$ Cost of rolls = $1.5 \times 4 = 6$ Total cost = $9 + 6 = 15$ **Final answer:** The smallest amount Miya could have spent is 15.