Is it possible to place 2015 different positive integers around a circle so that for any two adjacent numbers, the smaller integer always divides the larger integer and the ratio of the larger to the smaller is a prime number?
Give reasons for your answer.
Going from one number (position) to the next, one factor in the prime factorization is either added to or taken from that prime factorization. So the numbers alternate having an odd or an even number of prime factors (including duplicates so that 2 * 3^2, for example, has an odd number of prime factors).
This is possible only when the number of positions around the circle is even. Therefore such a circle with 2015 numbers (positions) is impossible.
Posted by Charlie
on 2015-09-11 11:04:22