A hard question on climbing out of a well for P3–P4, with a step-by-step solution, diagrams and the answer. Rarely set directly in competition papers, but a classic for learning to think about the last step on its own. Good for the younger years.
The question
A snail starts at the bottom of a well: each day it climbs 4 m, and each night it slides back 1 m.
During day 4, just as it finishes that day's 4 m, it reaches the top exactly.
If the well were 12 m deeper, on which day would the snail get out?
Show the hints (3)
💡Hint 1: First find the depth: how much does each of the first 3 days gain? And day 4?
💡Hint 2: The new depth is 13 + 12 m. The last day's climb is 4 m — how high must the snail be before it?
💡Hint 3: It must first reach 21 m or more; each whole day gains 3 m, and 21 ÷ 3 = 7 exactly.
Worked solution
Have a go first, then read the solution step by step.
1.Depth: the first 3 days gain 3 × 3 = 9 m, then day 4 adds 4 m: 9 + 4 = 13 m.
2.New depth: 13 + 12 = 25 m.
3.The last climb is 4 m, so before it the snail needs 25 − 4 = 21 m or more.
4.21 ÷ 3 = 7 exactly, so 7 whole days are needed (7 × 3 = 21 m).
5.So on day 8 it climbs 4 m and gets out.
⚠️Common mistake: 25 ÷ 3, answering day 9 — the last day has no slide.
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