Number Grid Puzzle

Because both 2 across and 1 down are perfect squares, by looking at the last digit of the 2-digit perfect squares, we know that these numbers are either $\{ 81, 16 \} , \{ 16, 64 \}, \{ 36, 64 \}$ .

Case 1: 1 down is 81, 2 across is 16. Let the unknown cell (top right corner) be $a$ .
Because 2 down is "twice of 1 across", we get that $2 \times (80 + a) = 10a + 6$ . This gives us $a = \frac{77}{4}$ which isn't a valid digit.

Case 2: 1 down is 16, 2 across is 64. Let the unknown cell (top right corner) be $a$ .
Because 2 down is "twice of 1 across", we get that $2 \times (10 + a) = 10a + 4$ . This gives us $a = 2$ , but $12$ is not a prime number.

Case 3: 1 down is 36, 2 across is 64. Let the unknown cell (top right corner) be $a$ .
Because 2 down is "twice of 1 across", we get that $2 \times (30 + a) = 10a + 4$ . This gives us $a = 7$ , and $74$ is not a prime number.

Hence, only case 3 is valid. Thus, the sum of digits is $3 + 7 + 6 + 4 = 20$ .

This problem can still be solved if we replace 1 across by "odd" instead of "prime"; and if we know nothing about 1 down.

Let the digits be $\begin{array}{cc} a & b \\ c & d \end{array}$ Then

• Since $\overline{ab}$ is odd, $b$ is odd.

• $\overline{bd} = 2\overline{ab}$ . With $b$ odd, this implies $b \geq 5$ , $b = 2a+1$ , and therefore $2 \leq a < 5$ .

• Since $d = 2(b-5$ ) and $b$ is odd, $d$ is a multiple of four.

• With $\overline{cd}$ a square, and $d$ a multiple of four, we must have $d = 4$ and $c = 6$ .

• Because $d = 2(b-5) = 4$ , we have $b = 7$ ; and from $b = 2a+1 = 7$ , it follows $a = 3$ .

The answer is $a + b + c + d = 3 + 7 + 6 + 4 = \boxed{20}$ .

There are only six 2-digit perfect squares, $16, 25, 36, 49, 64, 81$ . A quick glance at them shows that the only [reasonable--see reply below] one here which 2nd digit can be the 1st digit of another is $36$ . Once we have that, the rest follows almost immediately.