New answers tagged

1

I have a theory that This way But I'm confused by the 3rd question mark, as


3

I think the following grid will work Method


2

In the following, I've made quite a few assumptions, but to me they all seem somewhat sensible. Looks like there's a chance we can use the first 7 blocks to build a shape with the outline given by OP, or the mirror image of it, which works just as well for the purposes of building a DT gun. To see how this works, let's first improve OP's design by swapping ...


1

Here are some experimental results. The "heuristic algorithm", namely choosing smallest numbers one after the other, does not always produce the smallest possible sum. Examples: $[1, 9]$: heuristic: $[1, 9, 3, 13]$; optimal: $[1, 9, 4, 11]$; $[3, 5]$: heuristic: $[3, 5, 9, 16]$; optimal: $[3, 5, 11, 12]$; and many more. The example of $[3, 5, 11, ...


2

Here is a recipe: Interpretation of result vector A: Proof of correctness: Proof of optimality: Example: Total number of coins x places shifted:


2

An insight to get us started is that With that in mind,


5

Probably not the smartest approach but works: (Confession: I did check with computer but only afterwards.) Taken together: The final result is therefore:


3

Basically, at every moment, each column corresponds to a permutation in $S_3$, the symmetric group of three elements. The elements of $S_3$ are divided into two classes: the even permutations and the odd permutations. Any swap of two elements changes the parity. Now simply consider the number of even permutations in all six columns. Each operation changes ...


3

Perhaps it is worth pointing out that it isn't necessarily possible to tile a chessboard with 6 squares removed even if there are 3 of each color and the remaining squares are all rook-connected. ........ ........ ........ ........ ........ ........ X.X..X.X ...XX... So for any proof that it is possible with 4 squares removed, one should be able to point ...


0

which I think would give a score of infinity.


4

The given number is 1/27 times the cube of 9i-5. That Gaussian integer has norm 9^2+5^2=81+25=106=2.53 and accordingly 9i+5 factorizes into Gaussian primes as (1+i)(2+7i), and therefore we can factor your number in lots of different ways by grouping three lots of (1+i), three lots of (2+7i), and three lots of 1/3 in whatever way we prefer. (Your question ...


2

Pretty much all the work here is in deobfuscating the wilfully unhelpful writing. It's possible that I've done it wrong, but I think this is where we end up: That product of fractions raised to powers ends up being 3600, so the angle in question is 60 (it doesn't explicitly say degrees but we are clearly meant to assume this). The shape we start with is a ...


6

Again the short answer is with the example of The analysis and most of the heuristic from my 5x5 solution works here as well, but I introduced a little more heuristic which was unused in 5x5. First, prime factor analysis: 2 appears as a prime factor 15 times in total (2, 4, ..., 16) 3 appears 6 times (3, 6, 9, 12, 15) 5 appears thrice, 7 twice, 11 and 13 ...


2

If you don't need to use all the material and can rearrange material between cuts then you can construct at least 48 rods with only three cuts as shown below. So none of the optional answers appear to be correct. Alternately, if you want to use all the material to make unit length rods, I believe it can be done in cuts as shown in the following diagram.


4

It can be done in Picture contributed by @JaapScherphuis: Could it be done in fewer?


6

The short answer is with an example of The first step is to identify the prime factors in the range of 1 to 25. 13, 17, 19, and 23 appear once each. 11 appears twice (11, 22). 7 appears three times (7, 14, 21). 5 appears 6 times, 3 appears 10 times, and 2 appears 22 times. If a prime factor appears only once, it must necessarily be on the main diagonal. ...


1

A new solution using Wikipedia's full list of mathematical shapes: "Drawing" them (thanks again Wikipedia Commons): Process Old solution, depending on what your definition of "shape" is and what synonyms are allowed: Drawing the shapes (ie. copy-and-pasting from Wikimedia Commons):


1

I am assuming that we are looking for an integer solution. With this assumption, here is a top down approach that nobody else has used. The first question that needs to be asked is is that If a solution exists then what will be the last operator that would be used . The last operator cannot be ×3 or ^2 as no integer × 3 = 32 nor (any integer)^2 = 32 . ...


4

As long as we're not doing maths, my basic intuition about the matter is that should give such a line: there is definitely one bisecting plane with that line on it (cut along one edge, which is perpendicular to the other edge). Then, rotating that cutting plane about the axis of the special line seems to move symmetrical amounts of volume from one side to ...


1

Heavily based on Trenin's answer (in hexagons and heptagons). I got the final number to be 73. Trenin used one mirrored hexagon twice (the crossed on my image). Then there is: Triangles: 8 Quadrilaterals: 16 Pentagons: 23 Hexagons: 21 Septagons: 5 Octagons: 0 Sum = 73


-2

At 02:54:34 + 394/719 s and 09:05:25 + 325/719 s they'll be as far apart as possible (provided the clock isn't broken).


2

I think there is a unique answer for the squares: If there is a unique answer for the primes too, then I will say bravo!


-1

The number of solutions for squares are 77520 and the number of solutions for primes are 590080113475142976


1

I'm not sure if this is what you were thinking but here is one possibility


19

First Way Second Way Third Way


4

I used integer linear programming (and, sorry, a computer): The solution is unique up to rotation and reflection.


3

With some dynamic programming we can find all possibilities. Without exponents, we can go to exactly 75: 1: 1 * (2 / (3 + (4 - 5))) 2: (((4 + 5) - 3) / 2) - 1 3: 1 * (((4 + 5) - 3) / 2) 4: (2 + (3 * (5 - 4))) - 1 5: 1 * (2 + (3 * (5 - 4))) 6: 1 + (2 + (3 * (5 - 4))) 7: 1 * (2 + (5 * (4 - 3))) 8: 1 + (2 + (5 * (4 - 3))) 9: 1 + (2 * (3 - (4 - 5))) 10: 1 * (2 * ...


2

I'll use your notation of $b$ for the number of turns of the big wheel, and $s$ for the number of turns of the small wheel. You already deduced that $s=\frac{18}{10}b$. From the picture we can also deduce that: Now you can just substitute and see where it gets us: The total number of rotations of the small wheel is therefore:


5

"a guy" has covered the initial question here is the bonus We can even go a bit further, without much difficulty Here are a few more


8

1 = 1 2 = 2 3 = 3 4 = 4 5 = 5 6 = 5 + 1 7 = 5 + 2 8 = 5 + 3 9 = 5 + 4 10 = 5 + 4 + 1 11 = 5 + 4 + 2 12 = 5 + 4 + 3 13 = 5 + 4 + 3 + 1 14 = 5 + 4 + 3 + 2 15 = 5 + 4 + 3 + 2 + 1 16 = 4 ⋅ (1 + 3) 17 = (5 ⋅ 4) - 3 18 = (5 ⋅ 4) - 2 19 = (5 ⋅ 4) - 1 20 = (5 ⋅ 4) 21 = (5 ⋅ 4) + 1 22 = (5 ⋅ 4) + 2 23 = (5 ⋅ 4) + 3 24 = (...


5

EDIT: All solutions up to mirror symmetry, rotation, colour permutation found by brute force: End of EDIT. One possibility (I write X,O,+ for 0,1,2): A bit on how this was found:


2

It feels like this can be improved but the best I've been able to do so far is As follows


4

Using a modified version of Albert Lang's method on the previous question, the best I've managed so far is As follows


1

I don't know if this satisfies the intended problem but it follows the rules. I approached it by looking at the minimum multiple choice answer, and trying to exclude solutions of that size.


3

Here is a solution with six lines: It's difficult to tell how I found this, except that I already knew a solution for a 3 by 3 grid with 4 lines, which can be found e.g. here on our sister site Mathematics Stack Exchange. It's also possible that a solution with 5 lines exists. (By the way, the puzzle is missing the requirement that the lines must be ...


6

Short way to guess at the answer: Proving that this works: Now you asked about strategy:


7

One possible answer is It doesn't look very symmetric at first glance, but if you look more closely, I think it's likely that there isn't any answer which does not follow this pattern. I'm pretty sure this property holds for all answers, with the following reasoning: Let's classify the cells into four categories. A B C D C D A B B A D C D C B A And let's ...


1

I finally wrote a heuristic solver that finds good solutions. I found many solutions that achieve the longest chain length of Here are some example solutions


0

Since $x$ and $y$ are positive, $x+y$ is always greater than $x$ and $y$ - this means that undoing a move always involves decreasing the greater coordinate by some multiple of the lesser. In summary, the reachable points $(a,b)$ satisfy the following criteria:


19

To solve this, we need to examine more closely Grandpa's words when he says: Two books of International stamps. One book of 75 cent stamps. And I got – 20 cent stamps, 50 cent stamps and also 55 cent stamps. That is it! Knowing that Grandpa spent exactly \$50.00 and bought 13 books, what can we say for sure? If we take his words at face value then we run ...


4

Another proof of Albert's lemma (and one that I believe is much more elegant than the others presented): I will prove a stronger lemma instead. Namely: Proof: (A proof of the reduced statement follows. This proof is similar to Gareth's answer to the same question, but does not rely on an arbitrary choice of "leftmost".)


0

From (6,21) the robot can either go to (27,21) or (6,27). Another turn after that, the robot will be at either (48,21), (27, 48), (33, 27) or (6, 33). Because these numbers have no way of ever decreasing, the robot can go to any point (6a+21b,6c+21d) where a, b, c and d are all positive integers, b and c being larger than 0. Hope this answers your question!


4

Here's (what I think is) a simpler proof of Albert's lemma than the one in loopy_wall's answer. We'll find either a king-path of 0-squares connecting N and S sides, or a king-path of 1-squares connecting W and E sides. The basic idea is to walk along the boundary between 0-squares and 1-squares until we reach an edge of the board. So here's an example board; ...


2

I would like to present an argument that shows that In fact Proof


1

First of all, how do we find the winning probabilities given the agents' numbers of bullets and chambers? This is pretty straightforward if there are only two agents. In this case, whoever first fires an actual bullet will win. Suppose they have probabilities $p,q$ of doing so on a given turn (so these are the ratios bullets/chambers for the two agents) and ...


2

I think the following is probably what you are looking for


3

Proof of Albert's lemma (which solves the bonus question; please note that some repetitive details are omitted to keep down overall length to something reasonable):


1

A "systematic" solution. Bit tedious but no guessing no computer required. Only 4 cases are serious contenders: The remaining cases are easily dismissed: and we recover @hexomino's result


3

My attempt We can verify this is the best we can do mathematically because Edit: I just wrote a short program to check this and the following colourings also achieve the same result.


2

Every pair of zeros and every pair of ones are connected via some King chain is confusing. If you mean that every $1$ can be reached from every other $1$ by the traversal rules (similarly for $0$), then I think 8 is a minimum. If we can have disconnected pairs, gets you to 7. As for a proof, there might be fertile ground in looking at what can stop a path (...


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