By Titu Andreescu

The e-book is dedicated to the homes of conics (plane curves of moment measure) that may be formulated and proved utilizing in basic terms basic 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 homes of conics features a specified research of the polar correspondence, pencils of conics, and the Poncelet theorem. within the bankruptcy on metric houses of conics the authors talk about, particularly, inscribed conics, normals to conics, and the Poncelet theorem for confocal ellipses. The ebook demonstrates the benefit of basically geometric equipment of learning conics. It includes over 50 workouts and difficulties geared toward advancing geometric instinct of the reader. The e-book additionally includes greater than a hundred conscientiously ready figures, so that it will aid the reader to higher comprehend the fabric offered

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**Additional resources for 104 number theory problems. From the training of the USA IMO team**

**Sample text**

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 . Fermat’s little theorem provides a good criterion to determine whether a number is composite. But the converse is not true. , 11 | (a 561 − a) and 561 = 3 · 11 · 17). The composite integers n satisfying a n ≡ a (mod n) for any integer a are called Carmichael numbers.

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. Thus there are 12 choices for each of a and b; that is, 12 · 12 of the 100 · 100 divisors of 1099 are multiples of 1088 . Consequently, the desired 12·12 9 probability is 100·100 = 625 . 17. Determine the number of ordered pairs of positive integers (a, b) such that the least common multiple of a and b is 23 57 1113 .

Assume that {a1 , a2 , . . , am } and {b1 , b2 , . . , bm } are two complete sets of residue classes modulo m. Prove that {a1 + b1 , a2 + b2 , . . , am + bm } is not a complete set of residue classes. Proof: We approach indirectly by assuming that it is. Then we have 1 + 2 + · · · + n ≡ (a1 + b1 ) + (a2 + b2 ) + · · · + (am + bm ) ≡ (a1 + a2 + · · · + am ) + (b1 + b2 + · · · + bm ) ≡ 2(1 + 2 + · · · + m) (mod m), implying that 1 + 2 + · · · + m ≡ 0 (mod m), or m | even integers m. Hence our assumption was wrong.