By Derek Holton
The overseas Mathematical Olympiad (IMO) is an annual overseas arithmetic pageant held for pre-collegiate scholars. it's also the oldest of the foreign technological know-how olympiads, and festival for locations is especially fierce. This ebook is an amalgamation of the 1st eight of 15 booklets initially produced to lead scholars aspiring to contend for placement on their country's IMO crew. the cloth contained during this booklet offers an advent to the most mathematical themes coated within the IMO, that are: Combinatorics, Geometry and quantity conception. furthermore, there's a detailed emphasis on tips to process unseen questions in arithmetic, and version the writing of proofs. complete solutions are given to all questions. even though a primary Step to Mathematical Olympiad difficulties is written from the viewpoint of a mathematician, it truly is written in a fashion that makes it simply understandable to youngsters. This ebook is usually a must-read for coaches and teachers of mathematical competitions.
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Extra resources for A First Step to Mathematical Olympiad Problems
I'll point out generalisations of other results as they arise. We have already thought about this idea in Chapter 1. Problem 2. Students in a university lecture have black, brown, red, green, or blue and white hair. There are 101 students in the lecture. Show there are at least 21 students who have the same colour hair. Discussion. The pigeonholes here are the hair colours. There are 5 of these. The pigeons are the 101 students. In this question then, n = 5 and mn + 1 = 101. So m + 1 = 21. By the more general pigeonhole principle, there must be at least 21 students in the lecture who have the same colour hair.
The number 30 is best possible. Proof. 30 = 5 x 6, 31 = 4 x 6 + 7, 32 = 3 x 6 + 2 x 7, 33 = 2 x 6 + 3 x 7, 34 = 6 + 4 x 7, 35 = 5 x 7. Any n ≥ 30 can be written as 30 + 6k, 31 + 6k, 32 + 6k, 33 + 6k, 34 + 6k or 35 + 6k for some value of k. Hence any number greater than or equal to 30 can be written in the required form. Suppose 30 is not best possible. Then 29 = 6a + 7b, where a and b are not both negative. Hence 7b = 29 — 6a. But 7b ≥ 0, so 0 ≤ a ≤ 4. No matter which of these values of a we take, 29 — 6a is not a multiple of 7.
What could be simpler or more obvious? Problem 1. You can use the pigeonhole principle to come up with some startlingly trivial facts. For instance, to the nearest dollar, there are at least two wage earners in your country who earn precisely the same amount. Discussion. The easy way to see that is to observe there aren't too many people in the country earning more than $200,000 a year. ) But there must be more than 200,001 wage earners, earning less than $200,000. With the dollar amounts as pigeonholes and the wage earners as pigeons, the pigeonhole principle tells us that there are 2 wage earners at least, who earn the same amount of money in a year.