### An Enormous Number of Kilograms

For years the kilogram has been defined with respect to a platinum and iridium cylinder, but this is now no longer the case. Here's a puzzle about kilograms that's easy to state and understand, but the answer is very, very surprising.

I've always had a fascination with really large numbers. First 100 when I was really little, and as I got older and more sophisticated numbers like a googol and the smallest number that satisfies the conditions of the Archimedes cattle problem.

When I was an undergraduate I interviewed for a summer internship with an insurance company as an actuarial student. They gave me the following puzzle - what's the smallest number that when you move the last digit to the front it multiplies by 2? I calculated for a little while, then said "This can't be right, my answer has 18 digits!". It turns out that the smallest solution does, indeed, have 18 digits.

We can see this by letting our $$(n+1)$$-digit number $$x = 10 m + a$$, where $$m$$ is an $$n$$-digit number and $$0\leq a < 10$$. Moving $$a$$ to the front we get $$y = 10^n a + m$$, and our requirement is $$y = 2x$$. This gives: \begin{align} 20 m + 2 &= 10^n a + m \\ 19 m &= a(10^n-2) \\ m &= \frac{2a(5\cdot 10^{n-1} - 1)}{19} \end{align} The smallest $$m$$, if one exists, requires $$a,n$$ such that 19 divides $$5\cdot 10^{n-1}-1$$ (as 19 can't divide $$2 a$$) and the result has $$n$$-digits. It's easy to check that the smallest value of $$n$$ that satisfies the first condition is $$n=17$$. To get the smallest solution we try $$a=1$$, but this yields a value with only 16 digits. Setting $$a=2$$, however, yields the 17-digit $$m = 10526315789473684$$. The smallest solution to our puzzle is therefore the 18-digit number $$105263157894736842$$; that's surprisingly large.

Numbers with this type of property are known as parasitic numbers.

Later, I wondered if there were numbers with the slightly different, but equally interesting property, that moving the last digit to the front converted ("autoconverts") it from a value under one unit of measurement to an equivalent value under a different unit of measurement.

The first one I tried was moving the last digit to the front converts from Celsius to Fahrenheit. This is a fun puzzle that eventually made its way into the New York Times. The smallest such value is 275 C, which exactly equals 527 F. What's the next smallest temperature?

How about moving the first digit to the end? We'll need to use the little-known fact that, legally, a pound is exactly equal to 0.45359237 kilograms. Given this, does there exist a number such that moving the first digit to the end converts from pounds to kilograms? The answer is yes, but the smallest solution has 108,437,840 digits! The solution is similar to the above, but as it's computationally more involved I've written Sage code to solve it, which you can find in my GitHub puzzles repository.

The smallest number that autoconverts from gallons to liters, incidentally, is even bigger at 382,614,539 digits!

### A Bayes' Solution to Monty Hall

For any problem involving conditional probabilities one of your greatest allies is Bayes' Theorem. Bayes' Theorem says that for two events A and B, the probability of A given B is related to the probability of B given A in a specific way.

Standard notation:

probability of A given B is written $$\Pr(A \mid B)$$
probability of B is written $$\Pr(B)$$

Bayes' Theorem:

Using the notation above, Bayes' Theorem can be written: $\Pr(A \mid B) = \frac{\Pr(B \mid A)\times \Pr(A)}{\Pr(B)}$Let's apply Bayes' Theorem to the Monty Hall problem. If you recall, we're told that behind three doors there are two goats and one car, all randomly placed. We initially choose a door, and then Monty, who knows what's behind the doors, always shows us a goat behind one of the remaining doors. He can always do this as there are two goats; if we chose the car initially, Monty picks one of the two doors with a goat behind it at random.

Assume we pick Door 1 and then Monty sho…

### What's the Value of a Win?

In a previous entry I demonstrated one simple way to estimate an exponent for the Pythagorean win expectation. Another nice consequence of a Pythagorean win expectation formula is that it also makes it simple to estimate the run value of a win in baseball, the point value of a win in basketball, the goal value of a win in hockey etc.

Let our Pythagorean win expectation formula be $w=\frac{P^e}{P^e+1},$ where $$w$$ is the win fraction expectation, $$P$$ is runs/allowed (or similar) and $$e$$ is the Pythagorean exponent. How do we get an estimate for the run value of a win? The expected number of games won in a season with $$g$$ games is $W = g\cdot w = g\cdot \frac{P^e}{P^e+1},$ so for one estimate we only need to compute the value of the partial derivative $$\frac{\partial W}{\partial P}$$ at $$P=1$$. Note that $W = g\left( 1-\frac{1}{P^e+1}\right),$ and so $\frac{\partial W}{\partial P} = g\frac{eP^{e-1}}{(P^e+1)^2}$ and it follows $\frac{\partial W}{\partial P}(P=1) = … ### Solving a Math Puzzle using Physics The following math problem, which appeared on a Scottish maths paper, has been making the internet rounds. The first two parts require students to interpret the meaning of the components of the formula $$T(x) = 5 \sqrt{36+x^2} + 4(20-x)$$, and the final "challenge" component involves finding the minimum of $$T(x)$$ over $$0 \leq x \leq 20$$. Usually this would require a differentiation, but if you know Snell's law you can write down the solution almost immediately. People normally think of Snell's law in the context of light and optics, but it's really a statement about least time across media permitting different velocities. One way to phrase Snell's law is that least travel time is achieved when \[ \frac{\sin{\theta_1}}{\sin{\theta_2}} = \frac{v_1}{v_2},$ where $$\theta_1, \theta_2$$ are the angles to the normal and $$v_1, v_2$$ are the travel velocities in the two media.

In our puzzle the crocodile has an implied travel velocity of 1/5 in the water …