Download e-book for kindle: 104 number theory problems. From the training of the USA IMO by Titu Andreescu

By Titu Andreescu

The publication is dedicated to the homes of conics (plane curves of moment measure) that may be formulated and proved utilizing merely undemanding geometry. beginning with the well known optical homes of conics, the authors circulate to much less trivial effects, either classical and modern. particularly, the bankruptcy on projective homes of conics encompasses a certain research of the polar correspondence, pencils of conics, and the Poncelet theorem. within the bankruptcy on metric homes of conics the authors speak about, specifically, inscribed conics, normals to conics, and the Poncelet theorem for confocal ellipses. The e-book demonstrates the good thing about in simple terms geometric tools of learning conics. It includes over 50 routines and difficulties aimed toward advancing geometric instinct of the reader. The booklet additionally includes greater than a hundred conscientiously ready figures, so that it will aid the reader to higher comprehend the cloth awarded

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Additional 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.

Setting a = 1 in the last example leads to a property of the Fermat numbers, which will soon be discussed. 23. Determine whether there exist infinitely many even positive integers k such that for every prime p the number p 2 + k is composite. Solution: The answer is positive. First note that for p = 2, p 2 + k is always composite for all even positive integers k. Next we note that if p > 3, then p 2 ≡ 1 (mod 3). Hence if k is an even positive integer with k ≡ 2 (mod 3), then p 2 + k is composite for all all primes p > 3 ( p 2 + k is greater than 3 and is divisible by 3).

Assume that N = a1 a2 . . an is divisible by 5n and has only odd digits. Consider the numbers N1 = 1a1 a2 . . an = 1 · 10n + 5n M = 5n (1 · 2n + M), N2 = 3a1 a2 . . an = 3 · 10n + 5n M = 5n (3 · 2n + M), N3 = 5a1 a2 . . an = 5 · 10n + 5n M = 5n (5 · 2n + M), N4 = 7a1 a2 . . an = 7 · 10n + 5n M = 5n (7 · 2n + M), N5 = 9a1 a2 . . an = 9 · 10n + 5n M = 5n (9 · 2n + M). The numbers 1 · 2n + M, 3 · 2n + M, 5 · 2n + M, 7 · 2n + M, 9 · 2n + M give distinct remainders when divided by 5. Otherwise, the difference of some two of them would be a multiple of 5, which is impossible, because 2n is not a multiple of 5, nor is the difference of any two of the numbers 1, 3, 5, 7, 9.

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