By Derek Holton

The overseas Mathematical Olympiad (IMO) is an annual overseas arithmetic festival held for pre-collegiate scholars. it's also the oldest of the overseas technological know-how olympiads, and festival for locations is very fierce. This e-book is an amalgamation of the 1st eight of 15 booklets initially produced to lead scholars desiring to contend for placement on their country's IMO crew. the cloth contained during this e-book offers an advent to the most mathematical subject matters lined within the IMO, that are: Combinatorics, Geometry and quantity idea. additionally, there's a distinctive emphasis on how one can procedure unseen questions in arithmetic, and version the writing of proofs. complete solutions are given to all questions. although a primary Step to Mathematical Olympiad difficulties is written from the point of view of a mathematician, it really is written in a fashion that makes it simply understandable to young people. This booklet can also be a must-read for coaches and teachers of mathematical competitions.

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**Extra resources for A First Step to Mathematical Olympiad Problems**

**Sample text**

Can we do the same for s in general? If we can show that we can get 2s — 2, 2s — 1 and 2s using 3's and s's, then we can add on enough 3's and we can get any n. Well one of this triumvirate of numbers is easy. Surely you don't want me to prove that I can get 2s! So how do you get 2s — 1 and 2s — 2? Think about s for a minute. When you divide s by 3 you either get a remainder of 1 or a remainder of 2. This means that you can write s either as 3t + 1 or as 3t + 2 where t > 0. Let's have a look at the case s = 3t +1.

It means that when a certain symmetry exists (as here between solid lines and broken lines) we can argue on the assumption that one of them happens. This assumption does not alter the validity of the argument. Why not? In the present case suppose we dropped the “without loss of generality”. We could argue as we did, first assuming that there were at least three lines and we would get the result we wanted. However, to complete the argument we would need to consider the case when at least three broken lines came out of a.

Hmm. I can make 3, 6 or 9 litres just using the 3 litre jug and 5, 10 or 15 litres with the 5 litre jug. It's obvious, from those calculations that I'm going to have to use both jugs. Well, it's also pretty clear that 7 ≠ 3a + 56 if I keep a and b positive or zero. So I can't get 7 by just adding water from the two jugs in some combination. So what if I pour water from one jug into another? Let's fill up the 3 litre jug, then pour the water into the 5 litre jug. I can then fill up the 3 litre jug and pour into the bigger jug again until it's full.