To find a value to fill in the blank such that the product of \( 5 \sqrt{27} \) and that value yields a rational number, we first need to simplify \( 5 \sqrt{27} \).
We know:
\[ \sqrt{27} = \sqrt{9 \cdot 3} = \sqrt{9} \cdot \sqrt{3} = 3\sqrt{3} \]
So, we have:
\[ 5 \sqrt{27} = 5 \cdot 3 \sqrt{3} = 15 \sqrt{3} \]
Now, we want to find a number \( x \) such that:
\[ 15 \sqrt{3} \cdot x \text{ is a rational number.} \]
For this product to be rational, \( x \) must be a value that cancels out the \( \sqrt{3} \). Specifically, we can let:
\[ x = \frac{1}{\sqrt{3}} \quad \text{(which is the multiplicative inverse of } \sqrt{3} \text{)} \]
Now we can check:
\[ 15 \sqrt{3} \cdot \frac{1}{\sqrt{3}} = 15 \cdot \frac{\sqrt{3}}{\sqrt{3}} = 15 \cdot 1 = 15 \]
Thus, \( 15 \) is indeed a rational number.
Therefore, the blank can be filled with:
\[ \frac{1}{\sqrt{3}} \]
Alternatively, you could also multiply by \( \sqrt{3} \) to result in:
\[ 15 \sqrt{3} \cdot \sqrt{3} = 15 \cdot 3 = 45 \]
So, another option is:
\[ \sqrt{3} \]
To summarize, you can fill in the blank with either:
\[ \frac{1}{\sqrt{3}} \quad \text{or} \quad \sqrt{3} \]