Posts Some fun puzzles
Post
Cancel

Some fun puzzles

Introduction

In this post I thought I’d write about two fun mathematical puzzles I came across recently (in Martin Gardner’s book Mathematical Puzzles). Neither requires much mathematical sophistication, but they are both good examples of how the ability to think logically and model a problem technically are important problem-solving tools.

Round trip

Problem

An airline runs a round trip from city A to city B and back. The plane travels at a single constant speed. During one trip, there is no wind blowing during either leg. During a second trip, there is a constant wind that blows from A to B during both legs of the trip. Is the first trip longer than, shorter than, or the same length as, the second trip? (Take a minute to think about it before reading the solution.)

Solution

Let $d$ be the distance from A to B. If $r \cdot t = d$ (rate x time = distance), then $t = d/r$. What we are going to do is write equations for $t$ in each of the wind and no-wind cases and see if we can determine the relationship between them. In the no-wind case, if the plane’s constant speed is given by $r$, then $t_1 = 2d / r$. In the wind case, let’s say the wind speed is given by some $w > 0$. Then the plane travels at a rate of $r + w$ on the way from A to B and at a rate of $r - w$ on the way back. Thus, the total round trip time is given by

\begin{align*} t_2 &= \frac{d}{r + w} + \frac{d}{r - w} \\\\ &= \frac{d(r - w) + d(r + w)}{r^2 - w^2} \\\\ &= \frac{2dr}{r^2 - w^2} \end{align*}

Notice that because $w > 0$, $w^2 > 0$, so that

$$ t_2 = \frac{2dr}{r^2 - w^2} > \frac{2dr}{r^2} = \frac{2d}{r} = t_1, $$

showing us that the trip with wind takes longer.

Cornerless chessboard

Problem

You have an 8x8 chessboard and 32 2x1 dominoes. As is hopefully clear, you can cover all 64 squares on the chessboard with the 32 dominoes. Now suppose I remove two opposite corners from the board and take away one domino. Can you cover the 62 remaining squares with your 31 remaining dominoes? If so, show how. If not, prove it.

Solution

You cannot cover the remaining squares. To see why, the key observation is that each domino covers one black square and one white square. If you remove opposite corners, you are removing two squares of the same color. In order to cover what’s left, you would need to cover 30 black squares and 32 white squares, but per our observation, 31 dominoes can only cover 31 black squares and 31 white squares! Thus, covering the remaining squares with 31 dominoes is indeed impossible.

Conclusion

While delving into deep higher level mathematics is certainly fun, it’s fun to pause every so often and have a little fun with some less-involved puzzles; I hope you’ve enjoyed :)

This post is licensed under CC BY 4.0 by the author.