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By the way, I have a sequel.
Butera's solution divides the deck into two unequal parts. It's true that the condition is silent about the equality of the parts, so credit is deserved. // Hip hip etc. :)
However, I will not be silent.
How do you divide a deck into two EQUAL parts, each with the same number of upside-down cards?
The initial conditions are the same.
// there is a solution, and a pretty one at that
You're not confused, are you? That's not supposed to be the solution.
Bummer. You're right, it looks like my 'decision' turned out to be wrong. :(
I'll double-check.
Yeah, that's right, it's a hole-in-the-wall.
You got the cards dealt?
Bummer. You're right, it looks like my 'decision' turned out to be wrong. :(
I'll double-check it.
Your first action was to divide the deck into two of 26 ?
Nah. The idea was different. 1) Count 10 and turn over (like Buter's) 2) count from the first and second pile by half (21 and 5) and turn it over and turn it crosswise ;-)
Vapchetso the idea works under different starting conditions - given two decks, inverted X cards in each, decks shuffled. Further on in the text.
Then the solution is to count X cards from each deck, turn them over and put them on the opposite deck.
:)
Here's more. If a small deck (upside down) is returned to a large deck, it would seem that nothing can be said about the number of cards face down. However, this is not the case.
:)
Here's more. If a small deck (upside down) is returned to a large deck, it would seem that nothing can be said about the number of cards face down. However, this is not the case.
There will be 10 to 20 upside-down cards. and there will also be multiples of them.
0 to 40 ?