# The snail in the well: visual puzzle and solution | SwaVid

A snail climbs 3 m each day and slips 2 m each night. How long does it take to escape a 10 m well? See the complete step-by-step solution.

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# The snail in the well

## 8 days.

### Net gain per cycle

### Get within one climb of the top

### Climb out on day 8

SwaVid visual puzzle

A snail climbs 3 m each day and slips 2 m each night. How long does it take to escape a 10 m well?

Detailed solution

The snail gains 1 m per full day, but on the last day it climbs out before it can slip.

Climb 3, slip 2.

After 7 full cycles the snail starts day 8 at 7 m.

From 7 m it climbs 3 m to reach 10 m and leaves, so there is no eighth-night slip.

10 m ÷ 1 m per day = 10 days wrongly adds a night-time slip after the snail has already escaped.

- 1 Net gain per cycle Climb 3, slip 2. 3 − 2 = 1 m per day-and-night
- 2 Get within one climb of the top After 7 full cycles the snail starts day 8 at 7 m. 7 × 1 m = 7 m
- 3 Climb out on day 8 From 7 m it climbs 3 m to reach 10 m and leaves, so there is no eighth-night slip. 7 + 3 = 10 m

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