# Social Distancing at the Bar Riddle

I came across the following puzzle yesterday (edited for relevancy):

There is a bar with 25 seats in a line. People are social-distancing so when they walk into the bar, they always try to find a seat farthest away from others. To comply with government restrictions, no two people can sit next to each other. If a person walks in and there are no available seats, they will leave. Where should the owner tell his first customer to sit in order to maximize the number of people in the bar?

In order to solve this problem, we can utilize some wishful thinking. The best-case scenario is when every person is seated exactly two away from everyone else:

1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 23, 25.

How can we produce such a sequence? Well, if we had every fourth person, then the other people can sit in-between them (while still maintaining a distance of 2):

1, 5, 9, 13, 17, 21, 25.

To obtain this configuration, we take every eighth person, so the remaining people can sit a distance of four away from everyone else:

1, 9, 17, 25.

Finally, to obtain this configuration, we either need to put the first person on seat or 17. We used backtracking in order to determine the best starting seat.

For a visual solution, I created a MatLab program that would test out each starting location of the first customer and then generates how the other customers would fill out the bar.  For the original puzzle, this creates the following graph:

The y-axis is the location of the first customer seated, while the x-axis shows the seating configuration according to the bars’ rules. The starred values indicate the optimal configuration, which occurs exactly when the first person is seated at 9. The relative size of each Polkadot is the order the customer is seated in the configuration, which is why the line $y=x$ has the largest dots.

In this case, we were able to seat $\lceil \frac{25}{2}\rceil=13$ people effectively. Can we always obtain this bound for a bar with $n$ seats?

Unfortunately, the answer is no, but this isn’t obvious to see.  Plotting the number of seats against the maximum number of people seated gives the following graph:

The stars indicate when the maximum number of people seated equals $\lceil \frac{n}{2}\rceil$. It’s interesting to note the outliers here are when $n=15, 23, 27, 28, 29, 30, 31, 39$.

Let’s take one of these specific values and see what goes wrong in our backtracking approach. For $n=15$, the best possible scenario would be:

1, 3, 5, 7, 9, 11, 13, 15.

In order to obtain this, we take every fourth person as we did before:

1, 5, 9, 13.

To obtain this, we suspect we should seat the first customer at 9. However, plotting this shows this generates a different graph:

The sequence starting at 9 is 1, 3, 5, 7, 9, 12, 15. Why did we skip over 13? This is because our backtracking approach didn’t account for the selection mechanism.

Starting with 9, the second person selected will sit at 1. After this, the next person will sit at 15, then 5 to create the current configuration: 1, 5, 9, 15. The next person entering the bar will want to sit as far away from the other people as possible. Scanning all the seats, they’ll choose seat 12 since it’s 3 away from both 9 and 15. Therefore, 15 is a suboptimal starting value.

We noticed every value before this was optimal, which leads us to wonder if we can characterize all of the values of $n$ that give an optimal configuration.

Expanding the search up to $n=100$ gives the following values that are optimal:

n=1 2 3 4 5 6 7 8 9 10 11 12 13 14 16 17 18 19 20 21 22 24 25 26 32 33 34 35 36 37 38 40 41 42 48 49 50 64 65 66 67 68 69 70 72 73 74 80 81 82 96 97 98.

I found that every number in the list above is of one of the forms

$2^a, 2^a+2^b, 2^a+2^b+1, 2^a+2^b+2,$

where $a$ and $b$ are non-negative integers. This leads me to conjecture that these forms contain all the initial values of $n$ that produce an optimal configuration, however, this may be a case of The Strong Law of Small Numbers at work.

If how closely everyone is seated together makes you a bit uneasy in the graphs, we can expand the gap between people. For a fixed number of people in the bar, $n=25$, we vary the gap size, $g$ to produce the following seating configurations:

The starred values indicate the best seat configuration. Compare these with the graph of gap size 2 earlier. Notice the best first seat varies between 1 and 9.

We can now plot the number of people seated compared to gap size:

A value is starred if the maximum number of people seated equals $\lceil \frac{n}{g} \rceil$. In this case, not only is length 25 optimal for a gap size of 2, but it’s optimal for every gap!

We wonder when this pattern will remain. Of the values of n above that were found to be optimal for a gap size of 2, the following are optimal for every gap size:

n=1 2 3 4 5 6 7 8 9 10 11 12 13 14 16 17 18 19 20 21 24 25 26 32 33 48 49.

An example of a value of n which gives suboptimal gap sizes is n=68:

The gap sizes of 3, 6, 7, and 9 are all suboptimal. This is the same for n=69 and 70. Similarly, for n=80, 81, and 82, we have suboptimal gap sizes of 3, 6, 7, and 11.

Since the gap size of 3 seems to produce suboptimal behaviour for several values, we plot the number of seats compared to the maximum number of seats for $g=3$:

The starred values indicate when the max number of people seated equals $\lceil \frac{n}{3}\rceil$. We see the initial part is a step function in threes, similar to $g=2$.

There is also a long valley around n=70 and n=80, as predicted above.

The full list of values of $n$ that are optimal for $g=3$ are given below:

n=1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 24 25 26 27 28 29 30 31 32 33 36 37 38 39 48 49 50 51 52 53 54 55 56 57 60 61 62 63 72 73 74 75 96 97 98 99 100.

A few closing thoughts to the reader:

• Can we characterize the values of $n$ above similar to what we did for $g=2$?
• Are any of these sequences on the Online Encyclopedia of Integer Sequences?
• Has a full mathematical analysis been conducted? I’ve seen the XKCD Urinal Protocol Vulnerability, but this assumes the first person sits on the end.
• Stay safe, and continue to practice social distancing everywhere you go!

# Quick Divisibility Test for Primes up to 67

While there’s known divisibility tests for certain primes (see my notes), the following test very quickly determines if a number is divisible by any prime up to 67 by hand! The original method is due to John Conway, and I learned it from Arthur Benjamin.

Here’s a sample problem along with explanation:

63779 is the product of three primes, each less than $71$. Determine their sum.

We set up the equation $63779=pqr$, where $p, q,$ and $r$ are primes.

We begin by finding the factor of 2000 closest to 63, namely 64,000. We can therefore rewrite this as

$221=2000\cdot 32-63,779.$

We can quickly determine that since $2, 5\mid 2000$ and $2, 5\nmid 221$, $2, 5\nmid 63,779$. We now add $32$, the quotient in the divisibility above to both sides of the equation to produce:

$253=2001\cdot 32-63,779.$

Notice the factor of $32$ and the left hand side have both increased, while our original number, 63779 has remained the same. Now, as any good contest math student should have the years close to the current year memorized, we know $2001=3\cdot 23\cdot 29.$ Therefore $3\mid 63779\iff 3\mid 253$, however, $3\nmid 253$. Similarly $23\mid 63779\iff 23\mid 253$. This time, $253=23\cdot 11$, therefore $23\mid 63779$. We’ve found one prime divisor: $p=23$.

We could now divide $63779$ by $23$, however, this gives the equation $2773=qr$, and we don’t have any better ideas for how to factorize $2773$. Therefore, instead, we’ll use the prime factorizations of some other numbers close to $2000$:

$1998=2\cdot 3^3\cdot 37, 2002=2\cdot 7\cdot 11\cdot 13, 2006=2\cdot 17\cdot 59, 2009=41\cdot 49$

and $2010=2\cdot 3\cdot 5\cdot 67,2013=3\cdot 11\cdot 61, 2015=5\cdot 13\cdot 31, 2021=43\cdot 47.$

For instance, to determine if our number is divisible by $37$, we would take the original equation and subtract $2\cdot 32$ to produce $157=1998\cdot 32-63,779.$ Then, we know $37\mid 63779\iff 37\mid 157$, however, clearly $37\nmid 157$.

Similarly, to test if our number is divisible by $7, 11,$ or $13$, we simply add $2\cdot 32$ to our original equation to produce $285=2002\cdot 32-63779$. Therefore, again $11\mid 285\iff 11\mid 63779$, however, clearly $11\nmid 285$, hence the same is true for $63779$.

Next, to determine if the original number is divisible by $17$ or $59$, we add $6\cdot 32$ to our original equation (or more simply, $4\cdot 32$ to the equation in the last paragraph) to get

$413=2006\cdot 32-63779.$

Since $2006=2\cdot 17\cdot 59$, we see $17\mid 413\iff 17\mid 63779$, however, clearly $17\nmid 413$. Similarly, $59\mid 413\iff 59\mid 63779$. Now since $420=60\cdot 7$, we see $413=59\cdot 7$, hence $59\mid 413$, and $59\mid 63779$. We’ve therefore found a second prime factor: $q=59$.

At this point that we have two prime factors, we could continue on with our test, but we have a pretty good idea of what the third factor is by division: $63779=23\cdot 59\cdot r$. One way to quickly compute this in our heads would be to notice

$59\cdot 23=(60-1)(20+3)=1200+180-20-3=1357.$

Since we know $r$ is an integer, we attempt to bound it. We know $r=\frac{63779}{1357}$. We notice $40\cdot 1357=52480$, while $50\cdot 1357=67850$, therefore, $40.  We then subtract $40$ from both sides of our equation:

$r-40=\frac{63779-52480}{1357}.$

While we could compute the numerator mentally, since we know $1\le r-40\le 9$, we know it must be a single numerical digit! We therefore just consider the unit digit of the numerator which is $9$. Since we’re dividing by a number which ends in $7$, we know $r-40$ must be $7$, since $7\cdot 7\equiv 49\equiv 9\pmod{10}$. Therefore $r=47$.

Hence our desired sum is $p+q+r=23+59+47=\boxed{129}$.

Alternatively, we could verify $47\mid 63779$ using the prime factorization of next year, 2021. I’ll leave this as an exercise for anyone interested.

While this method isn’t super practical, I think it’s interesting that you can essentially compute prime factors up to 67 with a quick trick in your head. The most difficult part is memorizing the prime factorizations of the numbers close to 2000, but hopefully this is familiar for any seasoned contest mathematician (who might be interested in using this).

References: Arthur Benjamin, Factoring Numbers with Conway’s 150 Method
The College Mathematics Journal, Vol. 49, No. 2, pp. 122-125, March 2018.

# Similar to Fermat’s Last Theorem

Let $n$ be a positive integer. Prove that there exist distinct positive integers $x, y, z$ such that $x^{n-1} + y^n = z^{n+1}.$ (Source: 1997 IMO Shortlist N6)

The equation $x^{n}+y^n=z^n$ is known to have no nontrivial solutions for $n\ge 3$ due to Fermat’s Last Theorem (if you’ve not read Fermat’s Enigma I highly recommend checking it out). This very similar looking equation, however, has an infinite number of solutions. In this blog post, we’ll examine how this difficult looking problem can be solved with one simple observation: $1+8=9$.

For a concrete example of this, consider the case $n=3$ giving the equation $x^{2}+y^{3}=z^{4}.$ Assume we multiply the equation by a number of the form $2^{\alpha}\cdot 3^{\beta}$.  We then will have: $2^{\alpha}\cdot 3^{\beta}+2^{\alpha+3}\cdot 3^{\beta}=2^{\alpha}\cdot 3^{\beta+2}.$ Since we want the first part to be a perfect square, we should have $2\mid \alpha$ and $2\mid \beta$. Since the second part should be a perfect cube, we should have $3\mid (\alpha+3)$ and $3\mid \beta$. Finally, since the last part should be a perfect fourth power, $4\mid \alpha$ and $4\mid (\beta+2)$. Solving these system of linear congruences gives $\alpha=0$ and $\beta=6$! We get back the equation $3^6+3^6\cdot 2^3=3^8$, from which we read off the solution $(x,y,z)=(3^3, 3^2\cdot 2, 3^2)=(27, 18, 9)$

We now attempt to generalize this technique. We begin by multiplying $1+8=9$ by $2^{\alpha}\cdot 3^{\beta}$ again to obtain $2^{\alpha}\cdot 3^{\beta}+2^{\alpha+3}\cdot 3^{\beta}=2^{\alpha}\cdot 3^{\beta+2}.$ We then want the exponents on the leftmost term to be divisible by $n$, on the middle term to be divisible by $n+1$, and on the rightmost term to be divisible by $n+2$. This is equivalent to:

$\begin{cases} \alpha&\equiv 0\pmod{n} \\ \alpha&\equiv -3\pmod{n+1} \\ \alpha&\equiv 0\pmod{n+2} \end{cases}$   and    $\begin{cases} \beta&\equiv 0\pmod{n} \\ \beta&\equiv 0\pmod{n+1} \\ \beta&\equiv -2\pmod{n+2}. \end{cases}$

Notice that in the first case, the first and third conditions are equivalent to $\alpha\equiv 0\pmod{n(n+2)}$. Since $\gcd(n, n+2)=\gcd(n+1, n+2)=1$, we can use the Chinese Remainder Theorem to know that $\alpha$ has a unique solution mod $\text{lcm}(n, n+1, n+2)$

On the other hand, for $\beta$, in the case that $n$ is even, $\gcd(n, n+2)=2$, therefore, we have to use an extended version of the Chinese Remainder Theorem. Thankfully, since $\gcd(n, n+2)=2$ in this case and $0\equiv -2\pmod{2}$, we are still guaranteed a unique solution for $\beta$ mod $\text{lcm}(n, n+1, n+2)$.

Substituting these solutions in for $\alpha$ and $\beta$ gives a solution $(x,y,z)$ to the original equation. Once we have the congruences for $\alpha$ and $\beta$, we can then find an infinite parameterized version of solutions for $(x,y,z)$.

For instance, in the case of $n=3$ above, we had $\alpha\equiv 0\pmod{12}, \beta\equiv 6\pmod{12}$. If we in general write $\alpha=12\alpha'$ and $\beta=12\beta'+6$, we then find

$2^{12\alpha'}\cdot 3^{12\beta'+6}+2^{12\alpha'+3}\cdot 3^{12\beta'+6}=2^{12\alpha'}\cdot 3^{12\beta'+8}.$

This then translates into $(x,y,z)=(2^{6\alpha'}\cdot 3^{6\beta'+3}, 2^{4\alpha'+1}\cdot 3^{2+4\beta'}, 2^{3\alpha'}\cdot 3^{3\beta'+2})$. Notice for $\alpha'=\beta'=0$, this gives the solution triple $(x,y,z)=(27, 18, 9)$ found above.

It’s pretty interesting to me how when you change one small thing in the formulation of the problem (subtract $1$ from the exponent on $x$ and add $1$ to the exponent of $z$) gives an entirely new problem, where we go from no solutions to an infinite number!

# Puzzling Integral Solutions

At the falls clubs fair for the Mathematical Sciences Society at University of Alberta, we put up a poster board with two challenging integrals, offering free t-shirts to anybody who solved them. We put out several mini-whiteboards for people to work on, and to my surprise, there was quite a bit of interest (free tshirts are always a great motivator!). Here’s the solutions to two of these challenges. I really like how the first one looks symbolically challenging, but conceptually is not, while the second one looks more elementary, but involves a clever trick courtesy of Richard Feynman.

Link to Integral Solutions

# The Search for Mersenne Primes

To be written later…