By Titu Andreescu
This difficult challenge e-book via popular US Olympiad coaches, arithmetic lecturers, and researchers develops a mess of problem-solving abilities had to excel in mathematical contests and study in quantity idea. delivering idea and highbrow pride, the issues during the booklet motivate scholars to precise their rules, conjectures, and conclusions in writing. making use of particular thoughts and methods, readers will collect an exceptional realizing of the elemental options and concepts of quantity theory.Key features:* includes difficulties constructed for varied mathematical contests, together with the overseas Mathematical Olympiad (IMO)* Builds a bridge among traditional highschool examples and routines in quantity conception and extra subtle, difficult and summary strategies and difficulties* starts off through familiarizing scholars with ordinary examples that illustrate primary topics, by means of quite a few conscientiously chosen difficulties and vast discussions in their strategies* Combines unconventional and essay-type examples, workouts and difficulties, many provided in an unique style* Engages scholars in inventive pondering and stimulates them to precise their comprehension and mastery of the fabric past the classroom104 quantity concept difficulties is a helpful source for complicated highschool scholars, undergraduates, teachers, and arithmetic coaches getting ready to take part in mathematical contests and people considering destiny study in quantity thought and its similar components.
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Additional resources for 104 Number Theory Problems: From the Training of the USA IMO Team
51 [Ireland 1996] Find a positive integer n such that S(n) = Solution: Consider n = 1 33 . . 3 5. 5986 3’s Then 3n = 4 00 . . 0 5. 5986 0’s We have S(n) = 3·5986+1+5 = 17964 = 1996·9 = 1996S(n), as desired. 56. distinct digits. Determine whether there is any perfect square that ends in 10 Solution: The answer is yes. We note that 1111 ×1111 1111 1 111 1 1 11 11 1 1 12 3 4321 Likewise, it is not difﬁcult to see that 111111111112 = 12345678900987654321 is a number that satisﬁes the conditions of the problem.
Because of the symmetry, in order to prove (c) it sufﬁces to prove that S(n 1 n 2 ) ≤ n 1 S(n 2 ). The last inequality follows by applying the subadditivity property (b) repeatedly. Indeed, S(2n 2 ) = S(n 2 + n 2 ) ≤ S(n 2 ) + S(n 2 ) = 2S(n 2 ), and after n 1 steps we obtain S(n 1 n 2 ) = S(n 2 + n 2 + · · · + n 2 ) n 1 times ≤ S(n 2 ) + S(n 2 ) + · · · + S(n 2 ) = n 1 S(n 2 ). n 1 times To establish (d), we observe that by (b) and (c), h S(n 1 n 2 ) = S n 1 h bi 10i =S i=0 i=0 h ≤ n 1 bi 10i h S(n 1 bi 10i ) = i=0 h S(n 1 bi ) ≤ i=0 bi S(n 1 ) i=0 h = S(n 1 ) bi = S(n 1 )S(n 2 ).
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. For each n = 1, 2, 3, . . , 15, there are 100 − 6n suitable values of i (and j), so the number of solutions is 94 + 88 + 82 + · · · + 4 = 784. Euler’s Totient Function We discuss some useful properties of Euler’s totient function ϕ. 31. Let p be a prime, and let a be a positive integer.