104 Number Theory Problems: From the Training of the USA IMO by Titu Andreescu PDF

By Titu Andreescu

ISBN-10: 0817645276

ISBN-13: 9780817645274

ISBN-10: 0817645616

ISBN-13: 9780817645618

This hard challenge e-book via popular US Olympiad coaches, arithmetic lecturers, and researchers develops a large number of problem-solving abilities had to excel in mathematical contests and learn in quantity concept. supplying idea and highbrow satisfaction, the issues during the e-book inspire scholars to precise their rules, conjectures, and conclusions in writing. utilizing particular thoughts and techniques, readers will gather an outstanding knowing of the elemental suggestions and ideas of quantity theory.Key features:* includes difficulties constructed for varied mathematical contests, together with the foreign Mathematical Olympiad (IMO)* Builds a bridge among traditional highschool examples and routines in quantity concept and extra subtle, tricky and summary suggestions and difficulties* starts by means of familiarizing scholars with usual examples that illustrate valuable subject matters, by way of various conscientiously chosen difficulties and huge discussions in their recommendations* Combines unconventional and essay-type examples, routines and difficulties, many awarded in an unique model* Engages scholars in inventive pondering and stimulates them to precise their comprehension and mastery of the fabric past the classroom104 quantity idea difficulties is a worthy source for complicated highschool scholars, undergraduates, teachers, and arithmetic coaches getting ready to take part in mathematical contests and people considering destiny study in quantity concept and its comparable parts.

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Example text

1(9) in base 9 is triangular; that is, it is the sum of the first k positive integers for some positive integer k. Proof: Indeed 11 . . 1 (9) = 9n−1 + 9n−2 + · · · + 9 + 1 n 1’s 9n − 1 1 3n − 1 3 n + 1 = · · 9−1 2 2 2 3n − 1 = 1 + 2 + ··· + . 2 = Thus it is a triangular number. 48. Determine all positive integers n such that 11111(n) is a perfect square. Solution: The answer is n = 3. We have 11111(n) = n 4 + n 3 + n 2 + n + 1. If n is even, then n 2 + n2 and n 2 + n2 + 1 are two consecutive integers.

Cm } be a complete set of residue classes modulo m. 24, {ac1 − b, ac2 − b, . . , acm − b} is also a complete set of residue classes. Hence there exists ci such that ac1 − b ≡ 0 (mod m), or c1 is a solution to the congruence equation ax ≡ b (mod m). It is easy to see that all the numbers congruent to c1 modulo m also satisfy the congruence equation. On the other hand, if both x and x satisfy the equation, we have ax ≡ ax (mod m). 20, we have x ≡ x (mod m). 25 shows that if gcd(a, m) = 1, then there is x such that ax ≡ 1 (mod m).

Hence a = 44444444 < 104·4444 = 1017776 , and so a cannot have more than 17776 digits. Because each digit is at most a 9, A = S(a) ≤ 17776 · 9 = 159984. Of the natural numbers less than or equal to 52 104 Number Theory Problems 159984, the number with the largest digit sum is 99999, and so B = S(A) ≤ 45. Of the natural numbers less than or equal to 45, the number with the largest digit sum is 39. Hence S(B) ≤ 12. 45 (a), we have S(B) ≡ B ≡ S(A) ≡ A ≡ S(a) ≡ a ≡ 44444444 (mod 9). It suffices to show that 44444444 ≡ 7 (mod 9).

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104 Number Theory Problems: From the Training of the USA IMO Team by Titu Andreescu

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