104 number theory problems. From the training of the USA IMO - download pdf or read online

By Titu Andreescu

ISBN-10: 0817645276

ISBN-13: 9780817645274

ISBN-10: 0817645616

ISBN-13: 9780817645618

The e-book is dedicated to the homes of conics (plane curves of moment measure) that may be formulated and proved utilizing purely easy geometry. beginning with the well known optical houses of conics, the authors flow to much less trivial effects, either classical and modern. specifically, the bankruptcy on projective houses of conics encompasses a designated research of the polar correspondence, pencils of conics, and the Poncelet theorem. within the bankruptcy on metric homes of conics the authors speak about, particularly, inscribed conics, normals to conics, and the Poncelet theorem for confocal ellipses. The booklet demonstrates the benefit of in basic terms geometric tools of learning conics. It includes over 50 routines and difficulties aimed toward advancing geometric instinct of the reader. The ebook additionally comprises greater than a hundred rigorously ready figures, with a view to support the reader to raised comprehend the fabric offered

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Extra info for 104 number theory problems. From the training of the USA IMO team

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It follows that n has 9 · 5 · 5 · 5 = 1125 positive divisors. If d = 4202 , 4 4 then 420 d is also a divisor, and the product of these two divisors is 420 . We can thus partition 1124 divisors of n (excluding 4202 ) into 562 pairs of divisors of the form d, dn , and the product of the two divisors in each pair is 4204 . Hence the answer is 4204·562 · 4202 = 4202250 . Putting the last three examples together gives two interesting results in number theory. For a positive integer n denote by τ (n) the number of its divisors.

Xn ) of (1) satisfies the congruence a1 x 1 + a2 x 2 + · · · + an x n ≡ b (mod dn−1 ), which is equivalent to an x n ≡ b (mod dn−1 ). ϕ(dn−1 )−1 Multiplying both sides of (†) by an ≡ 1 (mod dn−1 ), we obtain ϕ(d ) an n−1 xn ≡ c ϕ(d (†) and taking into account that (mod dn−1 ), )−1 where c = an n−1 b. It follows that xn = c + dn−1 tn−1 for some integer tn−1 . Substituting in (∗) and rearranging yields the equation in (n − 1) variables a1 x1 + · · · + an−1 xn−1 = b − an c − an−1 dn−1 tn−1 . It remains to show that dn−1 | (b − an c − an−1 dn−1 tn−1 ), which is equivalent to an c ≡ b (mod dn−1 ).

16. [AIME 1988] Compute the probability that a randomly chosen positive divisor of 1099 is an integer multiple of 1088 . Solution: What are the divisors of 1099 ? Is 3 a divisor? Is 220 a divisor? We consider the prime factorization of 1099 , which is 299 · 599 . The divisors of 1099 are of the form 2a · 5b , where a and b are integers with 0 ≤ a, b ≤ 99. Because there are 100 choices for each of a and b, 1099 has 100 · 100 positive integer divisors. Of these, the multiples of 1088 = 288 · 588 must satisfy the inequalities 88 ≤ a, b ≤ 99.

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104 number theory problems. From the training of the USA IMO team by Titu Andreescu

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