Complete the six cells of the given table, placing a digit in each one, so that the two three-digit numbers formed horizontally and the three two-digit numbers formed vertically are perfect squares.

(a) What are the numbers?

(b) How many solutions exist?

![](https://cdn.mathpix.com/cropped/2024_05_01_666a1aec7e447b0a6cbeg-024.jpg?height=132&width=272&top_left_y=505&top_left_x=1189)
To solve this problem, it is convenient to list the two-digit perfect squares, which are

$$
4^{2}=16,5^{2}=25,6^{2}=36,7^{2}=49,8^{2}=64 \text { and } 9^{2}=81
$$

as well as the three-digit perfect squares, which are

$$
\begin{gathered}
10^{2}=100,11^{2}=121,12^{2}=144,13^{2}=169,14^{2}=196,15^{2}=225 \\
16^{2}=256,17^{2}=289,18^{2}=324,19^{2}=361,20^{2}=400,21^{2}=441 \\
22^{2}=484,23^{2}=529,24^{2}=576,25^{2}=625,26^{2}=676,27^{2}=729 \\
28^{2}=784,29^{2}=841,30^{2}=900 \text { and } 31^{2}=961
\end{gathered}
$$

In particular, we see that every three-digit perfect square is a number ending in $0,1,4,5,6$ or 9.

Thus, we establish that, of the two-digit perfect squares, 25, 36, and 81 cannot appear in the third column, marked with $X$. For this column, only the perfect squares 16, 49, and 64 remain,

![](https://cdn.mathpix.com/cropped/2024_05_01_666a1aec7e447b0a6cbeg-073.jpg?height=126&width=222&top_left_y=1793&top_left_x=1611)
therefore, we have three options, as follows.
(I)

![](https://cdn.mathpix.com/cropped/2024_05_01_666a1aec7e447b0a6cbeg-073.jpg?height=128&width=288&top_left_y=2015&top_left_x=473)
(II)

![](https://cdn.mathpix.com/cropped/2024_05_01_666a1aec7e447b0a6cbeg-073.jpg?height=131&width=303&top_left_y=2013&top_left_x=928)
(III)

![](https://cdn.mathpix.com/cropped/2024_05_01_666a1aec7e447b0a6cbeg-073.jpg?height=123&width=294&top_left_y=2017&top_left_x=1418)

(a) Let's examine each of the three options.

Option (I): The three-digit perfect squares ending in 6 are 196, 256, 576, and 676. Since no two-digit perfect square ends in 2 or 7, the perfect squares 256, 576, and 676 cannot appear in the second row, leaving only 196.

![](https://cdn.mathpix.com/cropped/2024_05_01_666a1aec7e447b0a6cbeg-073.jpg?height=115&width=203&top_left_y=2284&top_left_x=1623)
Now, the only two-digit perfect squares ending in 1 and 9 are, respectively, 81 and 49. We obtain the following solution, which is the only one within Option (I).

| 8 | 4 | 1 |
| :--- | :--- | :--- |
| 1 | 9 | 6 |

Option (II): The three-digit perfect squares ending in 9 are 169, 289, 529, and 729, so the second row can only be filled with the perfect square 169. In the first column, only the number 81 can appear, as it is the only two-digit perfect square ending in 1.

![](https://cdn.mathpix.com/cropped/2024_05_01_666a1aec7e447b0a6cbeg-074.jpg?height=115&width=212&top_left_y=565&top_left_x=845)

| 8 |  | 4 |
| :--- | :--- | :--- |
| 1 | 6 | 9 |

Now, we have two options to fill the last blank cell: 1 or 3. However, neither 814 nor 834 are perfect squares. Thus, Option (II) is impossible.

Option (III): The three-digit perfect squares ending in 4 are 144, 324, 484, and 784, so the second row can only be filled with the perfect square 144, and in the first column, only the number 81 can appear. Now, the only choice for the blank cell is the number 6.

| 8 |  | 6 |
| :--- | :--- | :--- |
| 1 | 4 | 4 |


| 8 | 6 | 6 |
| :--- | :--- | :--- |
| 1 | 4 | 4 |

However, 866 is not a perfect square. Therefore, Option (III) is also impossible.

(b) From what we have seen above, there is only one solution, found in the previous item, namely, | 8 | 4 | 1 |
| :--- | :--- | :--- |
| 1 | 9 | 6 |.