what is the least positive integer k for which 12 k is the cube of an integer?
We want 12k = m^3 for integers m and k
=> k = (m^3)/12
so we want the smallest m for which (m^3)/12 gives an integer. The smallest m for which this happens is 0, but that would mean k = 0, we want k positive. The next smallest m for which this happens is 6. So k = (6^3)/12 = 18
Hello, Schatzle06!
Sorry, JHevon . . . your reasoning was correct; your answer wasn't.What is the least positive integer k for which 12k is the cube of an integer?
If 12k = m³, then the left side must be cube.
The left side is: .2²·3·k
We want the exponents to be a multiple of 3 (smallest).
. . This happens when k = 2·3²
Then: .12k .= .(2²·3)(2·3²) .= .2³·3³ .= .6³
Therefore: .k = 18