104 number theory problems : from the training of the USA by Titu Andreescu

Number Theory

By Titu Andreescu

This difficult challenge e-book by means of popular US Olympiad coaches, arithmetic lecturers, and researchers develops a mess of problem-solving abilities had to excel in mathematical contests and in mathematical examine in quantity thought. delivering thought and highbrow pride, the issues through the ebook motivate scholars to precise their rules in writing to give an explanation for how they conceive difficulties, what conjectures they make, and what conclusions they succeed in. utilising particular ideas and techniques, readers will collect a great realizing of the elemental ideas and ideas of quantity theory.

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Extra info for 104 number theory problems : from the training of the USA IMO team

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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. 27. m such that m(m+1) , 2 which is not true for Let a be a positive integer. Determine all the positive integers {a · 1, a · 2, a · 3, .

Indeed, a 2s ≡ 1 (mod m), so d divides 2s. If d < 2s, then 22s−d ≡ −1 (mod m) violates the minimality assumption on s. Furthermore, if t is an integer such that a t ≡ −1 (mod m), then t is a multiple of s. Because a 2t ≡ 1 (mod m), it follows that d = 2s divides 2t, and so s divides t. It is then clear that t must an odd multiple of s; that is, at ≡ −1 1 if t is an odd mulptiple of s; if t is an even mulptiple of s. 36. [AIME 2001] How many positive integer multiples of 1001 can be expressed in the form 10 j − 10i , where i and j are integers and 0 ≤ i < j ≤ 99?

By Fermat’s little theorem, we have p 2 ≡ 1 (mod 3) and p 4 ≡ 1 (mod 5). Because a positive integer is relatively prime to 24 if and only if it is odd, ϕ(24 ) = 23 . By Euler’s theorem, we have p 8 ≡ 1 (mod 16). Therefore, p 8 ≡ 1 (mod m) for m = 3, 5, and 16, implying that p 8 ≡ 1 (mod 240). Note that this solution indicates that we can establish Euler’s theorem by Fermat’s little theorem. 24 (5)). Hence we can improve the result to p 4 ≡ 1 (mod 240) for all primes p > 5. 32. Prove that for any even positive integer n, n 2 − 1 divides 2n!

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