Consider the grandfather with the largest number of grandchildren, and suppose, for a contradiction, that that number is less than 14. There’s a child at the school who isn’t his grandchild, and each of his grandchildren shares a grandfather with that child. They can’t all share the same grandfather, because then that man would surpass our “maximal” grandfather. So there’s some third grandfather to divide that duty with him.

Now, every student we haven’t spoken about shares a grandfather with one that we have. And that leaves room for no fourth grandfather; every student in the school must count some two of these men as his grandfathers.

We have 40 grandfatherings to assign to these 20 students, and 40 = 3 × 13 + 1, so try as we might, we can’t assign 13 or fewer to each of the three men. One of them must have at least 14 grandchildren.

(This is all laid out more rigorously at the link below, page 371.)

(Olympiad Corner, Crux Mathematicorum 27:6 [October 2001], 357-373.)

Yes. The fact that Candidate 1 wrote no means that she can’t have seen more than four red sheets — if she’d seen five then she would have written yes, because a month cannot have more than five Sundays, and this fact would have told her that her own sheet was black.

Similarly, the fact that Candidate 2 wrote no means that she can’t have seen more than three red sheets, because if she’d seen four red sheets (sheets 3, 4, 5, and 6) then she could infer that sheet 2 must be black, as Candidate 1 could not have seen five red sheets, per the reasoning above.

And so on: Since Candidates 3, 4, and 5 all wrote no, they can have seen at most 2, 1, and 0 red sheets. And if Candidate 6 knows that Candidate 5 saw 0 red sheets, she can announce that her own sheet was black.

(From Roland Sprague, Recreation in Mathematics, 1963.)

Well, whose turn is it? If White has the move he can mate immediately with the pawn. But if it’s White’s turn, then Black must just have moved, and there’s no legal move he can just have made. So it must be Black’s turn to move, and after 1. … axb6 the black pawn can scurry down the board to the queening square, just beyond the reach of the pursuing king. Black wins.

The Germans call this the Vielväter (“many fathers”) position, because the idea has inspired literally thousands of problems.

(From Fairy Chess Review.)

Now try the same challenge with a horse.

First, three observations: (1) The number of dollars that the men received for the herd of cattle was a perfect square. (2) The number of sheep they bought was odd, because it was necessary to throw in the dog in order to divide the animals evenly. (3) Since they bought an odd number of sheep at $10.00 per head, the next-to-last digit in the square payment from observation (1) must be odd.

Now, as it happens, in any perfect square, if the next-to-last digit is odd, the last figure must be a 6. The simplest way to see this is to consider that any number can be expressed as 10a + b, where a comprises all the digits except the last and b is the last digit (0 to 9). The square of this number is 100a² + 20ab + b². But 100a² has zeros in the last two places, and 20ab has a zero in the last place and is preceded by an even number. So, in order for the square to have an odd digit in the next-to-last position, b² must have an odd first digit, and the only possibilities are 4² = 16 and 6² = 36.

In either case, the dog cost $6.00, and the rancher owed his partner $2.00 for accepting a $6.00 dog in place of a $10.00 sheep.

The next number is 9. This is Golomb’s sequence, a non-decreasing sequence of integers in which a_n is the number of times that n occurs in the sequence. We’re given a₁ = 1 to start with, and this indicates that the number 1 appears only once in the whole sequence. That means that a₂ is greater than 1. Setting a₂ to 2 immediately tells us that 2 appears twice in the sequence, so a₃ also must be 2. That tells us that 3 occurs twice in the sequence, and so on. In this way each number in the sequence is uniquely determined.

Yes — here’s one possibility:

(From E.G. Kozlova, Skazki i podskazki: Zadachi dlya matematicheskogo kruzhka [Fairy Tales and Hints: Problems for a Mathematical Circle], 2004 [in Russian], via Jennifer Beineke and Jason Rosenhouse, The Mathematics of Various Entertaining Subjects, 2017.)

1. e4 e6 2. Bb5 Ke7! 3. Bxd7 c6 4. Be8 Kxe8.

(Thanks, Jacques.)