The Frog and the Well: A Mathematical Puzzle Solved

Have you ever heard the age-old riddle about a frog trying to escape a well? The question often posed is, How many days will it take a frog to jump out of a well that is 50 feet deep when it can jump 2 feet up each day and slides back 1 foot each night? This classic puzzle has challenged and fascinated people for generations. Letrsquo;s break it down step by step and solve this intriguing problem using both common sense and mathematical analysis.

The Problem Analysis

Initially, one might think that the frog makes a net gain of 1 foot each day (2 feet up - 1 foot down). However, this simplistic approach misses a crucial detail: the frog will not slide back if it reaches the top of the well.

Day-by-Day Breakdown

Letrsquo;s analyze the situation day by day:

Beginning (day 1): The frog starts at the bottom of the well, 50 feet deep. Each day, it jumps up 2 feet but slides back 1 foot at night. This gives a net gain of 1 foot per day, except when the frog reaches the top. Day 2 through Day 27: The frog continues to make a net gain of 1 foot per day until it reaches 27 feet. At this point, the frog only needs to jump 3 feet up to escape the well. Since it can jump 2 feet during the day and will reach the top without sliding back the same night, it escapes on the 28th day. Day 28: The frog jumps 2 feet during the day, reaching the top of the well. It wonrsquo;t slide back because itrsquo;s already out.

Therefore, the total time required for the frog to escape the well is 28 days.

Mathematical Analysis

From a more mathematical perspective, we can set up the problem with an equation. Let the total height of the well be (H 50) feet, and the daily net gain be (1) foot per day. The frog needs to reach a certain height where, on the next day, it can jump 3 feet and escape the well.

If we denote the number of days by (d), the height the frog reaches at the end of each day can be represented by the equation:

[h(d) 49 - d]

The frog will escape on the day when (h(d) leq 28) because on the next day it can jump 3 feet and get out. Solving for (d):

[28 geq 49 - d implies d geq 21]

On day 27, the frog is 26 feet up. It gains 3 feet during the day, reaching 29 feet, and slides down 2 feet at night, ending the night at 27 feet. On the 28th day, the frog can jump the remaining 3 feet and escape.

Conclusion

By considering the day-by-day progression and the total height of the well, we can conclude that the frog takes 28 days to escape the well. This problem is a great example of how common sense and a step-by-step approach can lead to a clear solution, even for seemingly complex problems.

Further Exploration

For those interested in exploring similar problems, consider the following embellished versions:

What if the frog is already halfway up the well and needs to escape? How would the solution change if the well had a different depth? What if the frogrsquo;s jump had a different pattern each day?

These variations can provide a deeper understanding of mathematical problem-solving and logical reasoning.