Pure maths, physics, chemistry, etc.: brain-training tasks that have nothing to do with trade [Part 2] - page 8

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How easy it is when you know) It seems we've all abstracted too much from the immediate meanings of the cards.
 
First we lay the, er, "cyclically smaller" card
 
TheXpert:
First we put the, uh, "cyclically smaller" card

No, explain with an example, I don't get it at all ((
 
Mischek2:
No, explain with an example, I don't get it at all ((
For 9 and 10 we put 9. For K and 2 we put K. For K and 6 we put K. For K and 7 we put 7.
 
Mischek2:

No, explain with an example, I don't get it at all ((

Out of five, find two cards of the same suit, set one aside, and place the other first. This leaves us with three cards. Three cards can encode a number from 1 to 12. All the cards are pre-numbered (learned). We have 1, 2, 3. From these 1, 2, 3 it is possible to get 12 permutations.
 
Integer:

Out of five, find two cards of the same suit, put one away and the other first. This leaves three cards. Three cards can encode a number from 1 to 12. All the cards are pre-numbered (learned). We have 1, 2, 3. From these 1, 2, 3 it is possible to get 12 permutations.

Six. Set a cyclical displacement relative to the first card of the same suit laid out
 
Integer:

Out of five, find two cards of the same suit, put one away, the second card goes first. That leaves three cards. The three cards can encode a number from 1 to 12. All the cards are pre-numbered (learned). We have 1, 2, 3. From these 1, 2, 3 we can get 12 permutations.


got it ((

Leha is probably already looking for a new execution for his brain )))

 
And there are 13 cards of one suit.
 
Integer:
And there are 13 cards of one suit.

It's nothing, one is open, there are 12 left.
 
Not 12, but 6 in total.
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