If $n$ is a positive integer such that $n^2$ is divisible by 72, what is the smallest possible value of $n$?

Step 1 (Prime Factorization of 72): We find the prime factorization of 72:

Step 2 (Divisibility of n^2): We are given that $n^2$ is divisible by $72$. Let $n$ have the prime factorization $2^a{3}^b\dots$, so $n^2=2^{2a}{3}^{2b}\dots$
For $n^2$ to be divisible by $2^3\times 3^2$, the exponents of the primes in the factorization of $n^2$ must satisfy:

• $2a\ge 3⟹a\ge 1.5⟹a\ge 2$ (since $a$ must be an integer).

• $2b\ge 2⟹b\ge 1$

Step 3 (Minimize n): To find the smallest positive integer $n$, we choose the minimum values for $a$ and $b$:

Step 4 (Verify): If $n=12$, then $n^2=144$. Since $144=72\times 2$, it is divisible by 72. If $n<12$, no smaller positive integer squared is divisible by 72.

Step 5 (Conclusion): The smallest possible value of $n$ is exactly 12.