Two friends, Stewart and Ian, sat down to discuss what they had been working on, recently, in preparation for the upcoming puzzling competition in their school. Today's discussion centred on five closely related problems based on a well known set of puzzles but, for which, the boys had added their own slight variation.

Problem 1

"The answer to the first one is obvious, I think," said Stewart.
"Yes," agreed Ian, "even using a simple brute-force approach you can't help but stumble on the right answer."

Problem 2

"A brute force approach is also useful here, provided you proceed this way," said Stewart, indicating his strategy.
Ian scribbled down some numbers on his sheet of paper and, showing it to Stewart, declared, "My way of thinking of it was to represent the problem like so and only use those with two odd numbers."

Problem 3

"This one I found a bit trickier," admitted Stewart, "but I think the answer might be this which I deduced using a strategy I call 'topless left, topless right'."
"In fact, I can prove your answer is correct," declared Ian, rather smugly. "We can break our set up into equivalence classes like so," he said, as he wrote. "I can choose one member from each class but not simultaneously from both this one and this one."

Problem 4

"I thought this one would be much more difficult," said Stewart, "but then I realised the solution was staring me in the face".
"Egad, you're right!" exclaimed Ian.

Problem 5

"I have to be honest, I couldn't figure out this last one," said Stewart. "From the first problem, I know the answer must be less than or equal to $10$ and some rudimentary work has shown me it is at least $7$."
"I have thought a lot about this one but haven't finished it." said Ian. "My hypothesis is that the answer is $9$ and I stayed up all night trying to prove it."
Stewart proceeded to take out his laptop and said, "Let's ask our old friend, the internet, shall we?" After a few minutes of searching, Stewart closed the laptop again, looked sympathetically at Ian and said, "I'm afraid it's back to the drawing board on this one, buddy."

What are the problems the boys are trying to solve?
What are the answers in each case?

Hint 1

"For Problem 4, were you able to prove your answer?" asked Ian.
"Why yes," replied Stewart, "you can divide the whole set into duelling pairs and ..."
"Ah, I see," said Ian, "only one allowed in each pair."

Hint 2

"There seem to be many different configurations corresponding to the answer to Problem 1," remarked Stuart.
"That's true," said Ian, "more than three and half million. But I just chose a diagonal approach."

  • 2
    $\begingroup$ The paranoid part of me doesn't trust "Egad, you're right!" to just be flavour text. It sets off massive alarm bells xD $\endgroup$ – Joe Sep 23 '16 at 12:24
  • $\begingroup$ @humn Yes, I've clearly missed the mark on this one - too subtle, it seems. I'll add a hint in a little bit. $\endgroup$ – hexomino Sep 26 '16 at 9:17
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    $\begingroup$ Remark: "more than 3 and a half million" is a plausible way to describe the number 10!. $\endgroup$ – Gareth McCaughan Sep 30 '16 at 10:58
  • $\begingroup$ (and factorials count permutations, which can also be described by a particular kind of matrices, and if you were going to pick out a canonical one of those matrices you would pick one with 1s on its leading diagonal) $\endgroup$ – Gareth McCaughan Sep 30 '16 at 10:58

The problems are

How many non-attacking pieces of various kinds can you fit on a 10x10 chessboard.

Problem 1

Answer: 10
There are 10! distinct arrangements of 10 non-atacking rooks, including one where they are placed along a main diagonal (alluded to in hint 2).

Problem 2

Answer: 25
As Ian notes, you can place these in the squares that have both an odd column number and an odd row number.

Problem 3

Answer: 18
Stewart puts bishops in all squares in the first and last column except for the top row.

Problem 4

Answer: 50
looks tricky, but as they attack the opposite solout to the square they are currently in, you can put ones in every black square (which is staring you in the face as you look at a chessboard!).

Problem 5

Answer: 10
Can't be more the the number of rooks, but tricky to prove that it is in fact the same number.

  • $\begingroup$ Welcome to Puzzling, and great answer! $\endgroup$ – Deusovi Sep 30 '16 at 15:54
  • 1
    $\begingroup$ Awesome. Which clue gave it out ? $\endgroup$ – user27395 Sep 30 '16 at 15:55
  • $\begingroup$ @ArbitraryKangaroo Gareth's comments about permutations and matrices got me thinking about grids, and that was my main starting point. $\endgroup$ – bobajob Sep 30 '16 at 15:59
  • $\begingroup$ Yes, perfect. Sorry for the delayed tick. $\endgroup$ – hexomino Oct 3 '16 at 10:34

Partial answer

Solution to the first one:

Ceasar cypher (a cypher which you encript like this: let's say the number is 3 you will "move" the Abc by it for example a-c, b-d, c - e, ...) Brute force solution is that when someone tries all 23 different variants and therefor they are bound to find the right answer.


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