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 cuccēn 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 1u23, 1u23ēn, 1a2a3t, and i12ā3. 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, sbç, 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 1u23 being, say, a/n, 1u23ēn being b/n, and 1a2a3t i12ā3 being n1/n2.
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 1u23ēn is 2/n. Yes! We’ve finally broken into numbers – the rest follows logically. Another assumption follows – 1u23 is shorter than 1u23ēn, 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.
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
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.