n and p are two positive integes. If it is known the that 3n is a pefect square and `12n^(2)` p is a prefect cube. What is the smallest possible value of np ?

Answer

To solve the problem step by step, we need to analyze the conditions given and find the smallest possible values of \( n \) and \( p \).

### Step 1: Analyze the condition \( 3n \) is a perfect square.
We know that \( 3n \) must be a perfect square. Since \( 3 \) is a prime number, for \( 3n \) to be a perfect square, \( n \) must contain \( 3 \) in such a way that the overall exponent of \( 3 \) in \( 3n \) becomes even.

Let’s express \( n \) in terms of its prime factorization:
\[ n = 3^a \cdot 2^b \cdot k \]
where \( k \) is not divisible by \( 2 \) or \( 3 \).

Then,
\[ 3n = 3^{a+1} \cdot 2^b \cdot k \]

For \( 3n \) to be a perfect square, both \( a + 1 \) and \( b \) must be even.

- \( a + 1 \) is even implies \( a \) must be odd. The smallest positive integer \( a \) can take is \( 1 \) (i.e., \( a = 1 \)).
- \( b \) must be even, so the smallest value for \( b \) is \( 0 \).

Thus, the smallest \( n \) can be:
\[ n = 3^1 \cdot 2^0 \cdot k = 3k \]

### Step 2: Determine the smallest value of \( k \).
Since \( k \) must be a positive integer and not divisible by \( 2 \) or \( 3 \), the smallest value for \( k \) is \( 1 \). Therefore:
\[ n = 3 \]

### Step 3: Analyze the condition \( 12n^2p \) is a perfect cube.
Now, we need to check the condition \( 12n^2p \) being a perfect cube. First, we find \( n^2 \):
\[ n^2 = 3^2 = 9 \]

Now, substituting \( n \) into \( 12n^2p \):
\[ 12n^2 = 12 \cdot 9 = 108 \]

Next, we factor \( 12 \):
\[ 12 = 3^1 \cdot 2^2 \]
Thus,
\[ 12n^2 = 3^1 \cdot 2^2 \cdot 3^2 = 3^{1+2} \cdot 2^2 = 3^3 \cdot 2^2 \]

So,
\[ 12n^2p = 3^3 \cdot 2^2 \cdot p \]

For \( 12n^2p \) to be a perfect cube, the exponents of all prime factors must be multiples of \( 3 \):
- The exponent of \( 3 \) is \( 3 \) (which is already a multiple of \( 3 \)).
- The exponent of \( 2 \) is \( 2 \). To make this a multiple of \( 3 \), \( p \) must contribute at least \( 1 \) more \( 2 \) (i.e., \( 2^1 \)).

Thus, we can set:
\[ p = 2^1 = 2 \]

### Step 4: Calculate \( np \).
Now we have:
\[ n = 3 \]
\[ p = 2 \]

Thus, the smallest possible value of \( np \) is:
\[ np = 3 \cdot 2 = 6 \]

### Final Answer:
The smallest possible value of \( np \) is \( 6 \).

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