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

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By Titu Andreescu

This difficult challenge e-book by means of popular US Olympiad coaches, arithmetic academics, and researchers develops a large number of problem-solving talents had to excel in mathematical contests and study in quantity thought. supplying thought and highbrow pride, the issues through the ebook inspire scholars to precise their principles, conjectures, and conclusions in writing. using particular innovations and methods, readers will gather an outstanding knowing of the basic recommendations and ideas of quantity theory.Key features:* comprises difficulties built for varied mathematical contests, together with the overseas Mathematical Olympiad (IMO)* Builds a bridge among traditional highschool examples and workouts in quantity idea and extra refined, problematic and summary techniques and difficulties* starts by way of familiarizing scholars with average examples that illustrate relevant subject matters, by way of a variety of rigorously chosen difficulties and large discussions in their strategies* Combines unconventional and essay-type examples, workouts and difficulties, many provided in an unique type* Engages scholars in inventive considering and stimulates them to specific their comprehension and mastery of the cloth past the classroom104 quantity concept difficulties is a worthy source for complex highschool scholars, undergraduates, teachers, and arithmetic coaches getting ready to take part in mathematical contests and people considering destiny learn in quantity conception and its comparable components.

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

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But adding three cubes modulo 7 gives too many residue classes (since 7 is too small). We then consider n = 9 with ϕ(9) = 6. Because 2002 ≡ 4 (mod 9) and 20023 ≡ 43 ≡ 1 (mod 9), it follows that 20022002 ≡ (20023 )667 · 2004 ≡ 4 (mod 9). On the other hand, x 3 ≡ 0, ±1 (mod 9) for integers x. We see that x13 +x23 +x33 ≡ 4 (mod 9). 32 104 Number Theory Problems It remains to show that 20022002 is a sum of four cubes. Starting with 2002 = 103 + 103 + 13 + 13 and using 2002 = 667 · 3 + 1 once again, we find that 20022002 = 2002 · (2002667 )3 = (10 · 2002667 )3 + (10 · 2002667 )3 + (2002667 )3 + (2002667 )3 .

Defined by a n = 2n + 3 n + 6 n − 1 for all positive integers n. Determine all positive integers that are relatively prime to every term of the sequence. First Solution: The answer is 1. It suffices to show that every prime p divides an for some positive integer n. Note that both p = 2 and p = 3 divide a2 = 22 + 32 + 62 − 1 = 48. Assume now that p ≥ 5. By Fermat’s little theorem, we have 2 p−1 ≡ 3 p−1 ≡ p−1 6 ≡ 1 (mod p). Then 3 · 2 p−1 + 2 · 3 p−1 + 6 p−1 ≡ 3 + 2 + 1 ≡ 6 (mod 6), or 6(2 p−2 + 3 p−2 + 6 p−2 − 1) ≡ 0 (mod p); that is, 6a p−2 is divisible by p.

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 ! , 1 ≤ f m 2 ≤ m 2 , and 0 ≤ r2 < m 2 !. Keeping this process on r2 , and so on, we obtain a unique factorial base expansion of k. 43. Let F0 = 1, F1 = 1, and Fn+1 = Fn + Fn−1 for every positive integer n. ) Each nonnegative integer n can be uniquely written as a sum of nonconsecutive positive Fibonacci numbers; that is, each nonnegative integer n can be written uniquely in the form n= ∞ αk Fk , k=0 where αk ∈ {0, 1} and (αk , αk+1 ) = (1, 1) for each k.

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