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

By Titu Andreescu

This difficult challenge booklet by way of popular US Olympiad coaches, arithmetic academics, and researchers develops a large number of problem-solving talents had to excel in mathematical contests and learn in quantity concept. delivering notion and highbrow pride, the issues through the ebook inspire scholars to specific their rules, conjectures, and conclusions in writing. employing particular ideas and techniques, readers will gather a superb realizing of the elemental strategies and concepts of quantity theory.Key features:* comprises difficulties built for numerous mathematical contests, together with the foreign Mathematical Olympiad (IMO)* Builds a bridge among traditional highschool examples and workouts in quantity thought and extra refined, complex and summary thoughts and difficulties* starts via familiarizing scholars with average examples that illustrate imperative issues, by means of various rigorously chosen difficulties and large discussions in their options* Combines unconventional and essay-type examples, workouts and difficulties, many offered in an unique type* Engages scholars in inventive pondering and stimulates them to precise their comprehension and mastery of the cloth 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 idea and its comparable components.

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0 ≤ ak ≤ αk . Each divisor of n appears exactly once as a summand in the expansion of the product (1 + p1 + · · · + p1α1 ) · · · (1 + pk + · · · + pkαk ), from which the desired result follows, by also noting the formula for the sum of a finite geometric progression: r k+1 − 1 = 1 + r + r2 + · · · + rk. 19. Find the sum of even positive divisors of 10000. Solution: The even divisors of 10000 can be written in the form of 2a 5b , where a and b are integers with 1 ≤ a ≤ 5 and 0 ≤ b ≤ 5. Each even divisor of 10000 appears exactly once as a summand in the expansion of the product 56 − 1 5−1 = 242172.

H h Hence n ≡ k=0 ak 10k = k=0 ak ≡ S(n) (mod 9). For (c), we note that 10k = (11 − 1)k . Hence 10k ≡ (−1)k (mod 11), and so h n≡ h ak 10k ≡ k=0 ak · (−1)k ≡ s (n) (mod 11), k=0 from which the conclusion follows. For (d), the conclusion follows by the facts 1001 = 7 · 11 · 13 and n = ah ah−1 . . a3 · 1000 + a2 a1 a0 = ah ah−1 . . a3 · (1001 − 1) + a2 a1 a0 . For (e), the conclusion follows by the facts 999 = 27 · 37 and n = ah ah−1 . . a3 · 1000 + a2 a1 a0 = ah ah−1 . . a3 · (999 + 1) + a2 a1 a0 .

F m ), meaning that k = 1! · f 1 + 2! · f 2 + 3! · f 3 + · · · + m! · f m , where each f i is an integer, 0 ≤ f i ≤ i, and f m > 0. Proof: Note that there exists a unique positive integer m 1 such that m 1 ! ≤ k < (m 1 + 1)!. By the division algorithm, we can write k = m 1 ! f m 1 + r1 for some positive integer f m 1 and some integer r1 with 0 ≤ r1 < m 1 !. Because k < (m + 1)! = m! · (m + 1), it follows that f m 1 ≤ m. Repeating this process, we can then write r1 = m 2 ! f m 2 + r2 , with m 2 the unique positive integer with m 2 !

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