By Jiagu Xu
Olympiad arithmetic isn't really a suite of options of fixing mathematical difficulties yet a procedure for advancing mathematical schooling. This publication is predicated at the lecture notes of the mathematical Olympiad education classes performed through the writer in Singapore. Its scope and intensity not just covers and exceeds the standard syllabus, yet introduces a spread strategies and techniques in smooth arithmetic. In every one lecture, the techniques, theories and strategies are taken because the middle. The examples are served to provide an explanation for and increase their intension and to point their purposes. along with, applicable variety of attempt questions is offered for reader's perform and checking out objective. Their specified recommendations also are very easily supplied. The examples should not very complex in order that readers can simply comprehend. there are lots of actual festival questions integrated which scholars can use to make sure their talents. those attempt questions are from many nations, e.g. China, Russia, united states, Singapore, and so on. particularly, the reader can locate many questions from China, if he's drawn to knowing mathematical Olympiad in China. This publication serves as an invaluable textbook of mathematical Olympiad classes, or as a reference ebook for comparable lecturers and researchers. Congruence of Integers Decimal illustration of Integers Pigeonhole precept Linear Inequality and process of Linear Inequalities Inequalities with Absolute Values Geometric Inequalities options to checking out Questions
Read or Download Lecture Notes on Mathematical Olympiad Courses: For Junior Section, Vol. 2 (Mathematical Olympiad Series) PDF
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Extra info for Lecture Notes on Mathematical Olympiad Courses: For Junior Section, Vol. 2 (Mathematical Olympiad Series)
Let A2 = 2x9y1. Since 1412 = 19881 < A2 and 1752 = 30625 > A2 , so 1412 < A2 < 1752 . the units digit of A2 is 1 implies that the units digit of A is 1 or 9 only. Therefore it is sufficient to check 1512 , 1612 , 1712 , 1592 , 1692 only, so we find that 1612 = 25921 satisfies all the requirements, and other numbers cannot satisfy all the requirements. Thus, x = 5, y = 2, so that 3x + 7y = 15 + 14 = 29. Example 7. (CHNMOL/2004) Find the number of the pairs (x, y) of two positive integers, such that N = 23x + 92y is a perfect square number less than or equal to 2392.
Geonhole Principle, among them there must be at . . . • •. A ..... . . . A .. 2 4 . . .. least three with the same color. Without loss of ..... .... .. .. .... .. generality, it can be assumed that three are red, ...... • and they are A0 A1 , A0 A2 , A0 A3 . A3 Now consider three sides of the triangle A1 A2 A3 . e. its three sides are all red. Otherwise, the three sides of A1 A2 A3 are all blue, then the conclusion is also true. Testing Questions (A) 1. Prove that among any n + 1 integers, there must be at least two which are congruent modulo n.
When a = 1, then 145 = 1! + 4! , so 145 is a desired number. When a = 2, then b, c must be 5. But 255 = 2! + 5! , so no solution. When a = 3, then b, c must be 5. But 355 = 3! + 5! , so no solution. Thus, 145 is the unique solution. Example 8. (IMO/1962) Find the smallest natural number n which has the following properties: (a) Its decimal representation has 6 as the last digit. (b) If the last digit 6 is erased and placed in front of the remaining digits, the resulting number is four times as large as the original number n.