Step 1: Overview

The snail moves forward $4$ ft each hour, then a gust of wind pushes it back $2$ ft. Its net gain per full hour‑cycle is therefore $$\\text{net gain}=4-2=2\\text{ ft}$$. The snail eats the food the instant it reaches it, so the final wind‑push does not occur once the food is reached.

Step 2: Setup

Start distance is $30$ ft from the food.

Step 3: Distance after each full cycle

After each complete hour (forward + wind) the remaining distance shrinks by $2$ ft. After $n$ full cycles the distance left is $$d_n = 30 - 2n\\text{ ft}$$.

Step 4: When can the snail finish?

At the beginning of an hour the snail moves forward $4$ ft. If the distance at that moment is ≤ $4$ ft, the snail reaches the food before the wind blows. We need the smallest integer $n$ such that $$d_n \\le 4$$.

Step 5: Solve for $n$

$$30 - 2n \\le 4 \\;\\Longrightarrow\\; 2n \\ge 26 \\;\\Longrightarrow\\; n \\ge 13.$$ After 13 full cycles the snail is exactly $4$ ft away.

Step 6: Final hour

At the start of the 14th hour the snail is $4$ ft from the food, moves forward $4$ ft, and eats it instantly. No wind push follows.

Step 7: Total time

Total hours = $13$ (full cycles) + $1$ (final forward move) = $14$ hours.

Step 8: Quick sanity check

If we ignored the “instant eat” rule, the net‑gain view would give $30/2 = 15$ hours. Because the snail reaches the food before the last wind, we subtract one hour, confirming the answer of 14 hours.

261. Snail Against Wind

Similar Questions