Information lacking?

We can write the condition as follows:

$100 y + 70 + x = (10 a + b)^2 =100a^2 + 20 a b + b^2$ .

All possible values for $20ab$ will have even tens digit. We will need $b^2$ to have odd tens digit and that is possible only for values of $b=4, b=6$ giving $b^2 = 16, b^2 = 36$ . In both cases we get unit digit $\boxed{6}$ .

Note: That holds true for any odd tens digit. Any perfect square having odd tens digit has to have unit digit equal to 6.

Let $a \in \{0,1,..,9 \}$ . We want to solve the congruence $x^2 \equiv 70 + a \mod 100$ . This would typically entail finding both $x,a$ but there is an indirect way of obtaining $a$ . Remember that $x^2 \equiv 70 + a \mod 100 \implies x^2 \equiv 70 + a \mod 10 \implies x^2 \equiv a \mod 10$ . This means that $a$ is a quadratic residue $\mod 10$ . It is easy to list these, so we now know that $a \in \{0,1,4,9,6,5\}$ .

But also, $x^2 \equiv 70 + a \mod 100 \implies x^2 \equiv 70 + a \mod 4 \implies x^2 \equiv 2 + a \mod 4$ . Because the only quadratic residues $\mod 4$ are $0,1$ this implies that $a \equiv 2,3 \mod 4$ . So $a \in \{2,6,3,7\}$ .

Combining both statements, $a \in \{2,6,3,7\} \cap \{0,1,4,9,6,5\} = \{6\}$ . Thus, $a = 6$ .

By using trial and error method, you get ${24}^{2} = 576$ , So tens' digit is 7 and so unit digit is "6".

Other shortest way is using python.

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for x in range(8,300):
    if (str(x**2)[-2] == "7"):
        print x
        break

# Prints 24 :D