An algebra problem by Dhruv Singh
A frog is on the bottom of an empty well deep. It climbs up the well during the daytime and slips down 1m at nighttime. What is the number of days it takes for the frog to be out of the well?
The answer is 54.
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1 solution
At the end of the day before the frog can get out the well, the frog has to be 108 - 3 = 105 meters up from the bottom of the well (it is important to observe, that the frog won't slip back any more once it is out of the well (even if it doesn't make a difference in this case, it would if, say, the well would be 107 m deep).
On the days preceding the last day, the frog climbs 3 - 1 = 2 meters each day.
105 ÷ 2 = 52.5 ~ 53 days
If we add the last day, we will get our answer:
5 3 + 1 = 54 days