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# Download 104 number theory problems: from the training of the USA IMO by Titu Andreescu PDF

By Titu Andreescu

ISBN-10: 0817645276

ISBN-13: 9780817645274

This difficult challenge ebook through well known US Olympiad coaches, arithmetic lecturers, and researchers develops a large number of problem-solving talents had to excel in mathematical contests and in mathematical learn in quantity thought. supplying concept and highbrow satisfaction, the issues during the booklet inspire scholars to precise their principles 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 methods, readers will collect a great knowing of the basic ideas and ideas of quantity thought.

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C) Let s (n) = a0 − a1 + · · · + (−1)h ah (alternating sum). Then n is divisible by 11 if and only if s (n) is divisible by 11. (d) n is divisible by 7, 11, or 13 if and only if ah ah−1 . . a3 − a2 a1 a0 has this property. (e) n is divisible by 27 or 37 if and only if ah ah−1 . . a3 + a2 a1 a0 has this property. 1. Foundations of Number Theory 47 (f) n is divisible by 2k or 5k (k ≤ h) if and only if ak−1 . . a0 has this property. Proof: For (a) and (b), since 10k = (9 + 1)k , it follows that 10k ≡ 1 (mod 9).

AHSME 1973] In the following equation, each of the letters represents uniquely a different digit in base ten: (Y E) · (M E) = T T T. Determine the sum E + M + T + Y . Solution: Because T T T = T ·111 = T ·3·37, one of Y E and M E is 37, implying that E = 7. But T is a digit and T ·3 is a two-digit number ending with 7, and so it follows that T = 9 and T T T = 999 = 27·37, and so E+M+T +Y = 2+3+7+9 = 21. 42. [AIME 2001] Find the sum of all positive two-digit integers that are divisible by each of their digits.

For any positive integer n, d=n τ (n) 2 . 18. It is interesting to note that these three results can be generalized to the case that the powers of the primes in the prime decomposition are nonnegative (because if ai = 0 for some 1 ≤ i ≤ k, then ai + 1 = 2ai + 1 = 1, which does not affect the products). 16. For any positive integer n, τ (n) ≤ 2 n. √ Proof: Let d1 < d2 < · · · < dk be the divisors of n not exceeding n. , . d1 d2 dk √ It follows that τ (n) ≤ 2k ≤ 2 n. The Sum of Divisors For a positive integer n denote by σ (n) the sum of its positive divisors, including 1 and n itself.