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The Snail on the Wall

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A snail creeps 6 ft up a wall during the daytime. After all the labor it does throughout the day, it stops to rest a while... but falls asleep!! The next morning it wakes up and discovers that it has slipped down 4 ft while sleeping.

If this happens every day, how many days will the snail take to reach the top of a wall 18 ft in height?
Answer: 7

Solution:

On the first day, the snail climbs up 6 ft and slips down 4 ft while sleeping. So, next morning, it is 2 ft from where it started. The snail thus travels 2 ft upwards every day. Therefore, in 6 days, it has traveled a distance of 12 ft from the bottom.

Here lies the catch to the problem! On the last day, the snail travels 6 ft upwards and hence reaches the top of the wall in a total of 7 days.

Alternative Solution through Equation:

Let x be the number of days the snail takes to reach the top of the wall 18 ft in height.

On the last day, the snail will reach the top by traveling 6 ft upwards and there will not be any question of slipping down.
The number of remaining days excluding the last day are (x − 1). Since the snail climbs up 6 ft and slips down 4 ft while sleeping, it travels 2 ft upwards on each of these remaining days. Thus,

Distance traveled on last day + Distance traveled on remaining days = Wall height; or
6 + 2 (x − 1) = 18

On solving the above equation, we get

2 (x − 1) = 18 − 6 = 12; or
x = (12 / 2) + 1 = 7.


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