### 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 $\left( x_1\cdot \prod_{i=3}^{n} x_i - 1\right)\cdot \left( x_2\cdot \prod_{i=3}^{n} x_i - 1\right) = \left( \prod_{i=3}^{n} x_i \right)\cdot \left(\sum_{i=3}^{n} x_i \right) + 1.$ 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…

### Notes on Setting up a Titan V under Ubuntu 17.04

I recently purchased a Titan V GPU to use for machine and deep learning, and in the process of installing the latest Nvidia driver's hosed my Ubuntu 16.04 install. I was overdue for a fresh install of Linux, anyway, so I decided to upgrade some of my drives at the same time. Here are some of my notes for the process I went through to get the Titan V working perfectly with TensorFlow 1.5 under Ubuntu 17.04.

Old install:
Ubuntu 16.04
EVGA GeForce GTX Titan SuperClocked 6GB
2TB Seagate NAS HDD

New install:
Ubuntu 17.04
Titan V 12GB
/ partition on a 250GB Samsung 840 Pro SSD (had an extra around)
/home partition on a new 1TB Crucial MX500 SSD
New WD Blue 4TB HDD