We need to find all natural numbers $n$ such that $\sqrt{\frac{72}{n}}$ is a natural number.

1. Problem Description

We need to find all natural numbers $n$ such that $\sqrt{\frac{72}{n}}$ is a natural number.

2. Solution Steps

First, we find the prime factorization of

7

2. $72 = 2 \times 36 = 2 \times 2 \times 18 = 2 \times 2 \times 2 \times 9 = 2^3 \times 3^2$.

So, $\sqrt{\frac{72}{n}} = \sqrt{\frac{2^3 \times 3^2}{n}}$.

For $\sqrt{\frac{72}{n}}$ to be a natural number, $\frac{72}{n}$ must be a perfect square. Therefore, $\frac{72}{n} = k^2$ for some natural number $k$.

Then $n = \frac{72}{k^2}$. Since $n$ must be a natural number, $k^2$ must be a divisor of

7

2. The possible values of $k^2$ are perfect square divisors of

7

2.

The divisors of 72 are 1, 2, 3, 4, 6, 8, 9, 12, 18, 24, 36,

7

2. The perfect square divisors of 72 are 1, 4, 9,

3

6.

If $k^2 = 1$, then $n = \frac{72}{1} = 72$.

If $k^2 = 4$, then $n = \frac{72}{4} = 18$.

If $k^2 = 9$, then $n = \frac{72}{9} = 8$.

If $k^2 = 36$, then $n = \frac{72}{36} = 2$.

Now let us check our answers:

If $n=72$, then $\sqrt{\frac{72}{72}} = \sqrt{1} = 1$.

If $n=18$, then $\sqrt{\frac{72}{18}} = \sqrt{4} = 2$.

If $n=8$, then $\sqrt{\frac{72}{8}} = \sqrt{9} = 3$.

If $n=2$, then $\sqrt{\frac{72}{2}} = \sqrt{36} = 6$.

All the values of $n$ result in natural numbers.

3. Final Answer

The natural numbers $n$ are 2, 8, 18, and

7

2.