I never made any follow-up posts about my exams, but it’s partially because the discrete math tests took a while to get graded, and I wanted to know all the results before celebrating, hehe.

My hard work thankfully *did* pay off, as I aced all four of my exams! It feels so great to have lived up to my own expectations, and I can’t help feeling a bit proud of myself ^^

Also, the company we did our semester projects for might hire me! They really liked the solution I was responsible for, so I had a meeting with them last week - they talked about a lot of different challenges and optimization problems they had yet to solve and hinted that I could get a position and solve some of them. Nothing is certain yet, but I’ll be damned if it doesn’t feel amazing to get this kind of recognition already.

The Impossible Six Hat Riddle!

6 prisoners are given one hat each, 3 hats are black, 3 hats are white. They do not know the colour of the hats they are each wearing but can see the hats of everyone else around them.

They are told that if AT LEAST one of them figures out the colour of his hat, they are all free to go. If EVEN ONE of their guesses is wrong, they will all be executed.

4 prisoners are put in one room, and 2 in a different room. The prisoners cannot take off their hats and cannot talk to each other. There is also no way that prisoners from one room can communicate with prisoners in another room.

Is it possible for all prisoners to escape with 100% certainty in all situations?

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So this is a variant of a famous riddle I made myself, it’s not a comic but I wanted to try something different. If you figure out the answer comment below, reblog this riddle if you want to show off to your friends that you solved it, or reblog it if you’re too dumb and want your friends to feel dumb too!

Answer: **yes**, it is possible for the prisoners to escape in all situations

We have two possible situations,

- 3 of the 4 prisoners in the big room have the same colour
- 2 prisoners in the big room have black hats, 2 have white

*I am assuming that all prisoners are rational and capable of making logical deductions, and that they know there are 6 hats in total, out of which 3 are black and 3 are white.*

In *Situation 1*, 3 hats of the same colour is present in the big room. The 4th prisoner in this room will be able to see those 3 hats and deduce that because all 3 of the observed colour is present, his own hat must have the opposite colour.

In *Situation 2*, every prisoner in the big room will be able to see 2 prisoners wearing hats of one colour (let’s denote them Prisoners A and B) and 1 prisoner wearing a hat of the other (let’s denote them Prisoner C). They will each observe, after a time interval expected to make the deduction in *Situation 1*, that Prisoner C relative to them *does not* announce their hat colour. Therefore, they can deduce that Prisoner C *is not* facing 3 hats of the same colour. Their own hat colour will therefore be the same as Prisoner C.

I salute you for taking math/economics. You’re awesome, man. Good Luck :D

aww thanks :D I don’t understand all the math hate here on tumblr though, math is the best, and so useful too

you could say it

**M**akes **A**mazing **T**hings **H**appen

Studying for my exam in discrete mathematics this Wednesday, and I feel just about ready B)

- Camera: Sony Ericsson LT26i
- Aperture: f/2.4
- Exposure: 1/25th
- Focal Length: 4mm

fun math game: my current follower count is the sum of two squares, and also the sum of three squares. The difference between these two squares is the exponent of a Mersene Prime, is the larger number in a twin prime pair, and the product of its digits is not prime. Guess how many followers I have.

My guess is: **181 followers**

- 181 is the sum of the two squares (100, 81)
- 181 is the sum of the three squares (1, 36, 144) or (36, 64, 81)
- The difference between 100 and 81 is 19, the exponent of the 7th Mersenne prime
- 19 is the large number in the twin prime pair (17, 19)
- The product of the digits of 19 is 9, which is not prime

Did I guess correctly? ;D

Calculus exam in 15 minutes.

I can’t help being nervous, even though I feel like I got it. Nothing like an exam to make me doubt my grasp of the subject haha.

"Individually I hope you each do well on the exam, but collectively I hope you do about average."

— Abstract algebra professor (via mathprofessorquotes)

hOW DO YOU FIND THE PERIMETER OF THIS

View on blog for LaTeX formatting

Split it up in parts you know how to find the perimeter of. The shape consists of three half circles - two of diameter 8 and one of diameter 16.

The circumference of a circle is given as \( C = \pi \cdot d \), where \(d\) is the diameter. The perimeter of the shape can therefore be calculated as

\[ 2 \cdot \left( \frac12 \cdot \pi \cdot 8 \right) + \frac12 \cdot \pi \cdot 16 = 16\pi \]

As those of you who follow me on my main blog probably know, I am very busy with stuff, so there are many things I haven’t been able to do, and keeping up with my math blog was one of those things. I want to be active here too, but I probably won’t be able to update very often. I’ll try to post every now and then, but please don’t be upset if I fail to answer a question, because my priority list is very long :/

ANYWAY, on the brighter side, I’ve installed MathJax on my blog, which means I can now have LaTeX formatted equations rendered directly on my blog instead of having to screenshot, host and embed images.

For example, the following code:

\[ e = \sum_{n=0}^{\infty} \frac{1}{n!} = 1 + \frac11 + \frac{1}{1\cdot 2} + \frac{1}{1\cdot 2\cdot 3} + \dotsb \]

will look like gibberish on the dash, but render as a beautiful equation if you go to my blog. It makes things a lot easier for me, so I’ll be using that from now on (sorry mobile users).

to do in the future: put a ton of people into a room (one at a time) with a radio with a joke and make the punchline a variable amount of time from the actual joke. Graph average laugh loudness vs time and find the optimal time for a punchline.

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