## Each of the six faces on a cube has a different digit ($0$ to $9$) written on it; the same is done to a second cube. By placing the two cubes side-by-side in different positions we can form a variety of $2$-digit numbers.

For example, the square number $64$ could be formed:

In fact, by carefully choosing the digits on both cubes it is possible to display all of the square numbers below one-hundred: $01$, $04$, $09$, $16$, $25$, $36$, $49$, $64$, and $81$.

For example, one way this can be achieved is by placing $\{0, 5, 6, 7, 8, 9\}$ on one cube and $\{1, 2, 3, 4, 8, 9\}$ on the other cube.

However, for this problem we shall allow the $6$ or $9$ to be turned upside-down so that an arrangement like $\{0, 5, 6, 7, 8, 9\}$ and $\{1, 2, 3, 4, 6, 7\}$ allows for all nine square numbers to be displayed; otherwise it would be impossible to obtain $09$.

In determining a distinct arrangement we are interested in the digits on each cube, not the order.

$\{1, 2, 3, 4, 5, 6\}$ is equivalent to $\{3, 6, 4, 1, 2, 5\}$

$\{1, 2, 3, 4, 5, 6\}$ is distinct from $\{1, 2, 3, 4, 5, 9\}$

But because we are allowing $6$ and $9$ to be reversed, the two distinct sets in the last example both represent the extended set $\{1, 2, 3, 4, 5, 6, 9\}$ for the purpose of forming $2$-digit numbers.

How many distinct arrangements of the two cubes allow for all of the square numbers to be displayed?

### To form all the nine square numbers less than 100, you will need the digits 0, 1, 2, 3, 4, and then either 5 or 6, 8, and 9.

Let’s divide the problem into 2 cases:

Case 1: Both the cubes have the number 6 or 9:

From each cube, we choose 5 numbers from {0, 1, 2, 3, 4, 5, 8}. Two elements of 6 and 9 are already on both cubes so we don’t need to worry about them. There are 7 choices of 5 numbers or (7 choose 5 = 21) and we have two cubes, this gives a total of 21*21 = 441 arrangements. But, we have double-counted the cases that are similar, so we need to divide this number by 2 to correct the over counting, leading 441/2 = 220.5.

Case 2: Only one cube has the number 6 or 9:

From one cube, we choose 5 numbers from {0, 1, 2, 3, 4, 5, 8} and from the other one, we choose 5 numbers from {0, 1, 2, 3, 4, 8}. This gives a total of (7 choose 5)*(5 choose 5) = 21 arrangements.

The total number of distinct arrangements is the sum of the arrangements from the two cases:

220.5 (because we can’t have half an arrangement, it’s rounded up to 221) + 21 = 242 distinct arrangements.

It is important to note that there is a bit of number theory thinking involved in solving this problem and it is crucial to break it down into manageable cases and subtly use combinatorics to reach the solution.

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