By Titu Andreescu

This difficult challenge booklet through popular US Olympiad coaches, arithmetic lecturers, and researchers develops a large number of problem-solving talents had to excel in mathematical contests and in mathematical study in quantity thought. providing idea and highbrow pride, the issues in the course of the publication 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 achieve. utilizing particular strategies and techniques, readers will gather a superior realizing of the basic innovations and concepts of quantity theory.

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

**Example text**

On the other hand, there are ϕ(a) columns containing those elements in the table relatively prime to a. 24. Hence there are exactly ϕ(b) elements in each of those columns that are relatively prime to b. Therefore, there are ϕ(a)ϕ(b) numbers in the table that are relatively prime to ab. Hence ϕ(ab) = ϕ(a)ϕ(b) for relatively prime integers ab. 33. If n = p1α1 · · · pkαk is the prime factorization of n > 1, then ϕ(n) = n 1 − 1 p1 ··· 1 − 1 pk . 32. Second Proof: We employ the inclusion and exclusion principle.

We claim that there are no other answers. Since 11 . . 1 ≡ 11 ≡ 3 (mod 4), 11 . . 1 is not a perfect square. 24 k 1’s k 1’s (3)). (2) The answer is no. If n is a 5-digit number consisting only of distinct even digits, then the sum of its digits is equal to 0 + 2 + 4 + 6 + 8 = 20, which is congruent to 2 modulo 9; hence it is not a perfect square. ) 48 104 Number Theory Problems (3) The given number is not a perfect square because the sum of its digits is 6, a multiple of 3 but not of 9. 51. [AIME 1984] The integer n is the smallest positive multiple of 15 such that every digit of n is either 0 or 8.

If n = p1α1 · · · pkαk is the prime factorization of n > 1, then ϕ(n) = n 1 − 1 p1 ··· 1 − 1 pk . 32. Second Proof: We employ the inclusion and exclusion principle. Set Ti = {d : d ≤ n and pi |d}, for i = 1, . . , k. It follows that T1 ∪ · · · ∪ Tk = {m : m ≤ n and gcd(m, n) > 1}. Hence ϕ(n) = n − |T1 ∪ · · · ∪ Tk | k |Ti | + =n− i=1 |Ti ∩ T j | − · · · + (−1)k |T1 ∩ · · · ∩ Tk |. , |T1 ∩ · · · ∩ Tk | = n . p1 · · · pk Finally, n ϕ(n) = n 1 − i=1 =n 1− 1 p1 1 1 1 + − · · · + (−1)k pi 1≤i< j≤k pi p j p1 · · · pk ··· 1 − 1 pk .