The problem appears to involve identifying the number that can replace the blank square so that the entire number (read together as "321") becomes divisible by 3.

Divisibility Rule for 3:

A number is divisible by 3 if the sum of its digits is divisible by 3.

1. The digits given are $3$, $2$, and the missing digit $x$.
2. The sum of the digits is $3 + 2 + x = 5 + x$.
3. For $5 + x$ to be divisible by 3, it must satisfy $5 + x \mod 3 = 0$.

Now let's solve for $x$: $5 + x \mod 3 = 0 \quad \Rightarrow \quad x \mod 3 = 1$

Possible Values for $x$:

The smallest digit satisfying $x \mod 3 = 1$ is $1$. Therefore, the missing digit should be $1$ for the number to be divisible by 3.

Final Answer:

The missing digit in the blank square is 1.

Would you like further explanation or clarification on this?

Here are 5 related questions for practice:

1. What is the divisibility rule for 9?
2. How would the solution change if the number needed to be divisible by 6?
3. If the sum of the digits of a number is 18, is it divisible by 3? Why?
4. Can you determine a missing digit if a number must be divisible by 9 instead of 3?
5. What are all the possible values of the missing digit for divisibility by 3 if the sum starts at 5?

Tip: Always verify divisibility rules by applying them to simple examples!