**Firstly, Ben – tell us about yourself and what you do for a living.**

I’m a software consultant from the UK. I graduated university with a physics degree three years ago. In my spare time I work on my mobile games and spend time with friends and family.**Is it true you created Number Drop in your spare time? What inspired you to do this?**

That’s right! It took me about 6 months to create the first version.

I was looking for a new concept for a game. My Dad was addicted to WordBrain at the time and he suggested I should make a game like that.

I realised there was a gap in the market for a mathematical puzzle game that will challenge adults. There were so many word games but not many maths equivalents. And I created the concept from that thought.**How does the game work?**

A NumberDrop puzzle consists of a grid of numbers and some target values. The player must perform arithmetic operations on the grid of numbers in order to to match the target values, with no numbers left over.

This is really challenging for the player’s mental maths skills as they must perform multiple countdown style calculations with one set of numbers. There is also a logical component to the game as the numbers have gravity and fall in to spaces left by the performed operations. The order of the numbers that fall to the bottom of the grid must also match the order of the target values; that can be difficult.

The puzzles in the main game mode start easy and steadily increase in difficulty. We also have “Puzzles of the Week” and puzzles can be randomly generated.**Do you code in the numbers you’ll get on each level, or are the puzzles generated using an algorithm? How are the numbers chosen?**

The puzzles and numbers were originally generated using an algorithm. My Dad and I chose the more interesting and difficult puzzles to go in to the level packs.

You can randomly generate your own puzzles in “Random” mode and share any of the puzzles you discover with friends so they can play them too. If a user finds an interesting puzzle we would love to see it!

I have been working on an algorithm to automatically find more difficult puzzles. I have used the algorithm to help create some of the “Weekly” puzzles. The algorithm looks at the number of different operations used in the solution, as well as the nature of the target numbers, to decide whether a puzzle is “difficult”.**Did you test the game before release? Who did you get to help you test it?**

My friends and family very kindly helped me test the game. Testing is so important; there are always a couple of problems that need fixing before release. This really increases the quality of my games.**Who do you imagine will play Number Drop?**

The game is more suited to adults than children as the puzzles get quite challenging. But anyone who wants to improve their arithmetic and logical thinking should play Number Drop. I don’t want to limit the game to a certain type of person. Anyone can play it but those with a better grip of their times tables will likely move faster through the game.**We’ve had a go at playing the game – it starts off easy and gets much harder. What’s your personal best?**

I regularly delete my progress in the game for testing purposes so I don’t have a high score. A user who wants to get to the top of the leaderboards should try to complete the puzzles in one attempt, you get more XP for doing so.

**Is this the first time you’ve made a game app?**

NumberDrop is my second mobile game. My first mobile game RandomNation was a small project to practice my coding before starting work. It’s a game where you run a country and try to stay in power.**Do you have any other projects planned for the future?**

I’m currently concentrating my efforts on getting NumberDrop exposure. I believe the game can really help and entertain people and it is rated highly amongst the small user base. So no plans as yet.**Where can we download Number Drop?**

NumberDrop is available for on the Google Play Store and Apple App Store.

]]>I’m pretty sure that 158 isn’t prime. I reckon it’s $2\times 79$, and 79 is probably prime so that makes 158 a semiprime.

158 can be expressed as the sum of four or fewer squares. It is the sum of two primes. If you repeatedly halve it or multiply by three and add one depending on whether it is even or odd respectively you eventually reach 1.

158 is also the approximate length in minutes of the Cross Country train journey from Manchester Piccadilly to Cheltenham Spa that is currently providing me with no free wifi with which to determine interesting properties of integers.

Here are some interesting maths bits from the past month, which I had the foresight to open in tabs before I left the flat.

We celebrate the lives of three differently extraordinary mathematicians. Writing for the Smithsonian (yes that one), Evelyn Lamb recounts the life of Maria “Witch Of” Agnesi, and the curve that bears her name (and a mistranslated epithet). Edward Dunne at the AMS wishes happy 119th Birthday to Otto Neugebauer, who founded Mathematical Reviews and stood up to Nazism while doing it. And Futility Closet celebrates the very special ‘life’ of John Rainwater.

Of course, what you really come to the Carnival for is some abstract algebra. To this end, Rob J Low provides a crash course in equivalence classes and quotient spaces; Benjamin Leis shows us how to construct a model of the group $Q_8$ with just some card and three colours of dental floss (I have some in white you can borrow if you have two other colours…); and Mark Dominus recounts a (slightly one-sided) conversation with his daughter about mathsy card game Spot It!, known round these parts as Dobble, and explains how to construct some simpler versions.

If that whets your appetite for maths-based games/parenting, why not try the Function Machine Game? It’s a big hit with Brent Yorgey’s son, and eventually you might understand free theorems too. Meanwhile, if you want to let cardboard-based artificial intelligences play your games for you, Matthew Scroggs has worked out how much space you’ll have to clear out to have room for matchbox-based computers to play various games.

Two posts that are one way or another about convex geometry: Richard Elwes has had a fun afternoon constructing various convex polyhedra out of Magform — other magnet-based geometry contruction sets are available and apparently equally willing to post freebies to random internet mathematicians.

In somewhat more mindbending maths, David Orden explains some new results relating to Carathéodory’s theorem, to do with the properties of a point chosen inside the convex hull of a set of other points, including one excellently known as ‘Turkey depth’.

When he’s not tweeting maths facts from a terrifying number of Twitter accounts, John Cook occasionally dabbles in the longer form of maths exposition, and has an investigation into the robustness of the t-test — it works fine with normal distributions, but how does it cope with less pleasantly symmetric options?

Some mild maths frivolity to round things off. Richard Lipton ponders the implications of actor and crank-slash-master-interview-troller Terrence Howard’s assertion that one times one is in fact two. And Marquis de Geek has written about a Twitter bot created to incessantly churn out chunks of the decimal expansion of Pi.

*That’s it for this edition! Next month the Carnival will be hosted by Kartik at Comfortably Numbered. If you see any other good maths blog posts, YouTube videos or other new online content in the next month, send it in using the form linked to on the Carnival of Maths page.*

My first two posts are:

- a writeup of Robert Langlands’ work that won him this year’s Abel prize
- a post expanding on the recent chromatic number result by Aubrey de Grey

Keep an eye on the Spektrum blog, and the Aperiodical Twitter feed for news of further posts!

]]>For a while now I’ve been fascinated by the story of Claude Shannon, the pioneer of information theory and the originator of many fundamental concepts now used in all modern manipulation and transmission of data. Being sent a copy of this biography to review was a great chance to find out more about his work and life.

**A Mind At Play: How Claude Shannon Invented the Information Age
**

The authors, who describe themselves as biographers and writers foremost, have taught themselves the mathematics they need to explain Shannon’s work, and weave in some excellent and succinct explanations of the concepts amongst a fascinating human story. From his early years as an enthusiastic maker and tinkerer, through his various university courses and his placement at Bell Labs, to his later years at MIT and retirement, Shannon’s life is chronicled in detail, with a spread of well-chosen photographs to accompany the story.

Claude Shannon is described as the father of information theory – his seminal 1948 paper outlined concepts including the fundamental nature of binary numbers (coining the word ‘bit’, a binary digit), information density, communication channels, and the theoretical Shannon Limit of how quickly digital information can theoretically be transmitted in a noisy channel. These ideas predated even simple computing machines, and Shannon’s work was perfectly timed to provide a foundation for those creating early computers.

The story gives a real sense of how Shannon was well placed to create the mathematics he did – with a sharp intellect that was torn between his love of abstract mathematical theory and his fondness for hands-on inventing and engineering, he had just the right mindset to see what communication theory would become and how it could be made rigorous in a mathematical framework.

It’s also fascinating to learn about Shannon’s other passions in life – nothing he did before or since comes close to the major impact his work on information theory had, but it was far from his only passion. Other areas of mathematics and engineering, as well as pastimes including juggling, stock market predictions, and building robots all fell to his mighty intellect and he brought huge joy to the people around him with his stories and ideas.

The book is well written and lovingly put together (and has a frankly beautiful cover in the hardback edition). It was enjoyable to read, and full of interesting facts and stories. I didn’t realise until reading this book that a wooden box I have at home, which has a switch on top that when flipped, engages a robotic arm that pops out and flips the switch back again, is a modern incarnation of an invention of Shannon’s – he called it ‘the ultimate machine’, one which switches itself off. Knowing this was his original creation, and the joy I find in it, gives me a real sense of connection to this brilliant mathematician whose work changed the world for all of us.

A Mind At Play: How Claude Shannon Invented the Information Age by Jimmy Soni and Rob Goodman is published by Simon and Schuster.

]]>Here’s the final set of photos and video clips from the last week, and for the data fiends among you, a sneaky look at my spreadsheet of runs. With a graph, as requested by Hannah Fry.

Day 23 (aka @sportrelief day): stupid bloody GPS watch failed to log my distance, but I fixed it using the screen grab from my route planning map (which I won't share because it gives away where I live). A good pace! #pikmdotrun pic.twitter.com/6YWz03EeEl

— Katie Steckles (@stecks) March 23, 2018

Next up was this stroke of genius:

Day 25: had the genius idea of getting dropped of 3.14km away from the pub, where we're having a Sunday roast #winning #pikmdotrun pic.twitter.com/kndBgPY2gB

— Katie Steckles (@stecks) March 25, 2018

Augh! Just realised my tweet from yesterday didn't send! Day 27: ran out in the countryside near my nice hotel, missed the turning, had to just run 1.57km away and turn round. #pikmdotrun pic.twitter.com/62pzulsbUZ

— Katie Steckles (@stecks) March 28, 2018

Day 30: Final gym run (will be outdoors tomorrow). Accompanied by @aPaulTaylor & Waterbot. https://t.co/dtKU5Rj0qH pic.twitter.com/rm8AGaJqDQ

— Katie Steckles (@stecks) March 30, 2018

This final run video shows clips and photos from a bunch of the days, plus my triumphant final approach on Day 31:

There’s still time to chuck a final £3.14 on the pile at pikm.run, if you haven’t already. Thanks again to everyone! Now I’m going for a sit down.

]]>If I can make it to £1000 before the end of the month, I’ll be pretty pleased! Donate at pikm.run, or see below for my daily sweaty photos/videos/instagram posts.

Day 9: yep, I'm still doing this #pikmdotrun pic.twitter.com/MpFQvEImgr

— Katie Steckles (@stecks) March 9, 2018

On the 9th, I thought I’d make use of the mathematical properties of π to do a slightly silly one, and made a video:

Day 11: gym again. If you missed it yesterday, here's my video from Day 10: https://t.co/rdiPCJ534N pic.twitter.com/knyzH4L6f9

— Katie Steckles (@stecks) March 11, 2018

Day 13: still going. Thanks to everyone who's supported so far! https://t.co/dtKU5Rj0qH pic.twitter.com/p5qUUNncSD

— Katie Steckles (@stecks) March 13, 2018

I also managed to get in a few more joint runs with running companions:

Day 15: now officially kinda halfway! Buddy gym run again with @elsie_m_ #pikmdotrun pic.twitter.com/VhcVIsghll

— Katie Steckles (@stecks) March 15, 2018

Day 17: logistically complex. Managed to run πkm fully inside the B'ham NEC, as I've been working at @BigBangFair today. GPS watch gave up after 2.04km (no signal). #pikmdotrun pic.twitter.com/eebrhKjUgh

— Katie Steckles (@stecks) March 17, 2018

Day 19: staying in the gym due to the cold weather, on the world's shiniest treadmill #pikmdotrun pic.twitter.com/WO9xaPKaSV

— Katie Steckles (@stecks) March 19, 2018

I was also given an amazing gift by maths/knitting fan Linda Pollard, who came to see me perform at a show. She’s written up the knit of these magnificent π/sum gloves on a Ravelry page. I took the opportunity to test out their warmness on my next outdoor run:

This mild cry for help resulted in plenty of nice replies on Twitter, which has been a real boost – including Aperiodichum Colin Beveridge, who pointed out that my total is around $\pi^4$, a pleasing coincidence.

Day 21: Gym again (but went to a different gym for variety). Found it hard today. Encouragement please. #pikmdotrun pic.twitter.com/lIvuC7yBUZ

— Katie Steckles (@stecks) March 21, 2018

And of course, today’s effort:

Running continues. Watch this space for a final wrap-up and fundraising total at the end of the month.

Katie’s fundraising page at Sport Relief

More information about Sport Relief

This month I'm doing a completely irrational sponsored run for Sport Relief, aiming to raise £100π by running πkm per day, every day in March. I'm one week in, and here's the story so far.

Given the first few days of my challenge coincided with one of the most ridiculous periods of cold weather we've seen in a while, I wasn't quite willing to brave the outdoors yet, so my first few days were done on a treadmill in the gym.

On the fourth day, the weather broke, so I took advantage of the fact that my parents live near a lake whose perimeter works out to almost exactly πkm. I dragged my dad round it – and even made a short video:

On day 5, I dragged another of my family members for an outdoor run:

And finally, today I managed my first solo outdoor run, literally running an errand to fetch some workshop materials from the Museum of Science and Industry across town. By careful route choice, I managed to hit my target within sight of the front doors:

The best news of all is that on day four, I managed to hit my fundraising target! I reached £314.15 on 4th March, and promptly decided to extend the challenge by upping the target. My new goal is** £3141.59**, which I'm almost certainly not going to reach, but it gives me something to shoot for. I've got the rest of the month!

You can make a contribution (all to the brilliant cause that is BBC Sport Relief), and continue to follow along with my social media updates/proof, by heading to my fundraising page at pikm.run.

]]>

SIR – As one obsessed with prime numbers, I note that we have gone from 2017 (a prime) into 2018 (two times a prime), which will be followed by 2019 (three times a prime). I believe this sequence has only happened three other times in the past 1,129 years.

Keith Burgess-Clements

Maidstone, Kent

It’s lovely that newspapers will print this kind of letter, and a quick check verifies that Mr Burgess-Clements is indeed correct that these three numbers have the properties described:

- $2017$ prime, $2018 = 2 \times 1009$, $2019 = 3 \times 673$

His follow-up statement, that this sequence has only happened three other times in the last 1,129 years, takes a little more checking. But only a little – as we have a resident CL-P, who describes the necessary calculation as ‘a trivial amount of Python code’, and quickly came up with the following list:

- $13$ prime, $14 = 2 \times 7$, $15 = 3 \times 5$
- $37$ prime, $38 = 2 \times 19$, $39 = 3 \times 13$
- $157$ prime, $158 = 2 \times 79$, $159 = 3 \times 53$
- $541$ prime, $542 = 2 \times 271$, $543 = 3 \times 181$
- $877$ prime, $878 = 2 \times 439$, $879 = 3 \times 293$
- $1201$ prime, $1202 = 2 \times 601$, $1203 = 3 \times 401$
- $1381$ prime, $1382 = 2 \times 691$, $1383 = 3 \times 461$
- $1621$ prime, $1622 = 2 \times 811$, $1623 = 3 \times 541$

*Here’s that Python code, in case you’re curious. It uses Sage’s Primes() function.*

pr = set([p for p in range(2018) if p in Primes()])

double = set(2*p-1 for p in pr)

triple = set(3*p-2 for p in pr)

years = sorted(pr & double & triple)

This is a full list of all the cases below 2017, and hints at some nice more interesting patterns – $13$ and $541$ both occur as the prime year and the prime factor that's a third of another year. But we don't have time to dig into that now – we have to check if a person in the newspaper was wrong!

Keith's claim that this has happened thrice in the last 1,129 years is indeed correct – $2018 – 1129 = 889$, and three sets of years have occurred since then. I suspect this may have been a typo though, as if he'd said "the last 1,139 years", that would have included the tail end of the set starting in $877$. Maybe he was looking for the most impressive length of time with the fewest occurrences, to illustrate how rare it is (in which case 1139 would be your best bet, and probably what he meant). We favour "only four times since 1000AD" which still sounds pretty good.

One final question to answer is, how many of these will there be going forward? The next few will be:

- $2557$ prime, $2558 = 2 \times 1279$, $2559 = 3 \times 853$
- $2857$ prime, $2858 = 2 \times 1429$, $2859 = 3 \times 953$
- $3061$ prime, $3062 = 2 \times 1531$, $3063 = 3 \times 1021$

It's also worth checking when this pattern will continue for four years, so that the fourth year is four times a prime; that's $12721$, which is prime, while $12722 = 2 \times 6361$, $12723 = 3 \times 4241$ and $12724 = 4 \times 3181$.

What's your favourite number fact about 2018 so far? Answers in the comments.

]]>I’ll aim to run **π kilometres** (or as close as I can get, with the measuring instruments I have access to) each day during the month of March. This will either be on the treadmill at my gym – in which case I’ll try to get a photo of the ‘total distance’ readout once I’ve finished – or out in the real world, for which I’ll use some kind of running GPS logging device, to provide proof I’ve done it each day. Some days I’ll run on my own, and others I’ll be accompanied by friends/relatives, who’ll be either running as well or just making supportive noises. At the end of the month, I’ll post an update documenting my progress/success/failure.

**Serious request**: if you know of anywhere in the UK I can reasonably get to where there’s an established circle that’s exactly 1km in diameter, I can try to come and run round the circumference of it. Drop me an email if so.

If you’d like to support my ridiculous plan, you can follow my progress and donate on my fundraising page, or encourage others to do so by visiting pikm.run (I paid £4 for the URL, so now I have to do it). Sport Relief is the even-numbered-years-counterpart of Comic Relief, which together raise money for thousands of projects all over the UK and in the developing world, to help the vulnerable and those in need.

]]>I guess I’m alone in this, since Matt Parker’s ‘guess the mean of the digits in all the entries’ competition received the most entries of any competition meaning he also won a prize.

Bored with ‘guess the mean of all the entries’ style competitions, I decided to come up with something that flips this concept on its head.

My competition was to come up with a function for mapping a set of $N$ integers onto a single value, and whoever’s function takes a pre-determined set of integers I’m thinking of closest to a particular predetermined value I’m thinking of is the winner. I was slightly sloppy in the way I defined the question (I used $N$ to refer to both the set of numbers, and the size of the set) but everyone responded beautifully – with only some people feeling the need to point out my error – in what became a virtual festival of making up your own notation.

I received 32 entries to the competition, not all of which were actually valid entries – several had the bright idea of using ‘the number you’re thinking of’ as a variable in their function, which is 100% cheating. Here are some of my favourites, in increasing order of amazingness, culminating with the winning entry (scroll down to the bottom if you just want to see that).

The set of numbers I was thinking of was in fact The Numbers, from the TV series Lost – in which a seemingly random set of integers play a mysterious yet crucial role in the plot. The numbers are 4, 8, 15, 16, 23 and 42. As well as being one character’s winning lottery numbers, and popping up in various other places, for a whole chunk of season 2 the numbers must be typed into a computer every 108 minutes, or else… something bad will happen.

I actually quite enjoyed it but I have notoriously bad taste in TV and films.

If you haven’t seen Lost, and you have a spare 122 hours of your life you’re happy to not get back, go for it.

My set of numbers is therefore as follows: 4, 8, 15, 16, 23 and 42 (in that order). The target number for the function was chosen to be zero, for reasons I can’t explain (mostly panic).

On to the entries!

Several examples fell into this category – including:

$f(x_0, x_1, \ldots, x_N) = N$

Simply counting the number of numbers (minus one). While we appreciate your lovely and obfuscating zero-indexing, this gives the answer $5$.

$f(x,y) = x^2 – y$

Since this function is only defined on two inputs, we took the first two, giving $ 4^2 – 8 = 8$.

$f(x_1, x_2, \ldots, x_N) = x_N^2 – 1$

Taking the last of the numbers and squaring it, then subtracting one. This gives $42^2 – 1 = 1763$.

Various

statistical functions, including thesquare of the mean, theupper quartile, thelower quartileand the ‘anti-mode‘ (the least common value, and if there’s more than one value equally uncommon, the mean of these – since the set is all distinct, this is just the mean LOL!).

These gave the values $240.25, 19.5, 11.5$ and $18$.

“

The first prime not a factor of any of the numbers”

This is $11$.

“

Return a random element of the set”

Since the target value isn’t in the set, this won’t work but it’s a nice try!

\[f: x_1 \ldots x_N \rightarrow \frac{e^{\min(x_i)} \cdot \pi^{\max(x_i)}}{N}\]

Almost like Euler’s identity in how it effortlessly combines mathematical constants, but sadly returns a result in the region of $7 \times 10^{21}$.

Many entries fell into this category:

$f(x_1, x_1, \ldots, x_N) = \frac{\displaystyle\sum_{i=1}^N (-1)^i i x_N}{N}$

If this was meant to be an alternating sum of multiples of successive entries, which is how I initially interpreted it, it also doesn’t sum to zero. And you should be more careful with notation, because this is exactly what was written.

An alternating sum of multiples of the largest entry, which given $i$ runs from $1$ to $6$ means you end up with a total of three times that entry, divided by 6 – giving $x_N/2 = 42/2 = 21$.

$f(n_1, n_1, \ldots, n_N) = \frac{\displaystyle\prod_{i=1}^N n_i}{N}$

The product of all the entries divided by the number of entries, which is $1,236,480$ (very much not zero).

The $\big(\displaystyle\prod_{i=1}^N x_i\big)^{th}$ prime number, modulo $\displaystyle\sum_{i=1}^N x_i$

This one returns 67.

$\frac{1}{M} = \frac{1}{x_1} +\frac{1}{x_2} + \cdots +\frac{1}{x_N}$

For reference, $M$ was the name I gave to my target value, and this was described as the ‘parallel resistance sum’. It works out to give $M = 1.7499207463\ldots$.

$f(x) =\displaystyle\sum_{k=1}^n \sin (kx_k)$

I like it, but it gives $-1.168\ldots$ – close, but no cigar.

Several entries took a chance on me, and hoped they could guess what number I was thinking of, defining their function as the constant function giving a single value regardless of the inputs. Entries of this form, in increasing order of size, included $\pi, 4, 7, 20$ and $37$, sadly none of which win.

Ask a stupid question, and you’ll get the following answers. We tried to calculate some of these in the ten minutes we had to judge the competition, but for some of them we merely established they were definitely not zero and gave up.

$f(x_1, x_2, \ldots, x_N) = x_1 + \frac{1}{x_2+ \frac{1}{\cdots + \frac{1}{x_{N-1} + \frac{1}{x_N}}}}$

A continued fraction. We calculated the answer but all that’s written on here is “4.something”. Not equal to zero.

“Arrange the numbers in order and multiply the difference between them in pairs.”

I enjoy a good bit of tedious busy work, and I got $3724$.

$f(n_1, n_2, \ldots, n_x) = (x+2) \cdot \sqrt{n_1 \cdot n_x \cdot\displaystyle\prod_{\textrm{all } x} n_x}$

Interesting notation switcheroo from what everyone else seems to be using. This was a pain to calculate too and my bit of paper says ‘TOO BIG’.

“The median of all the answers to this competition + 1”

Given that we had 32 different entries, some of which we couldn’t properly calculate, but there was literally no chance the median answer was $-1$, the only thing written on the paper is the word ‘NO’.

“Sum the number of letters of the positive integers, minus the number of letters of the negative integers”

It’s lovely! Our bit of paper says “about 40ish?” but if anyone wants to work it out, go for it.

“For each element of N, go to the Wikipedia page for N(i) and find the N(i)th letter on the page; typecast this letter into I32 and take the sum of these values”

A bit of Googling has allowed me to establish that this is a way of converting letters into numbers, but since I assume none of the standard letters convert to a negative number, this will definitely not equal zero.

$f(x_1, x_2, \ldots, x_N) = e^{e^{\displaystyle\sum_{i=1}^N x_i}}$

We had a go at calculating this in Google Sheets, which was our terrible calculating tool of choice under pressure, and the value was actually too big for it to handle. In some ways, this could be the least correct answer.

Given that I named the number I was thinking of $M$, I received a small selection of entries from people who didn’t want to play like normal people, at least one of which just specified $f(N)=M$ – this is OBVIOUSLY not what you were meant to do. Others tried to define a function in terms of $N$ and $M$, including the following:

\[f(x_1, x_2, \ldots, x_N) = \begin{cases}

\Bigg(\frac{\displaystyle\sum_{i=1}^N \displaystyle\sum_{j=1}^{x_i} x_i^j}{N!}\Bigg)^{\displaystyle\frac{1}{x_{\max{3,N}}}}&& \textrm{if }N\textrm{ is coprime with }M\\

6 && \textrm{otherwise}

\end{cases}\]

For the record, while typing this in LaTeX I’ve hit a record of four consecutive close braces. Sadly $6$, the number of entries in my list, is not coprime with $0$, so this gives the answer $6$ (and is disqualified anyway for using $M$ in the definition).

$f(N,M) = 0. \big(\displaystyle\sum_{i=1}^{N-2} x_i\big) + M$

This was presumably hoping to get as small a number as possible in the first part, then add it to M and hope nobody was closer. Turns out not only is this cheating, it was also not close enough, because we did actually get an exact winner, and it was glorious. See below.

$f(n_1, n_2, n_3, \ldots, n_N) = n_2 + n_3 + \ldots + n_N \mod n_1$

This function takes all but the first number, adds them together and then takes the result modulo the first number, aka remainder on division. In the case of The Numbers, the result is:

$f(4, 8, 15, 16, 23, 42) = (8 + 15 + 16 + 23 + 42) \mod 4 \equiv 0$

Amazingly, through pure chance, the sum of these five numbers (104) is a multiple of 4, and so the answer is zero! We have a winner. Congratulations to the winner, who forgot to write their name on the sheet, but was identified and awarded their prize – a wind-up robot, as per the rules of the Competition Competition, worth strictly less than £1.

]]>