cipher - Translate into a sentence

111 221 241 151

222 543 421

233 161

166 222 331 353 444

The problem is to translate this into a sentence.

CLUES:

• 111, could also be 166, or 121


• It is a Homophonic Substitution Cipher


• The code he gave me before this was:
  
  
  
  • &!! "/ (?, }<€ \>!<* :~"& }?@! translates to
  
  
  • see if you can break this code
  
  
  
  


• It is for an interview for an assistant trader position in Chicago.

lateral thinking - A man is born in 1955

(I heard this in a movie and I think it’s pretty cool.)

A man is born in 1955. Today, the 14th of August, 2017, he is 18. How is this possible?

Answer

The answer is simply:

1955 is where he was born. So either hospital room number or home address if home birth!

I have heard this one before but I don't remember where, maybe the movie you mentioned.

mathematics - Rendering the number 10,958 with the string 1 2 3 4 5 6 7 8 9

Brazilian mathematician Inder Taneja has found a way of expressing every number between 1 and 11,111, except 10,958, by inserting mathematical operators in between the numbers 1 2 3 4 5 6 7 8 9 and evaluating the expression. He did so using the four basic arithmetic operations, exponentiation, concatenation, and brackets, but avoiding factorials, square roots, and decimals. If these last three operations are allowed, can 10,958 be likewise expressed?

Answer

Taking from

$(1 + 2 + 34) \times (5 × 6 + 7) \times 8 + 9 = 10961$

We have

$(1 + 2 + 34) \times (5 × 6 + 7) \times 8 + (\sqrt{9})! = 10958$

sudoku - Pseudoku: A Game of Finding Order in Chaos

You are Pat Doe, a lowly Sudoku puzzle-maker eking out a living by producing hand-made Sudoku puzzles.

After an exhausting eight hours, you've just completed your last Sudoku grid for the day. Unfortunately, it's only after you've completed the grid that you realize the usual $3\times 3$ blocks aren't legal (none of them contain exactly the digits 1 thru 9). To make matters worse, your hand malfunctioned and inserted several bold lines denoting boundaries of neighbouring blocks in unusual places.

                           

Rather than redoing the entire grid, you resolve that you will find out a way to partition the grid into 9 irregular blocks (nonominoes) in such a way that:

• each block contains the digits 1 thru 9 each exactly once


• the blocks respect the existing boundaries (i.e. a boundary may appear nowhere on a block except at its borders)

Can you recover from your error? Can you partition the grid and salvage a workable (if highly irregular) Sudoku grid?

Good luck! :)

---

The following is an example of a valid block. It is a nonomino containing the digits 1-9 each exactly once, and it respects established boundaries.

                           

---

The following is not a valid block because it is not a nonomino.

                           

---

The following is not a valid block because it doesn't respect existing boundaries.

                           

Answer

My solution:

Solution method:

I didn't do anything that systematic or logical. I made some vague approximate clusters that had most of the digits 1-9. Then, when a cluster needed a number that it wasn't adjacent to, I'd have it "trade" numbers with adjacent clusters. Often, a whole chain of trades was needed. Though I started on the right and went left, fixing a single disconnected cell in the last incomplete cluster caused a chain reaction that made me reconstruct the bottom right section of the board.

chess - How many bishops to walk across the board

N bishops are placed on a standard 8x8 chess board.

Each bishop can move normally but after touching or moving over an square, that square is considered marked. A bishop cannot pass over or land on a marked square again.

Each bishop may stop any number of times.

What is the minimum number (N) of bishops required to mark every square but follow these rules?

Answer

Preliminary remarks

It is quite obvious that visiting the black squares and visiting the white squares are two independent problems. No bishop can move from one color to the other and they don’t hinder each other in any way.

If N bishops can visit all squares of one color, then 2N bishops can visit all squares of the board.

The problem to solve

Because of the above, we will consider the problem of visiting all grey squares on the board on fig. 1 using N paths that each follows the movement rules of a bishop.

Fig.1 . . . Fig. 2

Fig. 2 illustrates one possible solution of visiting all squares with 4 paths. The direction of the movements is irrelevant.

If you try to solve the problem manually, you will probably find solutions in 4 paths, and you will experience that it resists being solved in less than 4 paths. It feels like, regardless of how you arrange the paths, there is a resource that gets exhausted and you need to add more paths to provide that resource.

A subproblem

To see what it is you are short of, consider another problem, which is to visit half of the board only. Let’s consider the upper-left triangular half (fig.3).

Fig. 3 . . . Fig. 4

Let’s add black and white dots on the board (fig. 4) in some kind of second-level chessboard pattern.

You can see that any bishop path must alternate between black and white dots. So, any bishop path will visit as many black as white dots, with maybe one extra black or white dot. But the imbalance between black and white dots for a single path will never be more than one dot.

On the other side, the imbalance of black and white dots in the triangle is important. There are 10 black dots and only 6 white dots, that is an excess of 4 black dots. This imbalance must be reflected in the paths that cover them. Since we have an excess of 4 black dots, and since one path can only have an excess of 1 black dot, no matter how we arrange the paths, we need 4 paths to reach the same imbalance.

This already proves that you need 4 disjoint paths to cover the triangular board.

(And to answer the question: you are short of white-dotted squares.)

Back to the problem

Let’s now consider the upper-left triangle as part of a square board. Let’s assume we have a set $P_S$ of paths that visit all gray squares of the square board. Let $P_T$ be the set $P_S$ restricted to the squares of the upper-left triangle. $P_T$ is made of pieces of paths in $P_S$.

$P_T$ covers the triangle exaclty so it must include at least 4 disjoint paths.

The fact that $P_T$ includes 4 paths doesn’t imply that $P_S$ also includes 4 paths, because the paths in $P_T$ can join each other via the other half. For example, the bottom right triangle on fig. 2 is completely covered with 3 paths.

The important thing is to see is that a set of paths also defines a set of path ends. (A single-square path counts as 2 ends, the start and the end). If $P_T$ includes 4 paths, it defines 8 path ends within the triangle. We can even assert that it defines 8 path ends located on black dots. That is because only a path with 2 black ends contributes to the imbalance of black dots over white ones.

Let’s take the solution on fig. 2 and see the set $P_T$ covering the upper left triangle (fig. 5).

Fig. 5

As expected, there are at least 4 paths and 8 black path ends.

Only 4 of the black path ends are located near the diagonal. So if we stick the 2 halves back together, only 4 of the black ends can be extended to the other side. The at least 4 remaining black ends are not affected by the other half. That means at least 4 black ends in $P_T$ are also path ends in $P_S$.

The result is that $P_S$ has at least 4 path ends in the upper-left triangle. Likewise, it has 4 path ends in the lower-right triangle. Altogether $P_S$ has at least 8 path ends. And this requires at least 4 paths in $P_S$.

So, to cover the square board, you need 4 bishop paths.

Conclusion

Visiting all squares of one color on a standard chess board requires 4 disjoint bishop paths. To do both colors you need twice as much, you need 8 bishops.

Note

In counting path ends, it is important to remember the multiplicity of path ends when the path starts and ends on the same square. This requires some precautions. For a single-square path, count 2 ends on that same square. This is compatible with the fact that one path has always 2 ends. Another point is that you could have a double path end on a square of the diagonal. This is not a problem since only one can extend to the other side. It doesn’t allow to extend more than 4 paths to the other side.

pattern - Had some real trouble with this logical sequence today

Reasonably sure I didn't get the job, but this particular part of their test really baffled me.

The numbers of angle brackets in each row and column should probably add up to ten, so the missing element is one of the last four options. I was thinking << because the first two columns have a 7/3 and 6/4 split, and it'd make the last one 5/5.

A friend says >> purely for the left to right << <> >> symmetry.

Anyone have any better ideas?

Btw, this question had a 75 second time limit, if you want to start a stopclock and stress yourself out :)

mathematics - Sub-puzzle: Master and Slave versus Bob the Violent but Honest Psychopath

[I created this as a sub-puzzle to "Strategy to beat the Casino reversed", that has been sitting unsolved for months. You may wish to read that puzzle first, but this is self-contained. My answer to that is rather long, and it occurs to me to break it up, into this partial answer, this puzzle here, and later my answer to this puzzle. EDITED: I've divided it into three puzzles, this one, and two followups: easy version and hard version.]

Below is a game that is played by a Master and Slave team versus Bob the Violent but Honest Psychopath. The Master and Slave team starts with a certain number of Mulligans. This number is decided by Bob, with certain restrictions, as we shall see. The main part of the game consists of four rounds in which all three players simultaneously play Red or Black on each round. Unless all three players play the same colour, the Master and Slave team loses one Mulligan... or else, if they have no Mulligans left, Bob will KIIIIIIILL THEEEEEEM BAHAHAAAHAH.

In addition to the four Red/Black rounds, the game has initial and final steps involving a certain Deck of cards, to be explained later. The point of these stages is that they allow us to chain strategies together to give a partial answer to the original puzzle (and also, a complete answer to "Strategy to beat the casino over the long haul").

The Master and Slave can agree upon strategies beforehand, but there is no communication allowed once the game starts, except as provided by the rules. It is for you to come up with strategies for them which guarantee their survival against Bob, who is bloodthirsty and very clever, but who will strictly follow the rules.

---

The game is this:

1. Bob selects a card from the Deck, which he reveals to the Master.


2. Bob reveals to the Master the Red/Black plays he will make in the four rounds. (And yes, Bob is Honest.)


3. The Master then selects a card from the Deck (possibly the same card), which he reveals to Bob.


4. Bob then decides to do one of two things:
   
   
   
   • He grants them one Mulligan, and tells the Slave a Large amount of information about the Master's card; or
   
   
   
   • He grants them two Mulligans, and tells the Slave a Small amount of information about the Master's card.
   
   
   
   


5. Then they play four Red/Black rounds.


6. Then the Slave decides to do one of two things:
   
   
   
   • He guesses a Large amount of information about Bob's card (and the guess must be correct, on pain of getting KIIIIIIILLED); or
   
   
   • He forfeits one of the team's Mulligans, then (correctly) guesses a Small amount of information about Bob's card.
   
   
   
   

To be quite clear: step (5) consists of four rounds, in each of the which, the Master, Slave and Bob simulatneously play either Red or Black. These are their two options in each round (and is not related to the Deck in steps (1),(3),(4) and (6)). Unless all three players play the same colour, the Master and Slave team loses one Mulligan. As seen above, the Master knows in advance what Bob will play, but cannot pass this information to the Slave directly. In the first round, the Slave only knows what Bob tells him in step (4). In subsequent rounds, the Slave also knows what was played in previous rounds.

In step (4), the Master also knows what Bob told the Slave.

And of course, in step (6), the Slave is only allowed to choose the latter option if they have a Mulligan left.

---

I have not yet said what the Deck is. And I have not yet said what Large and Small amounts of information mean. That is because it is up to you, dear puzzle solver! You can choose whatever Deck works for your answer. Of course, it must have the same meaning in all the steps. As I say, that is really the whole point of this, to "chain" answers to solve the original puzzle!

Example: in your answer, you might specify that the Deck is the standard deck of 52 cards; "Large amount" means you know exactly which card it is; and "Small amount" means you only know which suit the card is. This is just an example, if you use it, I'm pretty sure it will get them KIIIIIIILLED.

So, your task is: to come up with a (finite) Deck, and a definition for Large and Small amounts of information, and GUARANTEED (not probabilistic) survival strategies for this game.

Small hint:

You may wish to start by dividing the possibilities for Bob's four Red/Black rounds into odd and even parity.

Medium hint: the Deck I plan to use is

the top four ranks from a standard deck (16 cards). These correspond to the odd parity cases (3-1 split). The colour of the card is the majority colour, and the rank of the card is the position of the minority colour. The difference between clubs and spades, and between hearts and diamonds, is irrelevant (but becomes relevant in the followup puzzle).

And a larger hint

can be found by looking at the edit history of my linked answer

Answer

The Deck I am using is the one I gave in the hints,

A 16-card deck, top four ranks of the standard deck. As I say, only rank and colour matters here, there is no difference between clubs and spades, or between hearts and diamonds. So this is effectively an 8-card deck. A Large amount of information means knowing the rank and colour of the card (effectively complete information), and a Small amount means knowing the colour of the card.

The Master's strategy in step (3) is

first, to decide whether Bob's plays revealed in step (2) have "even or odd parity". "Odd parity" in this case means that the two colours are split 3-1.
- If it is odd, then the Master plays the corresponding card: the colour of the card is the majority colour, and the rank of the card indicates the position of the minority colour. I have chosen K-Q-J-A for rounds 1-2-3-4. So for example a black queen corresponds to black-red-black-black.
- If it is even (2-2 or 4-0 split), then the Master looks at the rank of the card Bob revealed in step (1), and 'flips' the play in the position corresponding to that rank to get an odd parity combination, then plays that card.

So for example, the Master plays a black queen in step (3) under the following circumstances:

(Actually this is part of my answer to the followup question (hard version), where the difference between clubs and spades will matter. Here it does not matter.)
The cases where there Master plays a black king follow the same pattern, but starting with red-black-black-black. Similarly for black jack and ace. This covers half the cases, and then reverse the colours to get the other half.

If Bob gives the Slave a Large amount of information in step (4), their strategy in step (5) is:

The Slave plays the odd parity combination corresponding to rank and colour of the Master's card. (e.g. if black queen, then play black-red-black-black.)
- If Bob's plays are even, the Slave will make one mistake. The position of the mistake tells the Slave the rank of Bob's card. The colour of the Master's play in that position tells him the colour of Bob's card. Thus he has a Large amount of information about Bob's card.
- If Bob's plays are odd, the Slave will make no mistakes.

- If Bob's card is the opposite colour as the Master's, the Master intentionally makes one mistake, in the position corresponding to the rank of Bob's card. The fact the Master made a mistake tells the Slave that Bob's card is the opposite colour as the Master's, and the position tells him the rank. Thus he has a Large amount of information about Bob's card.
- If Bob's card is the same colour as the Master's, the Master does not make a mistake. This tells the Slave that Bob's card is the same colour as the Master's. Thus he only has a Small amount of information about Bob's card, but they still have one Mulligan.

If Bob gives the Slave a Small amount of information in step (4), their strategy in step (5) is:

The Slave plays that colour four times.
- If Bob's plays are odd, the Slave will make one mistake. The colour of the Master's play in that position tells him the colour of Bob's card. Thus he only has a Small amount of information about Bob's card, but they still have one Mulligan.
- If Bob's plays are a 2-2 split, the Slave will make two mistakes. The positions of these mistakes will tell the Slave the two possible ranks of Bob's card. The colour of the Master's play in one position tells him which of these two ranks, and the colour in the other position tells him the colour of Bob's card. Thus he has a Large amount of information about Bob's card.
[To explain how the Slave knows that there are only two possible ranks for Bob's card: for example, suppose the Slave knows the Master played a Black card in step (3). Now suppose Bob's plays are red-red-black-black. The Slave deduces that Bob's card is a King or Queen. If Bob's card were a Jack or Ace, the Master would have played a red Ace or Jack in step (3).]
- If Bob's plays are all the same colour, the Slave will make no mistakes. As above, the Master either makes no mistakes, telling the Slave that Bob's card is one colour, or intentionally makes one mistake, telling the Slave that Bob's card is the other colour, and telling him the rank. This gives the Slave a Small or Large amount of information.

So they are guaranteed to survive.

cipher - Interesting email I received?

This email recently arrived in my inbox. I don't know how it avoided the spam folder, but it must be important. However, it looks like a dictionary in a blender... you'll see what I mean when you read it.

From: owttrapnisretteltsrif@gmail.com

1. maglev verdant jumper phi impossible ringing greater everyone excellent harbringer acute quicken never vengeance eternal rocky resolute llama car rebuke electric knockback kangaroo flute exquisite solution refinery yellow jubilant illuminati infinite carats striking salutation nocturnal samurai illogical evergreen north glade empathy


2. swimming wretched evoke xylophone unburdened quiz equalize heathen civilized impatient escalate environmental toxic zoology nightstand temporal unfold zigzag requirement earthquake quest xenophobic extravagance mettle zebras magical flummox salt attempt tracked language greater backwards crazed axiom xeroxed variation very tried ytterbium conundrum nova vanquished vaporize fallout bracket jumprope sanctuary enzyme locket vocalization warfare garbled erratic rambling revenge grandmother galvanize


3. important must learning help information virtuous explosive iambic farfetched trapped hierarchy inside desolate toughness distinct revolution weaponized titanium bloodlust desecrate quadruple cage seventy five bulwark freedom flight miles castle rhombus transpose naive blocked southeast neglected

(end of email)

If any of you can find a meaning or message in this, I'd love to hear it.

Partial meanings are welcome.

Update: I've recently received another weird, puzzle-seeming email. I've put it here in case it may have some relation.

From: maxhmaxktwwkxllvhfxl@gw.com

1. you may want to look for lbyuoxmsbmvolyyuwkbuc


2. make sure you know jpnxfiayzjrvnvqauigmfvg


3. No ccmdym can omfsydrhjqglbcgefbnvgqakcwvrg

Surprise, surprise. I got a third email. People have definitely seen codes in these, so I'm still not totally convinced they're a scam. Anyway, here it is, have fun.

From: ylfmgbglnliild@gmail.com

dsz gwlvhg svdl iwyllpnz ipsze vgs zg rhxrixf ozizmwsl dxz mr gyvy il pvm

Hi! One day left in the bounty, so I'm giving a really big hint. Hope someone can get it from this!

Look for double letters in part 1 broken between words.

Answer

First paragraph:

From the "Mega Hint", we are to look for double letters broken between words, i.e. a letter that comes at the end of a word and the beginning of the next word. (I don't know where this is hinted at in the puzzle besides the hint). Collecting those gives us:
VIGENERE KEY IS NINE

Second paragraph:

The email address backwards gives us "first letters in part two" (found by gabbo1092). This gives us:
SWEXUQEHCIEETZNTUZREQXEMZMFSATLGBCAXVVTYCNVVFBJSELVWGERRGG

Then, run that through a Vigenerer decode with key "NINE", which gives:
FOR THIRD PARAGRAPH READ PRIMES ONLY COUNTING UP FIRST WORD IS TWENTY

Third paragraph:

Setting the first word as 20 and selecting the prime numbered words (23, 29, 31, ...) gives us:
HELP TRAPPED INSIDE TITANIUM CAGE FIVE MILES SOUTHEAST

word problem - Half and Double Value

A 3 year old child tried to input or copy a number (written on the paper) to a calculator but he messed up with the digit places. All the digits are there but the value of the number he inputed becomes double. He tried to redo it. Unfortunately the digits are not on the right places again. But this time the value of the number he input is only half of the number he is trying to copy. When he grew older and learned about that early numeral experience the child wondered and asked: What is the smallest positive integer N where its value becomes half or double when its digits are rearranged?

Answer

285714­­­­­­­­­­­­­­­­­­­­­­­­

I used the following k program to compute it:

2*{~1=#?{x@

{...}(1+)/1 means start from 1 and increment (1+) while the condition in curly braces is true

1 2 4*x is a list of 1, 2, and 4 times the current value

$ convert to string

{x@ sort each

? unique

# count

1= equal to 1

~ not

2* multiply by 2

Arrange numbers 1 to 9 into the upsilon grid

Arrange numbers 1 to 9 into the octagon, so the operation is correct.
C is a constant.

Do the math operation in sequence, ($×$) and ($/$) is NOT HIGHER than ($+$) and ($-$).

Answer

You could deduce things using the equations.

As mentioned by @Bojan B (with different notation), we can write down some equations:
(a - b) * c = C
(d * e) / f = C
(g - h) + i = C
(a + d) / g = C

(b + e) / h = C
(c * f) - i = C

Then we can start writing down knowns:
1. a != b != c != d != e != f != g != h != i
2. C is positive, because (d * e) / f will not produce less than 1.
3. C is an integer because (g - h) + i will not produce a fraction.
4. a > b. Otherwise, (a - b) * c produces a negative value and contradicts #2.
5. a != 1. Otherwise, b is not in our set of values. b != 9. Otherwise, a is not in our set of values.
6. If a + d is a prime number, then g = 1.
7. If b + e is a prime number, then h = 1.
8. a + d and b + e both cannot be prime, because both g and h would be 1. Contradicts #1.

9. g > h and/or i > g - h. Otherwise, (g - h) + i produces a negative and contradicts #2.
10. f != 5 and f != 7 because d or e would need to be 5 or 7 to satisfy (d * e) / f and contradict #1.
11. C < 17 because the largest value of (g - h) + i is (9 - 1) + 8 = 16.
12. c * f < 25 because (c * f) - i = C or (c * f) = C + i and the largest possible value of C + i is 16 + 9 = 25 (#11). In order for cf = 25, c = f = 5 (contradicts #1).
13. c * f multiplies to 2, 3, 4, 6, 8, 9, 12, 16, 18, or 24 (no multiplies of 5 or 7 because #10).
14. c * f > i. Otherwise, (c * f) - i produces 0 or less.

One could keep going with this strategy and (hopefully) eventually deduce a non-brute force solution.

Edit: continuing my efforts.

Is it possible to add two equations together if we aren't following BODMAS?

(g - h) + i = C
+ (c * f) - i = C
= (g - h) + (c * f) = 2C
which would tell us both g - h and c * f are multiples of 2. So g and h are both even or both odd, and from #13, c * f no longer multiplies to 3 or 9.
Also c * f > g - h because c * f = (g - h) + 2i.

We also have 3 ratios that equal C: (d * e) / f, (a + d) / g, and (b + e) / h. Ratios that could potentially satisfy these are:
1 = 6/6, 8/8, 9/9
2 = 2/1, 4/2, 6/3, 8/4, 12/6, 16/8, 18/9
3 = 3/1, 6/2, 9/3, 12/4, 18/6, 24/8
4 = 4/1, 8/2, 12/3, 16/4, 24/6
6 = 6/1, 12/2, 18/3, 24/4

8 = 8/1, 16/2, 24/3
I've excluded ones where the numerator can only be factored by the denominator. Since we need 3 unique denominators in a set, I also excluded sets with less than 3 ratios.
So we at least know C is 1, 2, 3, 4, 6 or 8.

logical deduction - Ask about "Delay Creates a Conclusion"

In a Messenger group full of logicians:

"Does anyone here understand the concept of sequential logic in digital circuit theory?" said someone.

There is a long pause, no one replied.

And when all members on the group have read the message, one of them answered, "No."

---

Actually, I don't know whether there exists a logic concept/theory behind that story: a delay creates a conclusion. There is also a nice puzzle using the same concept which can be accessed here: https://tierneylab.blogs.nytimes.com/2009/03/16/the-puzzle-of-the-3-hats/

Does anyone in this community know about the theory? Is there any works related to it?

Update: How to interpret such concept on AIs? Let's say there are 3 AIs playing above puzzle of the 3 hats, or above Messenger group is full of IAs.

Answer

The term is called Argument from silence. It is when a person uses another person's silence (or absense of a statement) as an information in itself and not as just silence.

According to rationalwiki.org:

An argument from silence is an informal fallacy that occurs when someone interprets someone's or something's silence as anything other than silence, typically claiming that the silence was in fact communicating agreement or disagreement.

It is considered an informal fallacy because it is not a solid argument that support its conclusion with actual proofs. The silence could be due to various other reasons, thus the conclusion might as well be wrong as it might be correct. Take the three hats puzzle as an example, the prisoner's interpretation of the silent prisoner's silence could be wrong, the silence might be because:

• The prisoner is mute.


• The prisoner is blind, so he can't possibly deduce his hat color and give an answer even if he could.


• The prisoner might be slow-witted, so it should take him a longer time to deduce his hat color, thus the delay/silence is unreliable.


• ...

In all the above situations, the silence does not definitly indicate that the silent prisoner couldn't identify his hat color because of the arrangment of the hats alone.

But in the context of puzzles, one usually uses the best case scenario, where the argument from silence becomes a solid argument (the silence will only mean one thing, thus we use that thing in further deductions).

Note: The term, however, for @Bass comment is Arguments from ignorance. They're basically the same thing, except that AFS is interpreting silence, whereas AFI is interpreting ignorance of a matter/subject.

mathematics - The Devil's Brother

The 1933 movie "The Devil's Brother" (also known under the title "Fra Diavolo") takes place in the Northern Italy of the early 18th century. Stan Laurel and Oliver Hardy play the fierce robbers Stanlio and Ollio. They manage to steal 100 gold coins from the rich Lord Rocburg, and then discuss at length how to share the loot. Finally they decide to play the following game.

• In every step, Stanlio picks a handful of gold coins from the loot.
  
  Then Ollio decides, whether this handful should go to himself or to Stanlio.


• Once all coins have been assigned, the game ends.


• Once one of them has received 9 handfuls, the game also ends. In this case the other player (who has received at most 8 handfuls up to that moment) receives all the remaining gold coins.

Question: What is the highest number of gold coins that Stanlio can guarantee for himself?
(As usual, we assume that both players use optimal strategies.)

Answer

Ok, I'll have a stab. The highest number Stanlio can guarantee is:

46 coins

Optimal strategy:

An optimal strategy for Ollio is to claim a handful that has 6 or more coins, and to pass on handfuls of 5 or fewer coins.

So ...

Stanlio knows this, and so knows that producing a handful with less than 5 coins will result in a bigger pot for Ollio left after 9 rounds. Producing a handful with more than 6 coins just reduces the pot left for himself after 9 rounds.

Which means ...

If Stanlio produces 5 coins at a time, Ollio will let him keep them. This means after 9 rounds Stanlio will have 45 coins and Ollio can claim the remaining 55 coins. If Stanlio produces 6 coins at a time then Ollio will claim them. Meaning that after 9 rounds Ollio will have 54 coins and Stanlio can keep the remaining 46.

A Gathering of Number-Theorists

A certain number of the 5000 members of the World Arithmetical Society (each of which has a different membership number between 1 and 5000) got together to discuss a problem. Much to their surprise, when they were lining up for lunch they discovered that their membership numbers could be arranged to form a sequence of consecutive whole numbers and, moreover, that none of them was standing next to someone whose number was relatively prime to his own. (Remember that two numbers, such as 25 and 34, are relatively prime if they have no common divisor greater than 1.)

How many were the members of the Society who met and what were their membership numbers?

Answer

Here is a solution. I'm afraid I used a computer to help find it. Some explanation is below.

2197 13 2184 2 3 7 13 2191 7 2198 2 7 2194 2 2186 2 2188 2 2192 2 2196 2 3 2187 3 2193 3 2199 3 2190 2 3 5 2185 5 2195 5 2200 2 5 11 2189 11

This uses

17 numbers, from 2184 to 2200. Each row shows which "small" prime numbers divide the number on that row, and you can see that each pair of consecutive rows has a prime number in common.

To find this without requiring too much human brute force or computer cleverness, I

defined a set of numbers to be "plausible" if each shares a common factor with at least one other, and no more than two share a common factor with only one other,

a condition it's easy to check by computer, and then

conducted an exhaustive search for the shortest plausible string of consecutive numbers below 5000.

This might have yielded a non-solution, in which case I'd have continued the search. That would have happened if, e.g.,

you could arrange the numbers into a "chain" and some "rings", but there was no way to connect these together into a single "chain".

Is this solution unique?

This set of numbers is unique. (An earlier version of this answer had a sketch of how to prove it, with a remark to the effect that the details are fiddly enough to put me off doing it properly. Peter Taylor, in comments below, correctly observes that the details are much less fiddly than I'd thought.) Suppose you have a "working" set of numbers, and it includes a prime number $p$. This must be adjacent to at least one other number, which must therefore be a multiple of $p$. Therefore your range must extend at least as far as $2p$. But in the range $(p,2p)$ there is at least one prime number, by Bertrand's postulate. So if you have a prime number then you have a larger prime number; hence, any solution must include no prime numbers. The longest gap between prime numbers below 5000 has length 33; so brute force up to size 33 suffices. I've run my program further than that and found no viable sets of numbers other than the one above.

But

the exact sequence is not unique; e.g., we could swap 2187 and 2193.

geography - Introducing Gladys, an intrepid globetrotter

Gladys has always led an active life full of adventure. As a retiree, she has a lot of free time on her hands and loves to spend it on her two favourite pastimes – travelling and crossword puzzles. Today she embarks on a long journey that will take her halfway across the globe. She will send us postcards from some of her favourite destinations, each accompanied with a puzzle that reveals what place or landmark she is visiting. In the end, we may find out what she got out of her journey.

This is intended to be part 1 of 26 (assuming we all live that long). The answers will be needed in the final puzzle, all others are standalones. Note that while many of Gladys's destinations are obscure, each is notable enough to have its own separate article in the English Wikipedia.

Without further ado, let's hear what Gladys has to say:

Dear Puzzling,

As you know, today I started my long-awaited journey halfway across the globe. My trip has started very peacefully. I spent a quiet moment in a lovely church which has beautiful stained glass windows and just a wonderful atmosphere. I look forward to writing to you from the places I visit to let you figure out where in the world I'm travelling.

Wish you were here!
Love, Gladys.

---

_{Gladys will return in "Gladys unchained".}

Answer

I think the answer is

Heinz Memorial Chapel

Completed Nonogram

Traversed Maze

logical deduction - Different Teapot Riddle - fanmade

As inspired by @gabbo1092 in the first of this kind of riddle (see this for an example of how the answer works), I thought I'd come up with another riddle.

Rules:

• I will give you a number of clues.


• Each clue relates to a word.


• Each of these words will have a teapot (homophone/homonym) that has a different meaning.


• All of these teapots will be related in some way.


• Your task is to find each of the words, their teapots, and their connection.

The first word has a teapot that means you feel sad and depressed.

The second word has a teapot that is sweet and another teapot that belongs to William.

The third word has a teapot that helps you fall asleep and feel calm.

The fourth word has a teapot that both Indiana Jones and Ernest Shackleton would like versions of.

HINT ONE

The third clue refers to a plant's smell.

HINT TWO

The fourth clue refers to something edible.

HINT THREE

The William in the second clue is dutch.

HINT FOUR

The fourth word has potential prefixes. One of these prefixes is something to do with Indiana Jones, and a different prefix is something to do with Ernest Shackleton.

Answer

I think that they are all

Colours

The first word has a teapot that means you feel sad and depressed.

Blue - melancholy, sad, or depressed mood

The second word has a teapot that is sweet and another teapot that belongs to William.

Orange - which is a sweet fruit but also a feudal principality of which William III was prince.

The third word has a teapot that helps you fall asleep and feel calm.

Lavender - already solved by Naeem Shaikh

The fourth word has a teapot that both Indiana Jones and Ernest Shackleton would like versions of.

Cream - as in whipped cream (Indiana Jones had a signature whip) and ice cream (Ernest Shackleton was a polar explorer). Thank you to OP for the 4th hint.

wordplay - Yet another teapot riddle (No.37)

This riddle is probably very easy, I believe, but it's my first one (actually, it is my first riddle on the site).

The rules are as usual.

I have one word which has several (2 or more) meanings

Each of the meanings is a teapot (first, second ...)

Try to figure out the word with my hints.

1st hint

The first teapot is on the surface.
The second teapot always comes in a suit.

2nd hint

The first teapot is usually female.
The second teapot is usually male.

3rd hint

The first teapot can be less than 4 or more than 6.
The second teapot usually goes up to 11.

Answer

Is this a

Jack?

The first teapot is on the surface.

Like a headphone jack?

The second teapot always comes in a suit.

A playing card Jack

2nd hint The first teapot is usually female.

The female receptor end

The second teapot is usually male.

The card jack is usually male

3rd hint The first teapot can be less than 4 or more than 6.

The size is either less than 4mm or greater than 6mm

The second teapot usually goes up to 11.

A Jack’s value is usually 11 in cards, between the 10 and the Queen

geography - Gladys spins the wheel

_{This puzzle is part 5 of Gladys' journey across the globe. Each part can be solved independently. Nevertheless, if you are new to the series, feel free to start at the beginning: Introducing Gladys.}

---

Dear Puzzling,

I hope this message finds you well. Today I indulged in a bit of luxury. I booked a nice hotel room, had a massage at the spa, and watched an amazing magic show. I even tried my luck in roulette! But don't worry, I haven't spent my entire travelling budget here so the journey continues.

Wish you were here!
Love, Gladys.

Across
2. Social security
8. D.C. suburb

10. TV channel, sometimes unplugged
11. Chief officer or administrator, informally
12. Sister of Meg, Jo and Beth
14. Jay ____ Garage
15. Command for a dog
17. Illegitimately seize power
19. Unisex first name
20. Decorate anew
21. Prominent climate change activist
22. Name and password haver

23. Type of elephant
24. Creative work

Down
1. – mater
2. Buddhist temple
3. Jealousy
4. Cunning canines
5. Perpendicular to Manhattan's streets
6. Woodwind instrument
7. Psyche's partner

9. Of extremely high quality
13. Director Forman
15. Long epic story
16. Actress Hatcher
18. Long Beach or Busan, for example
20. Participate in an election as a candidate

---

_{Gladys will return in "Horribly broken and just plain wrong".}

Answer

She is at

Eldorado Resort Casino

Filled crossword

Grey parts

Taking the grey puzzle pieces and assembling it to a 5x4 rectangle gives the location.

logical deduction - Guess another's present

100 clever men receive presents from the president. Each man gets either a red or blue present, and only knows the color of his own gift.

Then each man must guess a colour of a gift of some man: he must chose a man (besides himself) and a colour, write these down on a piece of paper and give this paper to an organiser. Once all this is done the organiser counts amount of correct guesses out of 100.

The men know all the described procedure in advance and have time to develop a strategy, before receiving any presents. Once they receive the presents, they will be unable to communicate with each other.

Their task is to guarantee maximum amount of correct guesses. Your task is to say 1) what is this maximum amount, 2) what can be a strategy of the men and, the most important: 3) prove that there is no other strategy, which can guaranty a bigger amount.

P.S. "Guarantee" - means that this amount should be achieved independently of luck and what presents are. It can be that all 100 presents are blue, or all red, or a mix, distribution between men also is arbitrary.
P.P.S. It feels like 50 is right answer, it is easy to figure out a strategy to do this, but it is really hard to prove that this is the best result. Note that 1. several men can guess about one present, 2. man can chose who he is guessing about After he got his present.

Answer

Here is a proof that more than 50 is impossible.

No matter what strategy a player uses, they will be wrong exactly half of the time. Why? Let Alice be a particular player. For each present distribution $P$, define a corresponding distribution $f(P)$, where you change the color of the person that Alice is guessing and leave all other colors the same.

Since $f(P)$ doesn't change Alice's color, it doesn't change her guess, which means that $f(f(P))=P$. In addition, $f$ has no fixed points. Therefore, we can break up the set of all $2^{100}$ distributions into pairs $(P,f(P))$. Alice will be correct for exactly one distribution in each pair, so she is correct half the time.

This means that each person gets 0.5 guesses correct on average, so adding these up, the team gets 50 guesses correct on average. This proves that it is impossible to guarantee getting more than 50 (if they always get 51 or more, then the average would be 51 or more).

---

Put another way: from Alice's perspective, she is guessing the value of a coin flip she has no information about, so she has to be correct with probability 1/2.

word - Teapot Riddle no.8

Teapot riddle no.8: "rules as every year, James":
I have one word which has several (2 or more) meanings.
Each of the meanings is a teapot (first, second ...)

You try to figure out the word with my Hints.

• 
  

  First Hint:
  My first teapot seems to be made of everything
  My second teapot can be made of standing

  
  


• 
  

  Second Hint:
  My first teapot is a flavor of life
  
  My second teapot is a problem of life

  
  


• 
  

  Third Hint:
  My first teapot is made by your Grandma
  My second teapot is made by every other Person on the street

  
  


• 
  

  Final Hint:
  Have you tried Pancakes with it?
  
  Have you tried riding instead of it?

  
  

Good luck, have fun
more fun: last riddle

Answer

Is it

Jam (the first one is the food (Jam/Jelly) and the second is the traffic jam

First Hint

Vareity of Jams (Mixed fruit, orange, strawberries, etc.)
Not sure though

Second Hint

My first teapot is a flavor of life
- Jam is the sweet and the most delicious flavor (avoiding the sugar-free references here) :D My second teapot is a problem of life
- Traffic jams are a real problem almost everywhere

Third Hint

My first teapot is made by your Grandma
My second teapot is made by every other Person on the street

Final Hint

Have you tried Pancakes with it? Have you tried riding instead of it?

logical deduction - Consecutive ranks?

Tom, Paul and Joe have just learnt the ranks that they reached in a competition. None of them knows the ranks of the other two, but they all know that Tom and Paul ended up in consecutive positions. As they were curious to know each other's ranks, this little conversation took place:

Paul declares: I don't know Tom's rank.

Tom: Me neither. I also don't know your rank.

Paul: I already knew this.

Joe: Now I know that I am the last among the three of us.

Tom : Me too. Before I just knew that I'm not the first among the three of us.

What's everyone's rank?

NOTES:

• The ranks are strictly postive integers.


• No rank occurs twice.

HINT:

Every ones knows the maximum rank that could be obtained.

Answer

Note, my numbers are for if a higher rank is better..

Joe's rank is 4, Tom's rank is 5, and Paul's rank is 6

Paul: "I don't know Tom's ranking."

Paul knows that he and Tom have consecutive positions, so if his degree was either 1 or N (the max), he would know Tom's ranking.

Tom: "I don't know yours, either."

Tom now knows that Paul's ranking is neither at the top nor at the bottom, but since he still doesn't know Paul's ranking that means his ranking is not 2 or N-1.

Paul: "I already knew that."

This means Paul already knew Tom wasn't at 2 or N-1, so Paul's ranking must be greater than 3 and less than N-2. Note: I'm assuming here that Paul would have mentioned if they knew Tom's ranking at this point. If that's not the case, then Paul could have 2 or N-1 and know (although he hasn't said it) Tom's rank.

Joe: "Now I know that I am last among the three of us."

Right before Paul's second statement, Joe would only know that Paul's ranking must be at least 2, with Tom's possibly being 3. After Paul's second statement, Tom could still be at 3 with Paul being at 4. Joe's ranking must be 4 because he now knows he is last - up until Paul's second statement there was the possibility that Tom and Paul could be below Joe, but since Paul cannot be lower than 4, and Joe has 4, he knows Paul and Tom must be above him.

Tom : Me too I just knew that I'm not the first between the three of us.

Up until Joe's statement, Paul could have been in rank 4. Now that Tom knows he is below Paul, he must have rank 5, otherwise he would not have been able to tell. That leaves Paul to be rank 6.

cipher - (According to me.)

First thing

• My first three are the code for a port airport in Malaysia.


• Next you'll need one of the least-common Scrabble tiles.


• Finally, don't look in nature for my last two, they were first made in Germany in 1984.

Second thing

• My first three are broadcast by the USA, but not in the USA.


• Next, you’ll find the original ObiWan.


• My last two are where you want to be on a rainy day.

Third thing

Fourth thing

• I begin with two neighbors who together can make a beetle.


• My next two are a coin flip in ancient Rome.


• My last two are a guy on a baseball team. Or on an American football team.

Fifth thing

• My first three can be found by taking the 5th, 2nd, and 7th of a well-known sequence.


• Take half of that same sequence to get my last two.

Sixth thing

• My first four are a bit bloody.


• My last six are a place of peace and tranquility, or peril.

Final thing

According to all of you (↑), I am vyvjijljifvzzlrk

---

Hint for the third thing

Hail is used as a verb, but not in the sense of acclaim or glorification.
In the puzzle mentioned, we indeed had to do something, but "thin" is not the right verb for _ _ _ _ K.

Answer

After solving the sixth thing, Gareth proposed that ...

... the "things" are the names of transformations to which that transformation is applied.

This seems to be true. In particular, the first thing ...

... is ATBASH encoded with the Atbash cipher: ZGYZHS

ZGY is the IATA code of Kuching Port Airport. The Z tile occurs only once in a Scrabble set and scores 10. HS is the chemical symbol of Hassium.

The second thing ...

... is RAILFENCE encoded with the railfence cipher:

    R       F       E
A   L   E   C
I       N

Radio Free Europe or RFE is an US-govermnent funded broadcaster in Europe. ALEC Guinness played Obi-Wan Kenobi in the original Star Wars and when you are at home, or IN, you're dry in rainy weather.

The third thing ...

... is BACKWARD backward: DRAWKCAB.

You can hail a CAB and we had to DRAW K, i.e. follow a series of instructions to draw the letter K, in Alconja's puzzle. The clues to this thing are in backwards order, too, because they put the "cab" before the "draw K".

(This thing was solved by Gareth, with whom I agree that the cab clue should come last, so that the clues in order yield the whole word backwards.)

The fourth thing ...

... is VOWELLESS without vowels: VWLLSS.

Volkswagen or VW for short produced the Beetle. A coin toss is fifty-fifty or LL in Roman numbers. And SS could be a Shortstop in baseball or Strong Safety in American football.

The fifth thing ...

... is ROT13 encoded with Rot13: EBG13.

The sequence in question isn't any numerical sequence, but the English alphabet, whose fifth, second and seventh elements are E, B and G. The alphabet as 26 letters, so half of it is 13. (And it's quite embarrassing how long it took me to find this thing.)

The sixth thing ...

... was found by Gareth Mc Caughan. It is REARRANGED rearranged to RARE GARDEN.

The final answer:

At first I thought that the answer was a word that would be subjected to all transformations and that the final result of these transformations would be vyvjijljifvzzlrk. That can't be true, because when working backwards from that word, the last step is a transposition, which doesn't change the contained letters, and the seond but last step is Rot13, which yields a word with i's and e's. That can't be the result of the fourth step, which is to strip all vowels.

The procedure is to start with vyvjijljifvzzlrk and apply all transformations to it:

Atbash → ebeqrqoqrueaaoip
Railfence → errabqqquaopeoei

Backward → ieoepoauqqqbarre
Vowelless → pqqqbrr
Rot13 → cdddoee
Rearranged → decoded

So the final answer is:

According to all of you (↑), I am DECODED.

lateral thinking - Horde of bishops

This has got to be the coolest chess puzzle I have ever seen. Shamelessly stolen from here (Which is actually linked from the help center, which is how I found this) which links here. I don't believe anybody has posted this before as a puzzle though.

The question: Can white win in this position?

Answer

Try to solve this before reading below! (No, really!)

The answer is

Yes

Because

  1. B4f6  a6
  2. Ba1   a5
  3. Bdg5  a4
  4. Bgf6  a3
  5. Kc3   Kxa1   (or: 5. ...Kb1  6. Kb3 a2  7. Kc3 Kxa1  8. Kc2#) 
  6. Kb3+  Kb1
  7. Ba1   a2

  8. Kc3   Kxa1
  9. Kc2#

all Black's moves (other than 5) forced.
Thanks to Gareth for an enlightening comment in The Sphinx's Lair.

story - Uncle Sam's rebus - Clue Eleven

_{<---Previous clue}

---

After the easiness of the previous clue, you feel sure that the puzzles will start to get harder now. And the sign ahead seems to confirm that:

This is the last easy puzzle you'll get. After this, you'll really start to stretch your brain. But first, Uncle Sam has a rebus for you. Speak the answer, and proceed.

---

_{Next clue--->}

Answer

I think its

Uranium

1:

"You" = U

2:

"Ray" = Ra

3:

"Knee" = Ni (Thanks @PartyHatPanda)

4:

"Umm?" = Um

logical deduction - Help with the name of this puzzle

When I was cleaning my basement I found one puzzle that my father gave me when I was a child.

I really want to know the name of it! Could someone help me?

The puzzle consist in 4 disks, and each disk have 4 balls. To beat the puzzle you have to put balls with the same color in the same column.

here is some pictures:

Answer

That specific one looks like a Varikon tower to me. See

combinatorics - MPire coloring game basic strategy

The MPire coloring game is a game where several "empires" are laid out touching each other. Your task is to color each empire a different color, and not let two of the same color touch. You also have to use as few colors as possible.

I was wondering if anyone had specific strategies for this game? Is it better to start in the center or a corner? Plan a strategy around trick points in the map?

geometry - Probability of three segments forming a triangle

Consider a line segment of length L. This segment is cut in three segments at arbitrary points. What is the probability that these three segments could be rearranged as sides of a triangle?

Answer

It is possible to do so with any division where none of the three sections have a length that's over half the total.

This means:

a) first and second cut cannot be different by over half the maximum length

On the graph I colored the 'bad' areas: where the first is over half the maximum bigger then the second, and where the second is over half the maximum smaller then the first.

Simple math would show that's 1/4 that doesn't qualify

b) the first two cuts cuts cannot be both in the first half, nor can they be both in the second half.

This eliminates options colored in blue

Combining those, I think I end up with 1/4 odds for decent solutions. Putting this up now, I'll rethink it after.

language - Conspiracy Theories

^{An entry in Fortnightly Topic Challenge #35: Restricted Title 1. Title based on this xkcd.}

---

I think a friend of mine sent me a letter. However, I can't seem to decrypt his message. I'm worried about my potential friend, can you help me decrypt this message?

I think the writing is written in some form of English, but it's really confusing.

Answer

The message is:

written in mirror writing, and with shapes that resemble English letters. (e.g. the right-most three letters of the centred first line are mirrored characters for "Thanks to Riley for pointing out that including a reflected image would help the answer; I used one when writing but it totally slipped my mind to add to the answer.

Decrypted, it reads:

CONSPIRACY

MY DEAR FRIEND JOE-YOU-KNOW,
I NEED YOUR HELP. I HAVE DISCOVERED
A CONSPIRACY MOST HEINOUS AND BAD.
WHAT I HAVE FOUND WILL SHAKE THE
FOUNDATION OF WHAT YOU KNOW
ABOUT ICE CREAM. I HAVE LEARNED
THAT ICE CREAM COMPANIES ARE
TRYING TO TAKE OVER THE WORLD.

PLEASE HELP ME, OLD FRIEND.
I AM IN A MAXIMUM SECURITY
PRISON AT THE BLUE BELL

HEADQUARTERS. PLEASE, SAVE
ME FRIEND. YOU'RE MY ONLY HOPE.

YOUR FRIEND,
RILLOASRON

riddle - I'm yearning in grey

I have made films on wood.
I have stopped by a park.
Going outside late makes me happy.
I'm yearning in grey.

What is my name?

Answer

I have made films on wood.

Ed Wood was an American filmmaker.

I have stopped by a park.

Parked means stopped.

Going outside late makes me happy.

Elated means happy.

I'm yearning in grey.

Greedy as in having strong desires/hungering.

logical deduction - Next number in the following sequence?

In following sequence, Find the next number?

85, 79, 73, 69, ?

Answer

The values are in the reverse order of ASCII values of the Vowels
85, 79, 73, 69 equals to

U, O, I, E

So A's equivalent ASCII code is $65$

One more possible

The difference of 85 - 79 = 6
The difference of 79 - 73 = 6
The difference of 73 - 69 = 4
The difference of 69 - x = 4

So, x = 69 - 4 => 65

geometry - Tiling rectangles with W pentomino plus rectangles

Inspired by Polyomino Z pentomino and rectangle packing into rectangle

Also in this series: Tiling rectangles with F pentomino plus rectangles

Tiling rectangles with N pentomino plus rectangles

Tiling rectangles with T pentomino plus rectangles

Tiling rectangles with U pentomino plus rectangles

Tiling rectangles with V pentomino plus rectangles

Tiling rectangles with X pentomino plus rectangles

The goal is to tile rectangles as small as possible with the W pentomino. Of course this is impossible, so we allow the addition of copies of a rectangle. For each rectangle $a\times b$, find the smallest area larger rectangle that copies of $a\times b$ plus at least one W-pentomino will tile. Example shown, with the $1\times 1$, you can tile a $3\times 3$ as follows.

Now we don't need to consider $1\times 1$ any longer as we have found the smallest rectangle tilable with copies of W plus copies of $1\times 1$.

There are at least eight more solutions.

Answer

First, a generalizable solution for $1 \times n$, $n$ is even. By halving the rectangles, we can also obtain solutions for odd $n$, and the parts with just rectangles and no W-pentominos can be shortened.

These can be used for F pentominos as well.
Even $n$ fit in a $2n + 1 \times 3n$ rectangle (left)
Odd $n$ fit in a $2n + 1 \times 4n$ rectangle (right)

Computer research shows that this is optimal for $1 \times 8$, $1 \times 9$, $1 \times 10$, $1 \times 12$ and $1 \times 14$.

This is a way to tile

a 6x3 rectangle with Ws and 1x2s:

This is optimal for this $a \times b$ because

to fill the gap at the 'bottom' of the W, you need at least two 1x2s (using another W to fill it leads to at least a 5x4 rectangle which is bigger. That fixes the right half of the rectangle; that a 5x3 rectangle is not possible is easy to see with a checkerboard argument, and for a 4x4 we'd need another 7 squares, which is odd, so another W, and the only position to place that W leaves two corner squares empty.

And here is a way to tile

a 4x7 rectangle with Ws and 1x3s:

I've found two more, one for 1x4:

(11x8 = 88)

and a rather large one for 1x5:

(15x12 = 180)

Continuing down the line we have (AFAIK) minimal tilings for 1x6:

(8x18 = 144)

for 1x7 one which looks like a Star Wars fighter:

(10x37 = 370)

for 1x8:

(17x24 = 408)

for 1x10:

(21x30 = 630)

for 2x3:

(7x10 = 70)

for 2x5:

(20x20 = 400)

for 2x11:

(38x38 = 1444)

Notice the 8 by 8 'rounded square' in the center of the 2x11 solution, wrapped by two layers of W pentominos before moving on to the rectangles. This is the same layout as the 2x5 solution, but there the rounded square has dimension 2 by 2, so it vanishes. Whether this solution type is generalizable for other $2 \times n$ I don't know yet. It does not lead to solutions for $2 \times 7$; for example, a rounded rectangle of sizes $46 \times 74$ and $32 \times 102$ cannot be tiled by W pentominos.

Moving on to the $a = 3$ case, one for 3x4:

(19x28 = 532)

The solution for $3 \times 4$ is generalizable (but not necessarily minimal) to other $3 \times n$ ($n$ not divisible by 3). Here is a 'recipe' for this, a $3 \times 5$ solution, which also explains the parameters $x_i$ and $y_i$ necessary for the construction:

$28 \times 55 = 1540$

First, concentrate on the shape formed by the W pentominos alone. Note that it can extend indefinitely from the top left in two directions, and for each end, we have two choices for a 'tail'; a blunt end as seen at the bottom ($x_0 = 3$) and a sharp end at the right ($y_0 = 1$). It turns out that when the area formed by $x_1$ and $y_1$ is tiled by vertical rectangles, this only works when $y_0 = 1$ and $x_0 = 3$ (horizontal, it's the other way around); symmetry dictates it suffices to tackle just the vertical case. In the $3 \times 4$ case, $x_1$ is odd and $y_1$ is even; here, it's the other way around. They cannot have the same parity.

The only other rules for $x_i$ and $y_i$ are divisibility rules for both $a (= 3)$ and $b (= n)$. It seems to work just like magic:

$a |\, x_1$
$a |\, x_2$
$a |\, x_3 - 1$
$a |\, x_4 - 1$
$a |\, x_5$
$b |\, x_1 + 3$
$b |\, x_2 + 1$
$b |\, x_3$

$b |\, x_4$
$b |\, x_5 + 1$
$a |\, y_1 + 1$
$a |\, y_2$
$a |\, y_3 - 1$
$b |\, y_1$
$b |\, y_2 + 1$
$b |\, y_3$

The size of the solution is given by $(\sum_{i=0}^3 y_i + 3) \times (\sum_{i=0}^5 x_i + 2)$. To construct a solution for a given $a \times b$, just choose the smallest (non-zero) $x_i$ and $y_i$ which satisfy the equations; for $x_1$ and $y_1$ you have to check their parity.

For example, in the $3 \times 7$ case the smallest solution for $x_1$ is 18 and for $y_1$ = 14, but they're both even and we have to settle for $x_1 = 39$ (or $y_1 = 35$ but that will give a larger solution). This is the result:

$31 \times 70 = 2170$

The solution for $3 \times 10$ is rather small compared to the rectangle size and my program was able to verify (after a few weeks of calculation) that it's the minimal solution.

Interestingly, the $28 \times 55$ solution for $3 \times 5$ is not the smallest one. The one below is smaller and is generalizable for $3 \times n$ where $n$ is odd but not divisible by 3, but it's usually larger than the corresponding one for the previous family.

$35 \times 38 = 1330$