Problem
NT-B1-M02-P023 Cubes with Required Divisibility
#23
★★★★☆ Level 4 of 5
Three positive perfect cubes are divisible by \(18\). What is the smallest possible value of their common GCD?
Method inspiration: local number theory source
If a cube is divisible by \(2\cdot3^2\), what are the smallest possible exponents of \(2\) and \(3\) in it?
In a perfect cube, all prime exponents are multiples of \(3\). For a cube to be divisible by \(18=2\cdot3^2\), the exponent of \(2\) must be at least \(3\), and the exponent of \(3\) must also be at least \(3\). Hence every such cube is divisible by \(2^3\cdot3^3=216\). This value is attainable: take all three cubes equal to \(216=6^3\). Therefore the smallest possible GCD is \(216\).
Inspired by a local source, but the statement and solution are rewritten.