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An easy one for the middle of the week!

Jane and John were discussing a business idea. John wanted to make a little set to keep track of the date. It was to include two cubes with a single digit on each face that could be rearranged to show any day of the month (1-31).

"Look," he said, "there are 72 ways of arranging the cubes, so it must be possible to design it for just 31 days!"

"No," responded Jane, "there's less than 72. Presumably each cube has 6 different faces, so at least 2 are going to be overlapping. That's going to reduce it to... mmm"

[Question 1: how many different ways are there to put down the cubes to read out a two digit number in the condition that Jane just described?]

"Well, that should still be plenty." John replied.

"Hang on, there's another problem." Said Jane. "You have got 10 digits altogether. You will need 0, 1, and 2 on both cubes, since you need to make 11, 22, and 0n for all n between 1 and 9 inclusive. That means you need to pair 0 with at least 9 other numbers. So it's got to be on both. Amiright?"

"I guess so. The argument for 0 applies to 1 and 2 also. Either way, I agree. Both cubes are going to need 0, 1, and 2"

"That leaves 3 spaces on each cube for the other numbers, or a total of 6 spaces. But there are 7 numbers left: 3 to 9." Jane continued.

"I see what you mean. Maybe I should use dodecahedrons. I prefer them anyway." Said John thoughtfully.

"Wait!" Said Jane. "Here's one for sale on the internet. How do they do it?"

[Question 2: How do they do it?]

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    $\begingroup$ The awful thing is that my grandmother had them and I played with them endlessly as a kid, and STILL didn't realize the trick! $\endgroup$ – Francesco Dondi Jan 26 '17 at 10:59
  • $\begingroup$ It's fun, because the "proof" that it cannot be done is relatively easy! $\endgroup$ – Dr Xorile Jan 26 '17 at 19:45
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Question 1:

If you produce cube as Jane suggests, with minimal overlap:

Your cubes could be 012345 and 456789. If we ignore the trick used to resolve Question 2, we have 6*6 = 36 pairs, times 2 for the ordering, giving 72 in total as initially suggested. However, we now count 44, 45, 55, and 54 twice each. Removing those, we end up with 68 possible displayed numbers.

Question 2:

It turns out you do have enough spaces because, conveniently:

a 9 is simply an upside-down 6! Since they don't need to appear at the same time, that face can do double-duty.

One possible arrangement for the numbers is: 012345 on one cube, and 012678 on the other.

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First question:

$68$. Suppose you have one cube with $0$ to $5$ and one with $6$ to $9$ and say $1$ and $2$. You have $6\times 6\times 2 = 72$ total combinations but you need to subtract the repeated ones that are those with both 1s and 2s so $11$, $12$, $21$ and $22$.

Second question. You need two cubes with:

0 1 2 3 4 5
0 1 2 6 7 8

Because

You can use the 6 as a 9 by turning it upside down

My granddad had one of this type of calendars.

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Answer to Q2

6 doubles as a 9 since you never need to construct numbers like 66, 69, 96 or 99. So now you have 6 numbers for 6 remaining faces.

Answer to Q1. I hope I understood correctly.

There are 6 + 5 + 4.. +1 = 21 combinations. For 1 on one cube we get 6 combinations. for 2 we get only 5 because we already counted 2&1 as 1 & 2 and so on.

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