By Jiagu Xu

ISBN-10: 9814293555

ISBN-13: 9789814293556

Olympiad arithmetic isn't really a suite of options of fixing mathematical difficulties yet a approach for advancing mathematical schooling. This e-book is predicated at the lecture notes of the mathematical Olympiad education classes performed via the writer in Singapore. Its scope and intensity not just covers and exceeds the standard syllabus, yet introduces a range ideas and strategies in sleek arithmetic. In every one lecture, the ideas, theories and strategies are taken because the middle. The examples are served to give an explanation for and increase their intension and to point their purposes. in addition to, acceptable variety of try questions is accessible for reader's perform and trying out function. Their distinct recommendations also are very easily supplied. The examples will not be 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 skills. 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 attracted to realizing mathematical Olympiad in China. This ebook serves as an invaluable textbook of mathematical Olympiad classes, or as a reference e-book for similar lecturers and researchers. Congruence of Integers Decimal illustration of Integers Pigeonhole precept Linear Inequality and approach of Linear Inequalities Inequalities with Absolute Values Geometric Inequalities strategies to checking out Questions

**Read Online or Download Lecture Notes on Mathematical Olympiad Courses: For Junior Section, Vol. 2 (Mathematical Olympiad Series) PDF**

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**Extra resources for Lecture Notes on Mathematical Olympiad Courses: For Junior Section, Vol. 2 (Mathematical Olympiad Series)**

**Sample text**

10) yields 16x + 13y = 77. Since 16(x − 4) + 13(y − 1) = 0, then x = 4, y = 1 is a special solution. So the general solution is x = 4 + 13t, y = 1 − 16t, t ∈ Z. Since y ≥ 0, so t ≤ 0. But x ≤ 0 implies t ≥ 0, so t = 0 is the unique permitted value of t. 9). Lecture Notes on Mathematical Olympiad Testing Questions 1. 51 (A) (AHSME/1992) If k is a positive integer such that the equation in x kx − 12 = 3k has an integer root, then the number of such k is (A) 3; (B) 4; (C) 5; 2. (D) 6; (E) 7. (CHINA/1990) An integer solution of the equation 1990x − 1989y = 1991 is (A) x = 12785, y = 12768; (B) x = 12785, y = 12770; (C) x = 11936, y = 11941; (D) x = 13827, y = 12623.

Solution From assumption in the question, (1000)(def ) + abc = 6[(1000)(abc) + def ], (994)(def ) = (5999)(abc), (142)(def ) = (857)(abc). Therefore 857 | (142)(def ). Since 857 and 142 have no common factor greater than 1, so 857 | def . Since 2 × 857 > 1000 which is not a three digit number, so def = 857. Thus, abc = 142, and a + b + c + d + e + f = 1 + 4 + 2 + 8 + 5 + 7 = 27. Example 5. Prove that each number in the sequence 12, 1122, 111222, · · · is a product of two consecutive whole numbers.

N ai 1951 1951 1951 −1 + − 1 + ··· + − 1 < 1951, a1 a2 a1 1951 1951 1951 + + ··· + < 1951 + n < 2 · 1951, a1 a2 an 1 1 1 ∴ + + ··· + < 2. a1 a2 an Example 10. (CHINA/1992) Given that real number a > 1, the natural number n ≥ 2, and the equation ax = x has exactly n distinct real solution. Fine the range of a. Lecture Notes on Mathematical Olympiad Solution Since 43 From assumptions each solution x must be an integer and x ≥ 0. ax = a x + {a}x, so the given equation becomes x = ax = a x + {a}x .

### Lecture Notes on Mathematical Olympiad Courses: For Junior Section, Vol. 2 (Mathematical Olympiad Series) by Jiagu Xu

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