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

By Titu Andreescu

This hard challenge booklet by way of popular US Olympiad coaches, arithmetic academics, and researchers develops a mess of problem-solving talents had to excel in mathematical contests and in mathematical learn in quantity concept. delivering proposal and highbrow pride, the issues in the course of the ebook inspire scholars to specific their principles in writing to provide an explanation for how they conceive difficulties, what conjectures they make, and what conclusions they achieve. utilizing particular strategies and methods, readers will collect a high-quality knowing of the elemental strategies and ideas of quantity thought.

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1. Foundations of Number Theory 33 Solution: Because 10 j − 10i = 10i (10 j−i − 1) and 1001 = 7 · 11 · 13 is relatively prime to 10i , it is necessary to ﬁnd i and j such that the 10 j−i − 1 is divisible by the primes 7, 11, and 13. Notice that 1001 = 7 · 11 · 13; that is, 103 ≡ −1 (mod 1001). It is easy to check that ord1001 (10) = 6. 30, − 1) is divisible by 1001 if and only if j − i = 6n for some positive integer n. Thus it is necessary to count the number of integer solutions to i + 6n = j, where j ≤ 99, i ≥ 0, and n > 0.

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. There are also even Carmichael numbers, for example n = 2 · 73 · 1103.

And {4, 4+7, 4+2·7, . . } are two sequences satisfying the conditions of the problem. By noting that ϕ(9) = 6, we can also ﬁnd sequences of the form {a, a + 9, a + 2 · 9, . . }. We leave the details to the reader. 35. [IMO 2003 shortlist] Determine the smallest positive integer k such that there exist integers x1 , x2 , . . , xk with x13 + x23 + · · · + xk3 = 20022002 . Solution: The answer is k = 4. We ﬁrst show that 20022002 is not a sum of three cubes. To restrict the number of cubes modulo n, we would like to have ϕ(n) to be a multiple of 3.