### Learn One Weird Trick And Easily Solve The Product-Sum Problem

A tribute to Martin Gardner.

For which sets of positive integers does the sum equal the product? For example, when does $$x_1 + x_2 = x_1\cdot x_2$$? It's easy to see that this is only true when $$x_1 = x_2 = 2$$.

In the general case our equality is $$\sum_{i=1}^{n} x_i= \prod_{i=1}^{n} x_i$$. We can rearrange terms to give $x_1+x_2+\sum_{i=3}^{n} x_i= x_1\cdot x_2\cdot \prod_{i=3}^{n} x_i,$ and this in turn factors nicely to give us $\left( x_1\cdot \prod_{x=3}^{n} x_i - 1\right)\cdot \left( x_2\cdot \prod_{x=3}^{n} x_i - 1\right) = \left( \prod_{x=3}^{n} x_i \right)\cdot \left(\sum_{x=3}^{n} x_i \right) + 1.$ How does this help? Consider the case $$n=5$$, and without loss of generality assume $$x_i \ge x_{i+1}$$ for all $$i$$. If $$x_3=x_4=x_5=1$$ our factorized equation becomes $(x_1-1)\cdot(x_2-1)=4,$ with the obvious solutions $$x_1=5, x_2=2; x_1=3, x_2=3$$. The only remaining case to consider is $$x_3=2$$, as any other case forces  $$\sum_{i=1}^{n} x_i < \prod_{i=1}^{n} x_i$$. For this case our equality becomes $\left( 2 x_1 - 1\right)\cdot \left( 2 x_2 - 1\right) = 9.$ This gives us the remaining solution $$x_1 = x_2 = x_3 = 2$$ with the other $$x_i = 1$$.

This is quite efficient. For example, for the case $$n=100$$ the only equations we need to consider are \begin{aligned} (x_1-1)\cdot(x_2-1) &= 99\\
(2 x_1-1)\cdot(2 x_2-1) &= 199\\
(3 x_1-1)\cdot(3 x_2-1) &= 301\\
(4 x_1-1)\cdot(4 x_2-1) &= 405\\
(4 x_1-1)\cdot(4 x_2-1) &= 401\\
(6 x_1-1)\cdot(6 x_2-1) &= 607\\
(9 x_1-1)\cdot(9 x_2-1) &= 919\\
(8 x_1-1)\cdot(8 x_2-1) &= 809\\
(12 x_1-1)\cdot(12 x_2-1) &= 1225\\
(16 x_1-1)\cdot(16 x_2-1) &= 1633.
\end{aligned} Now 199, 401, 607, 919, 809 are all prime ruling them out immediately, and 301 and 1633 don't have factors of the right form. The other equations yield the five solutions $$(100,2), (34,4), (12,10), (7,4,4), (3,3,2,2,2)$$ with the other $$x_i = 1$$.

For $$n = 1000$$ you'd need to consider 52 cases, but most of these are eliminated immediately. I get the six solutions $$(1000,2), (334,4), (112,10), (38,28), (67,4,4), (16,4,4,4)$$.

Have fun!

### 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 …