As we saw in my post on part 1 of the GCHQ Christmas puzzle, the solution to the grid puzzle gives us a QR code which takes us to the following website:
Part 2 of the puzzle states:
"Part 2 consists of six multiple choice questions. You must answer them all correctly before you can move on to Part 3." We are then given our first question: Q1. Which of these is not the odd one out?
This is a strangely worded question... which of these is not the odd one out. We are more used to being asked which of these is the odd one out. What would it mean for something not to be the odd one out? My initial reading of the question, which turned out to be the correct one, is that each word except for one will turn out to be the odd one out for a different reason. For example, TORRENT is the only not starting with an S, therefore, it is the odd one out, and not the answer to the puzzle. SAFFRON is the only on ending in an N and is also there not the answer to the puzzle. As an aside, this question reminds me of the interesting number paradox. If we attempt to divide all numbers up into numbers that are interesting (such as 2 which is the only even prime) and numbers which are not interesting. Then the numbers that are not interesting must have a smallest member. However, since this is an interesting property (the property of being the smallest uninteresting number), this number cannot in fact be an uninteresting number. And we must conclude that there are no uninteresting number. Similarly, if we work out that one of the answers above is not the odd one out, then we have found a way in which the number is the odd one out. Due to the fact that it is the only one which is not the odd one out! Q2. What comes after GREEN, RED, BROWN, RED, BLUE, , YELLOW, PINK?
This turns out to be a cool question. We have a sequence of colours and we need to work out the next value. My first thought when I read this was that it perhaps relates to monopoly (possibly because I had played it just before trying to answer). I played around with this idea for a while but didn't get anywhere. I wasn't a million miles away as in the end my girlfriend gave me a hint that it related to snooker. Once you plug in the values of the snooker balls into this sequence you get the following sequence of numbers: 3 1 4 1 5  2 6 Which you will hopefully recognise as the decimal expansion of $\pi$. The next digit is 5 and therefore the answer is Blue. Q3. Which is the odd one out?
Q4. I was looking at a man on top of a hill using flag semaphore to send a message, but to me it looked like a very odd message. It began "ZGJJQ EZRXM" before seemingly ending with a hashtag. Which hashtag?
This should be quite a straightforward question to answer  my guess which turned out to be correct is that there is a fairly straight forward transformation that can be applied to how the man is using flag semaphore so that it makes sense. The first transformation I tried was to rotate the message by various degrees. I got bored of answering this question after a while so I decided to skip it. . Q5. What comes after 74, 105, 110, 103, 108, 101, 98, 101, 108, 108?
Here we have a sequence of numbers and we need to find the next number. Since the number all cluster around the 70110 area, my intuition was that they are some sort of encoding. The first thing I checked was the ascii values of these numbers. They come out as J i n g l e b e l l. This should be the answer then! The next value is s, and the next number is therefore 115. Ascii is a way for computers to represent letters and other characters using numbers. The numbers 0 to 255 (giving a total of 256 characters which is a power of 2) are each assigned a character here is a link to an ascii table: Q6. What comes next: D, D, P, V, C, C, D, ?
This was quite a fun question. I've got no idea how to help people solve it as I just stared at it for a few minutes then spotted the answer. These are the names of Santa's reindeers! I guess one possible hint was the use of jinglebells in the question above, and the fact that this is a Christmas puzzle. So we have the answer to 3 of the questions and we have narrowed down some of the options on the other questions. I noticed that the way the website stores your responses is to take the 6 answers and convert it to a web address. So by answering 1 on each question we will be taken to the following url. (Note the 6 As at the end which correspond to answering 1 in each question)
What I did next was set up a Spreadsheet with links to this url corresponding to the possible combinations of answers which I have not eliminated yet. For example, we know that the second digit is E and the second to last digit is C. The url of the next stage is the following:
LIBOR Bond Pricing3/1/2016
I was working through Hull's Options, Futures and other Derivatives when Hull states that the price of a Bond paying semi annual coupons in line with six month LIBOR, and discounted using a six month LIBOR discount rate is par. To me this statement seems like it's probably true but I wasn't 100%. I couldn't find the proof online, but it turns out that it is true, here is my derivation.
Proof by induction:
Base case: $n=1$ value(n year bond) $= $PAR $ = \sum\limits_{i=1}^{2n1}\frac{(PAR)R/2}{(1+R/2)^i} + \sum\limits_{i=2n}^{2(n+1)1}\frac{(PAR)R/2}{(1+R/2)^i}$ $+ \frac{PAR(1+R/2)}{(1+R/2)^{(2(n+1))}} + \left(\frac{(PAR)(1+R/2)}{(1+R/2)^{2n}}\frac{(PAR)(1+R/2)}{(1+R/2)^{2n}}\right)$ Rearranging:$ = \left(\sum\limits_{i=1}^{2n1}\frac{(PAR)R/2}{(1+R/2)^i} + \frac{(PAR)(1+R/2)}{(1+R/2)^{2n}}\right) + \sum\limits_{i=2n}^{2(n+1)1}\frac{(PAR)R/2}{(1+R/2)^i}+ \frac{PAR(1+R/2)}{(1+R/2)^{(2(n+1))}} \frac{(PAR)(1+R/2)}{(1+R/2)^{2n}}$ Then using the fact that a n year bond is priced at PAR:$ = PAR + \sum\limits_{i=2n}^{2(n+1)1}\frac{(PAR)R/2}{(1+R/2)^i}+ \frac{PAR(1+R/2)}{(1+R/2)^{(2(n+1))}} \frac{(PAR)(1+R/2)}{(1+R/2)^{2n}}$ No we just take the PAR outside and cancel out to 1.$ = PAR\left(1 + \sum\limits_{i=2n}^{2(n+1)1}\frac{R/2}{(1+R/2)^i}+ \frac{(1+R/2)}{(1+R/2)^{(2(n+1))}} \frac{(1+R/2)}{(1+R/2)^{2n}}\right)$ $ = PAR\left(1 + \frac{R/2}{(1+R/2)^{(2n)}} + \frac{R/2}{(1+R/2)^(2n+1)}+ \frac{1}{(1+R/2)^{(2n+1)}} \frac{1}{(1+R/2)^{2n1}}\right)$ $ = PAR\left(1 + \frac{R/2}{(1+R/2)^{(2n)}} + \left(\frac{R/2}{(1+R/2)^(2n+1)}+ \frac{1}{(1+R/2)^{(2n+1)}}\right) \frac{1}{(1+R/2)^{2n1}}\right)$ $ = PAR\left(1 + \frac{R/2}{(1+R/2)^{(2n)}} + \frac{1+ R/2}{(1+R/2)^(2n+1)}+ \frac{1}{(1+R/2)^{2n1}}\right)$ $ = PAR\left(1 + \frac{R/2}{(1+R/2)^{(2n)}} + \frac{1}{(1+R/2)^(2n)}+ \frac{1}{(1+R/2)^{2n1}}\right)$ $ = PAR\left(1 + \left(\frac{R/2}{(1+R/2)^{(2n)}} + \frac{1}{(1+R/2)^(2n)} \right)+ \frac{1}{(1+R/2)^{2n1}}\right)$ $ = PAR\left(1 + \frac{1 + R/2}{(1+R/2)^{(2n)}} \frac{1}{(1+R/2)^{2n1}}\right)$ $ = PAR\left(1 + \frac{1}{(1+R/2)^{(2n 1)}} \frac{1}{(1+R/2)^{2n1}}\right)$ $ = PAR$
When you order at 5 guys burgers (which you should if you haven't because they are amazing) you are given a ticket between 1 and 100 which is then use to collect your order. Last time I was there my friend suggested trying to collect all 100 tickets. The question is:
Assuming you receive a random ticket number between 1 and 100 inclusive on every visit. On average, how many visits would it take to collect all 100 tickets? Solution:
Define $S$ to be a random variable which counts the number of purchases required to get all 100 tickets. Then we are interested in $\mathbb{E} [S]$
To help us analyse $S$ we need to define another collection of random variables. Let $X_i$ denote the number of purchases required to get the $ith + 1$ ticket given we have already collected $i$ tickets. So if we have collected 2 tickets (say ticket number 33 and ticket number 51) and the next two tickets we get are 33 and 81, then $X_2 = 2$ because it took us 2 attempts to get a new unique ticket. We can actually view $S$ as the sum of these $X_i$, i.e. $S = \sum\limits_{i=1}^{100}X_i$ (to see that this is the case the number of visit it takes to collect all 100 tickets will be the number of visits to collect the first ticket plus the number of visits to collect the 2nd unique ticket, and so on) There are two further observations that we need to make before deriving the solution. Firstly, each $X_i$ is independent of $\sum\limits_{k=1}^{{i1}}X_i$. Secondly, each $X_i$ is actually distributed as a Geometric Distribution where:
$P( X_i = k ) = \frac{i1}{100}^{k1} \frac{101i}{100}$
Giving us: $\mathbb{E} [ S ] = \mathbb{E} [ \sum\limits_{k=1}^{100} X_i ] = \sum\limits_{k=1}^{100} \mathbb{E} [X_i ] = \sum\limits_{k=1}^{100} \frac{100}{101i} = 100 \sum\limits_{k=1}^{100} \frac{1}{101i} = 100 \sum\limits_{k=1}^{100} \frac{1}{i}$
This year GCHQ released a Christmas Puzzle, in this post I'll show you how to solve the series of puzzles, and also, the thought process that I used to get to the solutions.
If you would like to attempt to solve the puzzle yourself then here is the link to the first part of the puzzle. The solution to the first part will then lead to the second part of the puzzle. In total there are five parts. From the GCHQ website: "In this type of gridshading puzzle, each square is either black or white. Some of the black squares have already been filled in for you. Each row or column is labelled with a string of numbers. The numbers indicate the length of all consecutive runs of black squares, and are displayed in the order that the runs appear in that line. For example, a label "2 1 6" indicates sets of two, one and six black squares, each of which will have at least one white square separating them."
I decided to solve this puzzle in excel. The main reason for this was that I wanted to be able to fill in squares automatically rather than shading them by hand. I've posted a link to an excel file below which matches the start of the puzzle. In my version if you press Ctrl+a then the selected cell will turn black, and if you press ctrl+d then the selected cell will turn blue. I'll explain the use of blue cells later.
The first step to solving this puzzle is to count the number of rows and columns in the grid. In this case we have a 25 by 25 grid. In excel I then calculated the minimum number of cells required for each row and column. For example, the third column is listed as 1,3,1,3,1,3,1,3,1. Since there must be a blank space in between each block of black cells, this combination must take up a minimum of 25 cells but since there are only 25 cells in the column we now know which cells in this column need to be coloured black. We can now go ahead and fill in this column. We also know that the spaces in between the blocks of black cells need to be blank, to signify this we can colour these cells blue. Repeating this process allows us to complete 3 columns and 2 rows.
The next step is to look at rows and columns which have high minimum values but which are less than 25 but still high. For example, looking at the first row, it is listed as 7,3,1,1,7 with a minimum of 23. Since the first two blocks on either side are 7 long, cells 17:23 must be coloured in as a minimum, but as a maximum cells 19:25 must be filled in. Since there is an overlap between these two we can colour in the cells in the overlap. Here is an image demonstrating this. There are three copies of the first row, the first row contains the minimum cells and the second row contains the maximum cells. I have then coloured the overlap on the blocks of seven in black.
Filling in the first cell in a row is also useful as it lets us immediately use the information about that column. For example we know that the third cell in the first row is black, looking at the listing for this column we see that the first block of blacks is a 1, therefore the cell below it in the second column must be blank and can be filled in blue. Using this these techniques we eventually complete the grid. Giving the following image.
You may recognise this kind of pattern as a QR code. A QR Code is a 2 dimensional bar code. I downloaded a QR reader to my phone and scanned the image which takes you to the following url:

AuthorI work as an actuary and underwriter at a global reinsurer in London. Categories
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