1. Split them in to 2 piles of N (pile 1) and 52-N (pile 2) cards.

2. Let us say that in pile 1 there are M up cards, then

pile 1 will have N-M down cards;

pile 2 has N-M up cards (because pile 1 has M up cards and there are only total N up cards)

3. Flip the cards in pile 1. pile 1 has N-M up cards now and so does pile 2.

0|0|0|0|1|1|1|1

0|1|1|0|0|1|1|0

1|0|1|0|0|1|0|1

0|0|1|1|1|0|0|1

1|1|0|1|1|0|0|0

1|1|0|1|0|1|0|0

0|0|1|1|0|0|1|1

1|1|0|0|1|0|1|0

Cube 2: 0, 1, 2, 6 or 9, 7, 8

I found this problem in “The Colossal Book of Short Puzzles and Problems” by Martin Gardner (1.3)

There is a US patent issued for the calendar cube.

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