Thread: Algebra2 homework urgent please help.

what is the least positive integer k for which 12 k is the cube of an integer?

Originally Posted by Schatzle06

what is the least positive integer k for which 12k 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!

What is the least positive integer k for which 12k is the cube of an integer?

Sorry, JHevon . . . your reasoning was correct; your answer wasn't.


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

Originally Posted by Soroban

Hello, Schatzle06!

Sorry, JHevon . . . your reasoning was correct; your answer wasn't.


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

Thanks for looking out Soroban. I saw that my answer was wrong a little before your post and corrected it. Apparently I was thinking of something else at the time not sure what