31

This seems to fit: The initial step was to replace all 1's (red) with a 2 (black) and all 9's with an 8: Then, whenever a red number was +/- 1 of a black number which was on the same row, column or box, the red number was changed to its other possible value. E.g. if a red 6 was on the same row as a black 5, the red 6 was changed to a black 7 as it couldn't ...


27

I have been asked by a couple of people to show the creation process for this puzzle, so here we go: Also if people want to see more of these strange, sudoku mash ups then I'll be more than happy to combine some new types :) Wrap-up: The Making Of This Samurai Pseudoku This is not a solution to the puzzle, but provides notes from its poster. This type of ...


12

Let's trace through the code! So, after setup, we're here: So, what is that "something"? Armed with this knowledge, we can finally check out lines 200-230: A sample output:


12

Here's the first one with Here's the second one with Here's the third one with No idea if these are optimal though


8

Observe that every 4th power is If one of $a,b,c,d$ is $9$, then If there is a $7$, then we must have $a=9$, so the other two digits are at most $4$ (from the RHS). Here there is If there is no $7$, then we must have either $a=6$ or $a=9$; since $6^4+9^4>7000$, it must be $a=9$. The LHS is over 9000, so the other three numbers must be $6,6,?$ or $6,5,5$...


7

Let us denote the ages of Person 1, Person 2, Person 3 by $x,y,z$ respectively. We'll assume that $x,y,z$ are positive throughout. The product of the 1st person's and the 2nd person's ages is $311 \frac{2}{3}$ plus the 3rd person's age. The sum of the 1st person's age and the quotient of the 3rd person's and the 2nd person's ages is $41 \frac{17}{24}$ ...


7

I think the best we can do is The following tries will cover all possibilities Proof that this is the best possible Alternative Proof (courtesy of Jaap Scherphuis in the comments) Bonus Not totally sure but the best I've managed to do is As follows


7

I'll refrain from giving the answer I know from before, since I didn't originally figure it out for myself, so giving that answer would feel a bit like cheating. Instead, here's how you might find the balance point using one hand only: Now you can actually see the centre of balance: (If you are very agile, you may even be able to catch the stick by ...


6

This sounds to me like Because


6

Seems a valid solution is Less than 7 digits and more than 3 digits ALL digits in the number are Prime numbers-- some repeated. All individual digits in the number add up to a Prime Number-- whose digits also add up to a Prime number The last 2 digits of the number are same as the first 2 digits. And, it exhibits another property described by a ...


6

The answer is that Yzarc thinks that the universe is: As pointed out in the Hint, the first step to solving this puzzle is to 'unscramble the 8 word anagram'. The 8 words in question are 'Nature's jaunty, purer music ruptures man, returns heaven' and these anagram to: Next, turn our attention to the ciphertext: SsARrMRrrrVRrrEeeMESU. The way we need to ...


5

Summary so far: Oof, that wasn't pretty, but I think that's them all:


5

Pictorial analysis There are $27$ different possibilities to enter into the lock. Let's say two of them are connected if they have two common digits (in the same positions); so every possibility is connected to exactly $6$ others. We can show them on a diagram like this: We can also see here that two triple-digit possibilities which are cycled versions of ...


4

I observed the following: This is because This observation immediately excludes many numbers from consideration. It remains to be shown that the numbers that were not excluded do all end at $153$. For completeness, here is my working out of the remaining cases. Rand al'Thor already did this first in his answer. Like him, I do not see any clever way that ...


4

Considering cycles The largest number such a chain can ever reach is $1486$ (every number between $2001$ and $2100$ gives at most $8+0+729+729=1466$ at the first step, and the largest possibility resulting from any number up to there is $1+27+729+729=1486$). So we have an upper bound, which means every chain must eventually end in a cycle. In the OP you ...


3

The most elegant solution I could find was this one: let the matrix be \begin{equation*} \begin{pmatrix} A & B & C \\ D & E & F \\ G & H & I \end{pmatrix} \end{equation*} Let the sum of each row/column/diagonal be $S$. Then \begin{eqnarray} A+B+C + D+E+F = A+E+I + C+F+I = 2S &\to& I = \frac{B+D}{2} \\ A+D+G = G+H+I + S &\...


3

First I'll prove a property of $3\times3$ magic squares. Using this property you can use a similar proof to find the central cell in this case: The rest of the magic square then follows: I originally used a less elegant more general method by finding a generic solution: Now it is just a matter of applying that to this particular problem.


3

From the QA Department (solver-led improvements) – Community Wiki, feel free to edit Test Ø.   A puzzle.txt that could be produced by the posted program were it set to use more rows and columns. It’s ASCII,,, but is it art? Good enough, at least, to solve by hand and test its simplistic puzzle-producing algorithm. Test D.   puzzle.txt ...


3

I'm fairly sure the answer is perform handshake install es on both ends pause process in exce swap chang.exe combine previous and modify terminate communication


3

There are My process: At this point, The numbers, and divisors are:


2

Partial answer (without the proof of uniqueness):


2

Partial Answer (almost done; can't figure out the last part) Warning: The design of this puzzle pretty much requires some mild profanity, so consider yourself warned. Riddle Note: I'd already figured this part out before @bobble posted her answer. Looking at the code, I recognized it as Looking at the riddle, My prefix is something that controls you. My ...


2

Partial answer (to Riley Riddle) warning: spoilers contain profanity My prefix is something that controls you My suffix is something that students mustn't do My infix is one over nothing And it's a programming language, as noted by a hint I have no idea how to go fro here.


2

As a starting point, the number of digits must be The first rule tells us that it's Given that, the sum of the digits must be If it's a 5-digit number However at that point I'm out of steam because there's multiple combinations there, and even more for 6 digits.


1

I have no idea about the "hyphenated word" part, which IMO doesn't fit the style of the rest of the problem. What if that wiki page changes its title in the future? What if another "hyphenated word" is added to wiki, ruining the uniqueness? Without this part, I just ignored the no-computers tag and found the following list (assuming that ...


1

I think that this is not but With the first move being Analyzing all of black options: After this move, black cannot play bacause in these cases One possibility for black is the move In this case, white cannot play but has the option to play Looking at the other possible first moves for black: Leaving just


1

I know there are plenty of correct answers, but here is a super-simple one.


Only top voted, non community-wiki answers of a minimum length are eligible