68. Bob planted some lettuce seedlings in his garden. A few days later, after one-third of these seedlings had been eaten by a rabbit, Bob planted 15 new lettuce seedlings. A week later, again one-third of the seedlings had been eaten, leaving 22 seedlings unharmed. How many lettuce seedlings had Bob planted originally?

Let's break down the problem step by step! After the rabbit had its first snack, one-third of the seedlings were eaten, leaving Bob with two-thirds. Then he planted 15 more seedlings. When another one-third got eaten the following week, he ended up with 22 seedlings. Let's denote the original number of seedlings as \( x \). After the first incident with the rabbit, the number of seedlings left is \( \frac{2}{3}x \). After planting the 15 new seedlings, the total becomes \( \frac{2}{3}x + 15 \). Then, after another one-third were eaten from this new total, we have: \[ \text{Remaining seedlings} = \left(\frac{2}{3}x + 15\right) - \frac{1}{3}\left(\frac{2}{3}x + 15\right) = \frac{2}{3}x + 15 - \frac{1}{3}\left(\frac{2}{3}x\right) - 5 = 22. \] Solving \( \frac{2}{3}x + 15 - \left(\frac{2}{9}x + 5\right) = 22 \): \[ \frac{2}{3}x - \frac{2}{9}x + 10 = 22. \] Combining like terms, we can convert \( \frac{2}{3} \) to \( \frac{6}{9} \): \[ \left(\frac{6}{9}x - \frac{2}{9}x\right) + 10 = 22, \] which simplifies to: \[ \frac{4}{9}x + 10 = 22. \] Subtracting 10 from both sides gives: \[ \frac{4}{9}x = 12, \] and multiplying both sides by \( \frac{9}{4} \) gives: \[ x = 27. \] So, Bob originally planted **27 lettuce seedlings**!