# Challenging mathematical problems with elementary solutions by Gordon, Basil; Âglom, Isaak Moiseevič; McCawley, James;

By Gordon, Basil; Âglom, Isaak Moiseevič; McCawley, James; Âglom, Akiva Moiseevič

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Additional resources for Challenging mathematical problems with elementary solutions Volume I : Combinatorial Analysis and Probability Theory

Example text

But three of these four points, A, D, C, are the vertices of an equilateral triangle of side a, and in fig. 39c the only equilateral triangle has side b > a. So A, D, C, E must be situated as in fig. 39a. Since E does not coincide with D, the only 56 SOLUTIONS possible configuration is that illustrated in fig. 41a. But in this configuration DE = 2a has length neither a nor h. So we have shown that it is impossible for four of our five points to be arranged as in fig. 39a. Similarly, we can show that the points A, B, C, D cannot be situated as in fig.

41 XIII. The theory ofprimes where 010 O 2• 0a••••• a .. and bI> b2• ba, •••• b.. are any two sequences of numbers. Denote the sums bl , b 1 + b2 • b1 + b2 + ba• bi + b2 + ... + b.. by Bh B2 , B a, •••• B .. , respectively. _ 1 - ... )Bn _ 1 + anB... h. Using Abel's formula, calculate the value of + 2q + 3q2 + ... + nq"-l. 1 + 4q + 9q2 + ... + n2qn-l. 1. I 2. *** Mertens' second theorem. a. Let 2,3,5,7,11, ... ,p be the primes not exceeding the integer N. Show that for all N > 1, the expression -1+ -1+ -1+ 1 - + -1+ ...

See above, under problem 161). 171. *** Mertens' first theorem. 16 Let 2, 3, 5, 7, II, ... ,p be the primes not exceeding a given integer N. Show that for an N, the quantity I log 2 + log 3 + log 5 + log 7 + log 11 + ... + log p _ log N 2 3 5 7 11 p I is bounded, in fact < 4. As N increases indefinitely. so does its logarithm, for log N is greater than any given number K as soon as N is greater than 10K • The sum log 2 + log 3 + log 5 + ... + log p 235 P where 2, 3, 5, 7, II, ... ,p are the primes;;:;; N, also tends to infinity with N.