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But this isn't about the undead.

Imagine that we are talking about the undead and the corpses of their children.

Let's say one of them is from kindergarten, meaning they are under 5 years old (okay, let's say up to 6). At what age do they start taking children to kindergarten, considering the nursery group (let's say from age 3)? There is an older and a younger child (so, they are not twins). Let's round the age difference down to a minimum of one year. The concept of a "long meeting" is too vague to use it to exclude certain options. Let's assume that counting the number of pigeons is problematic if there are more than twenty, so the product should be 20 or less. I'm left with 3 and 4, 3 and 5, 4 and 5 years old – I can't narrow it down further. Perhaps I should choose 3 and 4 because counting 12+ pigeons is also difficult at a glance. Although... it turns out that the desired product could have been obtained in at least two ways, one of which is x^2. Then, the possible products are: 4, 9, 16, 25, the rest are clearly outside the range. Of these, 4 and 16 theoretically could be obtained without using x^2, but for 16, it would have to be 2 and 8, and 8 years is too much. 9*1, 16*1, and 25*1 obviously don't work. This leaves the product 4, and the ages 1 and 4 years old. (yes, I made a mistake with the nursery group).

And we have a winner!

I wanted to write about the fact that, in theory, you can have non-twins with a small age difference (less than a year). And in general, the fact that the wife was alone all the time is not mentioned in the problem.

The magician took a deck of cards (52 cards), all facing down, counted out 20 cards, turned them face up, and handed them to the spectator. The spectator mixed the face-up cards with the entire deck and thoroughly shuffled the deck so that the face-up cards were randomly distributed within it. Then, holding the deck under the table so that the cards were not visible, he counted out 20 cards from the top and handed them to the magician.
The magician, taking the stack and still holding it under the table, says: "Now I will try to equalize the number of face-up cards in my part of the deck and in yours by touch. To do this, I will need to turn over a few more cards." Then, after some manipulations, still under the table, he takes out his cards, lays them out on the table, and counts the face-up cards. There are exactly as many as there are among the 32 cards of the spectator.

What is the secret of this trick?

I wanted to write that, in theory, it's possible to have non-twins with a small age difference (less than a year). And in general, the fact that the wife was alone all the time is not mentioned in the problem.

The magician took a deck of cards (52 cards), face down, counted out 20 cards, turned them face up, and handed them to the spectator. The spectator mixed the face-up cards with the entire deck and thoroughly shuffled the deck so that the face-up cards were randomly distributed within it. Then, holding the deck under the table so that the cards were not visible, he counted out 20 cards from the top and handed them to the magician.
The magician, taking the stack and still holding it under the table, says: "Now I will try to equalize the number of face-up cards in my part of the deck and in yours by touch. To do this, I need to turn over a few more cards." Then, after some manipulations, still under the table, he takes out his cards, lays them out on the table, and counts the face-up cards. There are exactly as many as there are among the 32 cards of the spectator.

What is the secret of this trick?

The number of face-up cards with the spectator can be determined by subtracting the number of face-up cards in his (the magician's) deck from 20. Either he could really determine the face-up cards by touch, or there was someone under the table who took the deck and left a number of cards face up equal to 20 minus the number of face-up cards.

Added 4 minutes later
Let's say there are 5 face-up cards left in the spectator's deck, and the magician or the person under the table counted 15 face-up cards in his deck. This means that 10 face-up cards need to be turned back face down.
If, for example, there are 15 face-up cards left with the spectator, then the magician or assistant will count 5 in his deck, and another 10 face-down cards will need to be turned face up.

No, there are no special tricks like having a friend under the table, the audience is not in cahoots with the magician, and there are no extrasensory abilities like being able to identify face-down cards by touch.

I think I understand! He flipped all the cards.

Added 2 minutes later
The number of flipped cards the magician has will be 20 minus the number of flipped cards the spectator has. And since the spectator gives him 20 cards, after he flips all of them, the number of flipped cards he has will be:

20-(20-Q_{spectator's cards})=Q_{spectator's cards}

Yeah, Hermit, you’re right.

Hermit, let's brainstorm some undead-themed ideas. You'll get a badge.

I just remembered a riddle that Dirty Player posed, and Forest solved. The riddle was: what is the minimum number of peasants needed to kill a sapphire dragon? As far as I remember, it took 15 peasants with a level 99 Tazar. That made me think of posing a similar riddle: how many skeletons would it take to kill a bone dragon? But for that, I need to have a good understanding of the physics engine. :D So, I don't have anything to pose right now. Let someone else come up with a riddle. Sorry.

In Heroes of Might and Magic V, it takes 77 skeletons to kill a Bone Dragon. I calculated that a long time ago.

I can do it alone.

Okay, I'm not entirely sure about one, but I guarantee the other two. And the dragon doesn't necessarily have to be skeletal; any dragon will do.

A 1st-level hero with no magic? In the middle of nowhere?