MathDB

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Part of 1989 Bundeswettbewerb Mathematik

Problems(2)

Bundeswettbewerb Mathematik 1989 Problem 1.4

Source: Bundeswettbewerb Mathematik 1989 Round 1

9/23/2022
Let nn be an odd positive integer. Show that the equation 4n=1x+1y \frac{4}{n} =\frac{1}{x} + \frac{1}{y} has a solution in the positive integers if and only if nn has a divisor of the form 4k+34k+3.
egyptian fractionsnumber theorydivisor
Every integer on a circle divides its neighbor sum

Source: Taiwan 2014 TST3 Quiz 3, P1

7/18/2014
Positive integers x1,x2,…,xnx_1, x_2, \dots, x_n (n≥4n \ge 4) are arranged in a circle such that each xix_i divides the sum of the neighbors; that is xi−1+xi+1xi=ki \frac{x_{i-1}+x_{i+1}}{x_i} = k_i is an integer for each ii, where x0=xnx_0 = x_n, xn+1=x1x_{n+1} = x_1. Prove that 2n≤k1+k2+⋯+kn<3n. 2n \le k_1 + k_2 + \dots + k_n < 3n.
inductionstrong inductioncombinatorics proposedcombinatoricsTaiwan TST2014