1. **State the problem:** We need to find the smallest number of bags, each containing one pack of pens (20 pens), one pack of pencils (12 pencils), and one pack of highlighters (5 highlighters), such that there are no leftover packs. 2. **Identify the key concept:** The problem is asking for the smallest number of bags so that the total number of pens, pencils, and highlighters are all whole multiples of their pack sizes without leftovers. 3. **Use the Least Common Multiple (LCM):** Since each bag contains one pack of each item, the total number of bags must be a multiple of the pack sizes' LCM to avoid leftovers. 4. **Calculate the LCM of 20, 12, and 5:** - Prime factors: - 20 = $2^2 \times 5$ - 12 = $2^2 \times 3$ - 5 = $5$ - LCM takes the highest powers of all primes: - $2^2$ (from 20 or 12), $3$ (from 12), and $5$ (from 20 or 5) - So, $$\text{LCM} = 2^2 \times 3 \times 5 = 4 \times 3 \times 5 = 60$$ 5. **Interpretation:** The smallest number of bags is 60. 6. **Calculate packs needed:** - Pens: 1 pack per bag, so 60 packs. - Pencils: 1 pack per bag, so 60 packs. - Highlighters: 1 pack per bag, so 60 packs. **Final answer:** - Smallest number of bags = 60 - Packs needed: 60 packs of pens, 60 packs of pencils, 60 packs of highlighters