New answers tagged

1

Well, the first thing I noticed is that From here, there's not really enough information to get a 100% definitive answer, but the most reasonable one to me is that Therefore, I propose that your solution is


7

There's no no-computers, and it was easy enough to hack my Kyoto CNF generator to produce a SAT instance for the problem: a 9×9 bit array where each row, column and box has two 1s and no two adjacent variables may be both 1. no-computers proof:


10

As long as the triangles are allowed to overlap, I think I can just Zorro through like this: Counting: Triangles with all 3 sides made of grid lines: 0 Triangles with 2 sides made of grid lines: 12 (4 per each drawn line) Triangles with 1 side made of grid lines: 8 (4 for each intersecting red line pair) Triangles with 0 sides made of grid lines: 0 As for ...


0

Second Player will always : Winning Strategy + Reason + Explanation : Comment: With 4 Dots on each side, the Winner will be the Other Player.


0

This is a difficult puzzle, especially for an interview. It seems that there ought to be a clever solution, rather than a brute-force or trial and error kind of solution. Typically, I would expect something hinted at by @FlorianF where the puzzle is equivalent to something that has an obvious solution. Or that there would be some simple algorithm that yields ...


4

The fruits of my labor: Extension to larger grids: Improvement, possibly optimal:


6

Proof without words:


3

Let $$(4x^4+8x^3+7x^2+3x+\frac{1}{2}) = (i^2-j^2)$$ Then $$(4x^4+8x^3+7x^2+3x+\frac{1}{2}) = (i+j)(i-j)$$ Assuming i and j are polynomials in x of degree no greater than 2, we could say $$i = ax^2 + bx + c$$ $$j = dx^2 + ex + f$$ Substituting into $(i+j)(i-j)$ and expanding we can then group like terms and compare with the original equation to get the ...


8

(I don't think I've seen this before, but having figured it out and written it down I have a bit of a sense of déjà vu...) Suppose we A bit of notation: Now So Therefore In other words, Here's a fairly crappy diagram that may help follow the above:


8

To kick this thing off, I've got: EDIT: I've come across useful patterns that provide good answers for even $n$ greater than 6: Imgur Link


-2

I'm not really sure I understand the question, especially the meaning of "such that the sums of the prefixes of lengths 1,2,… of the 6 permutations are distinct" - as it is ambiguous as to whether or not 'distinct' carries across all sums of all prefixes of all permutations, or just the sums of the prefixes of each permutation. An additional clue ...


0

Consider a Single "guess row": It can have 5 Wrong entries - 1 Possibility It can have 4 Wrong entries - (5/1) x 2 Possibilities (the remaining 1 is either Correct or Partially Correct) It can have 3 wrong entries - (5x4/2x1) x 2x2 Possibilities (the remaining 2 are either Correct or Partially Correct) It can have 2 wrong entries - (5x4x3/3x2x1) x ...


-1

[EDITED to add:] No, this is wrong at present. For instance, my analysis assumes that you can have a row looking like GGGGY, which of course isn't possible because there's no other place for that last letter to go. That might possibly be the only way in which it goes wrong; I will think about that later, if someone else hasn't posted a more correct answer by ...


4

This puzzle of mine appears to have languished with no solution for some time. I don’t remember the solution I initially had in mind, but I thought of one recently that might be of interest, so answering my own puzzle here. It suffices to show that for any $n$, there exists an $n$-gon inscribed in the unit circle with pairwise-distinct rational distances, ...


4

Black can stalemate in move White's job here is moving their pieces such that black can capture almost every move while bringing their pieces into the final position. Final position:


3

Daniel Mathias has solved this question in a comment on the MSE post, so for completeness I will summarize the entire strategy here. X starts by claiming the center. Up to symmetry, O's response is one of the nine squares below: $$ \begin{array}{|c|c|c|c|c|c|c|} \hline &&&&&& \\\hline &&&&&& \\\hline &&&...


0

First of all, let's put this into mathematical language: We're looking at the action of the group $S_5$ on a set $X$ of 5 elements $A,B,C,D,E$. Specifically, we're interested in the subgroup $H\leq S_5$ generated by three 3-cycles $(A,B,D)$, $(A,C,D)$, $(B,D,E)$. I will prove exactly what subgroup $H$ is, and thereby answer the two questions you gave. Can ...


4

Turns out that the optimal solution has been known since 1991. The answer is we need $$\left\lceil\frac{7n}{6}\right\rceil$$ gloves to have $n$ doctors operating on $n$ patients without the risk of infecting each other. This number is optimal. Ilan Vardi published it in chapter 10 of Computational Recreations in Mathematica (Addison-Wesley). Both the proof ...


0



2

Here is a solution What I have not considered


6

I suggest we break the post down into 2 challenges: introduce the outer automorphism w/o any explicit reference to group theory or to modern algebra; find a good puzzle (emphasis by me) whose solution requires this peculiar automorphism. This early in the year, I can only answer to the 1st: I do not expect it to be accepted, only to tickle your brains into ...


0

A double-n set of dominos (for n > 1) can be represented by a graph with n+1 nodes and (n + 1)(n + 2)/2 arcs. The acs represent the individual dominos. A traverse of the graph using each arc exactly once would be equivalent to laying out a domino chain that obeys the rules of dominoes. Inspection of the graph for n = 3 show that no domino chain using all ...


8

I think you are Reasoning


5

I just want to point out that


4

Answer (spoiler, I was wrong): Reasoning: Getting the correct answer:


23

It seems likely that they have Reasoning: Edit for detailed reasoning:


3

13 moves (25 ply): 1. e3 a5 2. Qh5 Ra6 3. Qxa5 h5 4. h4 Rah6 5. Rh3 f6 6. Qxc7 Kf7 7. Rf3 d5 8. Qxb8 Bf5 9. Qxb7 Bh7 10. Qxd5+ Qxd5 11. Rf5 Qc4 12. Bxc4+ Kg6 13. Be6 1/2-1/2 I started off with the well-known stalemate (without queen exchange) in 10 moves, composed by Sam Loyd (see this chess.com post). In that final stalemate position, the role of the queen ...


1

What we find from this recursive process is Now In conclusion


1

By visual inspection of the diagram below it can be seen that a 3x3 square is divided into 7 rectangles whose sides are in 2:1 ratio.


9

Is it possible? Why?


4

A word is in the set $\Pi$ if and only if Examples The image of Earth Hint 1 Hint 2 What do "homo", "cell", "mega" and "micro" have in common? The crown picture Astonishing coincidence


3

The answer is: We can work out the values of each icon (GIRL, BOY, HEART) by solving the first 3 equations: The trick to solving the final equation is the same as for all internet puzzles of this type - you have to notice the additional icons hidden in the image. In this case:


0

357 or 36 or 51


1



5

(Dupe, but with pic) You have an X-wing of ones at the bottom right, so you can place the 1 in the 12 clue:


6

In the bottom right corner, in the 6-clue pointing right, you know: This forces: The bottom right corner falls to easy deductions: Next: Finishing up:


6

You can start in the area near the orange cursor:


6

Neither of $X,Y$ can be zero, because So we can assume $X,Y$ are both non-zero digits. The following is true: The above relation therefore shows that there will remain at least two possibilities for $X,Y$ unless From this it follows that the sum of $X$ and $Y$ is This doesn't actually show that the situation described in the question can actually happen, ...


4

The answer is


5

Two solutions: And (Verified both with online character counters) Explanation: And through this analysis, I can also be confident there are no other solutions.


4

Jaap Sherpuis's comment hints at a perhaps more intuitive explanation.


4

This problem doesn't really have anything to do with the integers from 1 to 2n. Fix any set consisting of 2n real numbers (repetitions allowed); then split them into a_1 <= a_2 <= .... <= a_n and b_1 >= b_2 >= ... >= b_n. Then the sum |a_1 - b_1| + |a_2 - b_2| + ... + |a_n - b_n| is constant, that is, it doesn't depend on the partition.


10

This is just an "optimised" version of Gareth's answer which I couldn't resist making. Please keep upvoting him and also prefer his answer over mine for acceptance. For any index $i$ let us denote $M_i$ the larger of $a_i,b_i$ and $m_i$ the smaller. Then the given sum of absolute differences can be rewritten as We will now prove that for any pair ...


10

So, first of all, let's In view of But why? I offer first a rather routine, prosaic solution, the sort of thing someone who's done this sort of thing before knows they will be able to get to work. (This is the first solution I found.) Then something slicker. (This is the second solution I found.) Then a way to streamline the second part of the slicker ...


6

The easiest way to count is


0

The smallest natural number that fits the bill is but so I guess the other answer is better...


1

Here is a Starting Point: That was the crux of my earlier comment: Does the slash indicate closeness or separation ? Meaning : Is "a paid day" one unit or "day/cOmes" is one unit ? Does the comma indicate separation ? Or it is not significant ? Assuming slash is separation, ignoring comma: Assuming slash is closeness, ignoring comma: ...


4

The answer is Reasoning Then In particular The 2nd number And since


30

First observe that the first $6$ digits of happen to multiply to $1792$. Now all we need is to remember the bounds to find that the number does the trick. Indeed, setting $n = 1\ 000\ 000\ 000\ 000$ we get Multiplying through with $1\ 000\ 000\ 000\ 000^{1\ 000\ 000\ 000\ 000}$ we confirm that the first $10$ or so digits of the $1\ 000\ 000\ 000\ 000$th ...


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