I woke up this morning feeling rather good, and thus decided to approach a problem that has proven particularly irksome (or “compelling”, if you prefer). It’s taken from the first I.O.L.; it’s the second in the set of individual problems for that year. Because I can no longer link to the .pdf (it seems to have been removed, else the web page has somehow disintegrated), I will reproduce the question below, in bold text.

**Problem 2 (25 marks)**

**Below you see arithmetic equalities written in Egyptian Arabic. All summands, as well as all sums except the last one, are represented by fractions in which neither the numerators nor the denominators are greater than 10, nor is any denominator equal to 1:**

** tumn + tumnēn = talatt itmān **

**sabaçt itlāt + suds = çašart irbāç **

**tusçēn + tusç = sudsēn **

**xamast ixmās + subç = tamant isbāç **

**subçēn + xumsēn = 24/35**

**Assignment 1. Write these equalities in figures.**

**Assignment 2. The equality ****rubç + çašart itsāç = sabaçt isdās ****is missing a sign. Which one?**

**Note: The letter ****š ****is pronounced as English ****sh****, ****x**** as the ****ch**** in loch; ****ç ****is a specific Arabic consonant. A bar above a vowel indicates length. **

Please note that as inaccurate as it may be, I’ve resorted to utilizing a *c* with a cedilla in the place of the aforementioned Arabic consonant. Additionally, I’ve written “24/35” as a shilling fraction not because I enjoy that notation, but by dint of my attempting to make it at least moderately reminiscent of the other sums and summands. This will later prove useful, I’m thinking.

I was looking at this rather quizzically; I can safely say that I had absolutely no idea as to where to begin. If it were a person, then I would most certainly have concluded that it was out of my league. In either case, I looked it up and down and began by noting that several words appear to be of the same breed – that is, several pairs of words are similar in the same way. You can see how inarticulate this is going to be, I presume.

Anyway, noting that, we’ve *tumn- tumnēn, suds-sudsēn, tusç-tusçēn, and subç-subçēn. This phenomenon is inherently evident; as such, the solver may think ēn to harbor some form of linguistic significance. It doesn’t, though – its function is purely mathematical (there is a less obvious allusion here – you may laugh now, but you won’t later). Notice also that all of the words end in consonants. None of this, however, appears to be at all helpful at this point. *

*I thus proceeded to write out all of the words in frenzied frustration; I noticed something interesting about the distribution of consonants and vowels. There are only four patterns present – certainly, that indicates something? We’ve cvcc, cvccvc, cvcvcc, and vccvc; writing out all of the words that fall under each category, we observe that cvcc is actually cucc, cvccvc is cuccēn, cvcvcc is cacact*, and

*vccvc*is

*iccāc*. The fact that a profound relationship exists where

*cucc*and

*cucc*

*ēn*are concerned is becoming increasingly evident. Also, note that cvcvcc is always followed by*vccvc*.* Moving forwards, we shouldn’t leave c to denote “consonant” – that’s rather vague, as we’ve many consonants. Let’s number them, at least within the context of the words. The patterns thus become 1u*23, 1

*u*23

*3*

*2a**ēn*, 1a*t*, and

*i*12

*ā*3. Note that one word always harbors precisely three consonants. This appears to be amounting to something.

Recall the question and its specifications – it was noted that no numerator aside from that which was given can exceed ten – *ergo*, we’ve ten possible numerators. A denominator can neither equal one, nor can it exceed ten, a constraint which results in nine possible denominators. So, we know that we are, in some context, looking for the natural numbers from one to ten inclusive. We’ve noted that the consonants always appear in triples: *tmn*, *tlt*, *sb ç, sds, çšr, rbç, tsç, and xms being the possibilities. These, incidentally, never differ – the order of the consonants is always identical. Why not let the consonant triples, then, represent the set of numbers for which we are searching? * The logic behind that is at least considerably evident – whilst the consonants remain fixed, there are more of them than there are vowels, which increases the probability of their representing numbers.

That done, we can basically conclude that the positioning of the vowels within the patterns dictates functions, with 1*u*23 being, say, *a/n*, 1*u*23*ēn being b/n, and 1a2a*3

*t*

*i*12

*ā*3 being

*n*

_{1}/n_{2.}Furthermore, it becomes evident that *xamast ixmās *is *n/n*, or one.

The fifth example, evidently, was given to be of the most immediate aid. Recalling that the denominator must be lesser than or equal to ten, we can conclude *b/7 + b/5 = 24/35*. Evaluating further, *b* = 2, and 1*u*23*ēn* is 2/n. Yes! We’ve finally broken into numbers – the rest follows logically. Another assumption follows – 1*u*23 is shorter than 1*u*23*ēn*, and should thus be more basic, which leaves it to be 1*/*n (I’d noted that you wouldn’t laugh).

So, reverting to the triples at hand – *tmn*,* tlt*,* sbç*,* sds*,* çšr*,* rbç*,* tsç*, and *xms*. We saw *sbç *and *xms *in the fifth example – one of them, therefore, equals five, whilst the other is seven. We cannot know which is which yet.

We can also speculate that numerators come before denominators – this is confirmed by the first example, where *tlt* is thereby equal to three.

Hereafter, I was whole-heartedly confused – I then proceeded to enable my rather feeble mind to wander. As a result of this, I stumbled upon an interesting etymological assumption – I presume that, as I proceed, *rbç* will be shown to equal four. I don’t wish to divulge why, as it’s entirely a speculation, and I’m not even sure that my reasoning is sound (I’m not largely familiar with Arabic roots). I’m going by a lot of phonology, some intuitive anagramming, and the information given with the question (that *ç* is an Arabic consonant).

So, cheating entirely too blatantly and assuming that *rbç *is four, and knowing that *tlt* is three, we proceed.

-It is evident that *tmn *is greater than *sbç* by precisely one; this is shown to be true in the fourth example.

–*tsç *is divisible by three, as we may readily extract from the third example.

The latter is a big give-away, but what now? We’ve the *tsç*–*sds* pair to evaluate, essentially, but *tsç *could be either six or nine, given that 3(*sds)* = 2(*tsç)*. I’m going to assume that *sds *is six, given the previously-mentioned etymology; *tsç *is consequently nine. Now that we have six and know it to *not* be *tmn*, we can conclude that *sbç *is not five but seven, and that *tmn *and *xms *are therefore eight and five respectively. Finally, substitute six into the second equality alongside the other numbers that we’ve unearthed; get 10/4 to be the *çšr*–*rbç *pair, which proves my assumption.

Assignment 1.

tumn + tumnēn = talatt itmān –> 1/8 + 2/8 = 3/8 sabaçt itlāt + suds = çašart irbāç –> 7/3 + 1/6 = 10/4 tusçēn + tusç = sudsēn –> 2/9 + 1/9 = 2/6 xamast ixmās + subç = tamant isbāç –> 5/5 + 1/7 = 8/7 subçēn + xumsēn = 24/35 –> 2/7 + 2/5 = 24/35

Assignment 2.

Transcribe *rubç + çašart itsāç = sabaçt isdās*; arrive at 1/4 + 10/9 = 7/6. Add a radical symbol.

So, to summarize, that was a lot of assumption where it wasn’t necessary. The only concept that has been thoroughly proven is my not being cut out for this form of thing – I completely suck at producing coherent work. In any case, I wrapped up this day, which has thus far been brimming with logic (or lack thereof), by rabidly tearing a box of Nestle “Favorites” open at the wrong end.

Hmm, I’m fairly surprised a question with a relatively common dialect like Egyptian Arabic was on the IOL at all–they seem to usually use very uncommon, nearly isolated languages so that it’s unlikely any particular country will have an advantage.

Ah, yes – Cyril and I were speaking of that being an issue where the French puzzle is concerned. I’m not certain as to why they enable such languages to be integrated into the puzzles. Nevertheless, I believe that knowing the language may, on occasion, inhibit your explaining the solution logically – that’s occurred several times in my case. However, this is me; my case should not be examined in proving anything, as I tend to be rather fail.

On an unrelated note, what do you think of this one? I found it far more irritating than the first, and certainly more challenging than all of the others in the same set.

Additionally, would you like me to send you a copy of the .pdf? I don’t believe that it is available presently.

EDIT: Lastly, how do you think that I may prove the

rbç-related point? I just assumed that it was the root for the Arabic word for “quatrain”; I’ve verified that. However, that’s not a satisfactory explanation, speaking logically.I must admit I was preoccupied when commenting on this so I hadn’t even really looked at the full question itself. 🙂

Summer makes me lazy and not the most productive partner!

Haha, all right, well, that’s entirely fine. I’m certain that you’re not lazy; it is the summer, you’ve entirely the right to relax free of linguistic puzzles. One who can’t go without inserting diacritics in French writing is most definitely productive where I’m concerned 🙂

SOPHIA I CANT BELIEVE YOU “BLATANTLY CHEATED”. *shames you*

I continue conformity to my hopeless pursuit of commenting on small observations completed within 2-3 minutes (and before reading the entire post). Anyway, is the denominator of these fractions written before the numerator?

Haha, well, I enjoy rather spontaneous, unplanned commenting – it often gifts us the chance to pursue interesting debates, and venture into realms of thought we wouldn’t otherwise have braved. Please, continue at your leisure!

A good idea, but not quite – the numerator is written first.