Grades 11-12 Video Solutions 2011-11&12 Video Solutions 2011 problem28

11&12 Video Solutions 2011 problem28

Video Transcription

Video Summary

The problem involves finding positive integers \(a\), \(b\), and \(c\) such that \(a^2 = 2b^3 = 3c^5\), aiming to determine the smallest possible number of divisors of their product. By factoring \(a\), \(b\), and \(c\) with powers of 2 and 3, and analyzing the exponents, minimal solutions for \(k\), \(l\), and \(m\) are derived. Setting \(m1 = 2\), \(l1 = 3\), \(k1 = 5\) solves the equations. The total number of divisors is calculated as \((10 + 1)(6 + 1) = 77\). Thus, the smallest number of divisors is 77.

Keywords

positive integers

divisors

exponents

factorization

minimal solutions