Make perfect squares!

The answer is $0$ . Since any perfect square ends with only the digits $0, 1, 4, 5, 6, 9$ , the ending digit as stated in the problem must be one of $2, 3, 7$ or $8$ to be a perfect square because those are the only numbers we have. Since any perfect square number cannot end with the digits we have, we cannot make any perfect square hence the answer is $0$ .

Note: if you want to know why a perfect square can never end in $2, 3, 7$ or $8$ read below.

Let us observe if it is true or not.

From the table, we can see that perfect squares less than $100 = 10^2$ , do not include any number ending with digit $2, 3, 7$ or $8$ .

Next, any number perfect square more than 100 can be expressed in the form of

$(10x + y)^2, \forall x \in \N, \forall y\in {{0, 1, 2, 3, 4, 5, 6, 7, 8, 9}}$

$= 100x^2 + 20xy + y^2$

Since $100x^2$ and $20xy$ do not affect the unit or the ending digit, it all depends on y^2

As we stated the condition that $0 \leqslant y \leqslant 9$ , $y^2 \notin 2, 3, 7, 8$

And we are done!

Perfect squares can only end in one of the digits $0,1,4,5,6,9.$ (Look at the possible values of $x^2$ mod $10$ for $0 \le x \le 9.$ ) So the answer is $\fbox{0}.$