If `12y=x^(3)` and ` x` and `y ` are positive integers, what is the least possible value for y?

Options

A. `6`
B. `18`
C. `144`
D. `216`

Answer

To find the least possible value of \( y \) given the equation \( 12y = x^3 \) where \( x \) and \( y \) are positive integers, we can follow these steps:

### Step 1: Rearranging the Equation
We start with the equation:
\[ 12y = x^3 \]
From this, we can express \( y \) in terms of \( x \):
\[ y = \frac{x^3}{12} \]

### Step 2: Understanding Divisibility
For \( y \) to be an integer, \( x^3 \) must be divisible by \( 12 \). Therefore, we need to find the smallest integer \( x \) such that \( x^3 \) is divisible by \( 12 \).

### Step 3: Prime Factorization of 12
Next, we perform the prime factorization of \( 12 \):
\[ 12 = 2^2 \times 3^1 \]
This means that \( x^3 \) must contain at least \( 2^2 \) and \( 3^1 \) in its prime factorization.

### Step 4: Finding Minimum \( x \)
To ensure \( x^3 \) is divisible by \( 12 \), we need to determine the smallest \( x \) such that \( x^3 \) has at least \( 2^2 \) and \( 3^1 \).

Let’s denote \( x \) in terms of its prime factors:
\[ x = 2^a \times 3^b \times k \]
where \( k \) is a product of other primes.

Then:
\[ x^3 = 2^{3a} \times 3^{3b} \times k^3 \]
To satisfy the conditions:
- \( 3a \geq 2 \) (for \( 2^2 \))
- \( 3b \geq 1 \) (for \( 3^1 \))

From \( 3a \geq 2 \), the smallest integer \( a \) can be is \( 1 \) (since \( 3 \times 1 = 3 \geq 2 \)).
From \( 3b \geq 1 \), the smallest integer \( b \) can be is \( 1 \) (since \( 3 \times 1 = 3 \geq 1 \)).

### Step 5: Calculating \( x \)
Thus, the smallest \( x \) is:
\[ x = 2^1 \times 3^1 = 2 \times 3 = 6 \]

### Step 6: Finding \( y \)
Now we substitute \( x \) back into the equation for \( y \):
\[ y = \frac{x^3}{12} = \frac{6^3}{12} = \frac{216}{12} = 18 \]

### Conclusion
The least possible value for \( y \) is:
\[ \boxed{18} \]