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

By Titu Andreescu

This demanding challenge e-book by way of well known US Olympiad coaches, arithmetic lecturers, and researchers develops a mess of problem-solving abilities had to excel in mathematical contests and learn in quantity thought. supplying thought and highbrow satisfaction, the issues through the publication inspire scholars to precise their principles, conjectures, and conclusions in writing. making use of particular innovations and methods, readers will collect an excellent figuring out of the elemental strategies and concepts of quantity theory.Key features:* includes difficulties constructed for numerous mathematical contests, together with the foreign Mathematical Olympiad (IMO)* Builds a bridge among traditional highschool examples and routines in quantity thought and extra refined, complicated and summary ideas and difficulties* starts off by way of familiarizing scholars with normal examples that illustrate vital issues, by way of various rigorously chosen difficulties and huge discussions in their recommendations* Combines unconventional and essay-type examples, routines and difficulties, many awarded in an unique type* Engages scholars in inventive considering and stimulates them to specific their comprehension and mastery of the fabric past the classroom104 quantity thought 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 study in quantity conception and its similar parts.

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Solution: The answer is 207. Note that digits 4, 6, and 8 cannot appear in the units digit. Hence the sum is at least 40 + 60 + 80 + 1 + 2 + 3 + 5 + 7 + 9 = 207. On the other hand, this value can be obtained with the set {2, 5, 7, 43, 61, 89}. 44. Write 101011(2) in base 10, and write 1211 in base 3. Solution: We have 1010011(2) = 1 · 26 + 0 · 25 + 1 · 24 + 0 · 23 + 0 · 22 + 1 · 2 + 1 = 64 + 16 + 2 + 1 = 83. Dividing by 3 successively, the remainders give the digits of the base-3 representation, beginning with the last.

For each n = 1, 2, 3, . . , 15, there are 100 − 6n suitable values of i (and j), so the number of solutions is 94 + 88 + 82 + · · · + 4 = 784. Euler’s Totient Function We discuss some useful properties of Euler’s totient function ϕ. 31. Let p be a prime, and let a be a positive integer. Then ϕ( pa ) = pa − pa−1 . 32. Let a and b be two relatively prime positive integers. Then ϕ(ab) = ϕ(a)ϕ(b). Proof: : Arrange the integers 1, 2, . . , ab into an a × b array as follows: 1 a+1 .. 2 a+2 .. ··· ··· ..

But n = a0 + a1 b + · · · + ak bk ≤ (b − 1)(1 + b + · · · + bk ) = bk+1 − 1 < bk+1 , a contradiction. If h = k, then a 0 + a 1 b + · · · + a k b k = c0 + c 1 b + · · · + c k b k , and so b | (a0 − c0 ). On the other hand, |a0 − c0 | < b; hence a0 = c0 , Therefore a1 + a2 b + · · · + ak bk−1 = c1 + c2 b + · · · + ck bk−1 . By repeating the above procedure, it follows that a1 = c1 , a2 = c2 , . . , and a k = ck . Relation (∗) is called the base-b representation of n and is denoted by n = ak ak−1 .

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