### Instruction: A number $x$ is equal to $7\cdot24\cdot48$. What is the smallest positive integer $y$ such that the product $xy$ is a perfect cube?### Response: I need to find a way to factor $x$ into prime factors, so that I can see what powers of each prime are in $x$.
I know that $7$ is already a prime factor, so I can leave that alone.
For $24$, I can use the divisibility rule that says if a number is divisible by both $2$ and $3$, then it is divisible by $6$.
So I can divide $24$ by $6$ to get $4$, and then divide $4$ by $2$ to get $2$.
This means $24$ is equal to $2\cdot2\cdot2\cdot3$, or $2^3\cdot3$.
For $48$, I can use the same rule, and divide it by $6$ to get $8$, and then divide $8$ by $2$ to get $4$, and then divide $4$ by $2$ to get $2$.
This means $48$ is equal to $2\cdot2\cdot2\cdot2\cdot3$, or $2^4\cdot3$.
Now I have $x=7\cdot24\cdot48=7\cdot2^3\cdot3\cdot2^4\cdot3=7\cdot2^7\cdot3^2$.
This means $x$ has one factor of $7$, seven factors of $2$, and two factors of $3$.
To make $xy$ a perfect cube, I need to have each prime factor raised to a power that is a multiple of $3$.
For $7$, I already have one factor, so I need two more to make it a perfect cube. This means $y$ must have $7^2$ as a factor.
For $2$, I already have seven factors, so I need two more to make it a perfect cube. This means $y$ must have $2^2$ as a factor.
For $3$, I already have two factors, so I need one more to make it a perfect cube. This means $y$ must have $3$ as a factor.
Therefore, the smallest possible $y$ that satisfies the condition is $7^2\cdot2^2\cdot3=588$.