New answers tagged

22

The volume filled is:


9

Here is the answer: $2=2$ $3=3$ $5=2+3$ $7=\frac{43-3}{2}-13$ $11=43-2\cdot(13+3)$ $13=13$ $17=43-13\cdot2$ $19=13+3\cdot2$ $23=43-2\cdot(13-3)$ $29=43-13-3+2$ $31=43-13+3-2$ $37=43-3\cdot2$ $41=\frac{43+3\cdot13}{2}$ $43=43$ $47=2\cdot43-3\cdot13$ $53=43+13-3$ $59=43+13+3$ ...


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4

Is it this one? Found at this link:


1

I think it's unlikely that this is really the solution, since you could make up several similar ways of "solving" the puzzle, but maybe it's worth a try.


5

Using 3, 5, 7, 11 2 = 5 - 3 3 = 3 5 = 5 7 = 7 11 = 11 13 = 7 + 3! 17 = 11 + 3! 19 = 3 × 5 + 11 - 7 23 = 11 + 7 + 5 29 = 5!/3 - 11 31 = 11 × 3 + 5 - 7 37 = 7 × 3! - 5 41 = 5 × 3! + 11 43 = 7 × 5 + 11 - 3 47 = 5!/3 + 7 53 = 7!/5! + 11 59 = 11 × 5 + 7 - 3 61 = 11 × 5 + 3! 67 = (3+5) × 7 + 11 71 = 7 × 11 - 3! 73 = 11 × 3! + 7 79 = 11 × 7 + 5 - 3 83 = 11 × 7 ...


2

This was probably not the most exciting question. Anyway there are many solutions. Here are some examples: Interestingly one cannot add a single other bishop of any color.


19

Using 2,3,7,11: $2 = 2$ $3 = 3$ $5 = 11 + 3 - 7 -2$ $7 = 7$ $11 = 11$ $13 = 2 + 11$ $17 = 3! + 11$ $19 = 2^3 + 11$ $23 = 3 \cdot 7 + 2$ $29 = \frac{(7-2)!}{3} - 11$ $31 = 3 \cdot 11- 2$ $37 = (11-3!) \cdot 7 + 2$ $41 = 7^2 +3 - 11$ $43 = 2 \cdot 11 + 3 \cdot 7$ $47 = 3 \cdot 11 + 2 \...


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83 is probably not allowed - I swear I'll find a legitimate solution soon... arrgh... Using 2,3,5,7: $2, 3, 5, 7 = 2, 3, 5, 7$ $11 = 7 + 5 + 2 - 3$ $13 = 7 + 5 + 3 - 2$ $17 = 7 \cdot 2 + 3$ $19 = 7 \cdot 2 + 5$ $23 = 7 \cdot 3 + 2$ $29 = 7 \cdot 5 - 3!$ $31 = 7 \cdot 5 - 3! + 2$ $37 = 7 \cdot 5 + 2$ $41 = 7 \...


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I believe I found a grid that requires 19 steps to solve, but I cannot prove it. The grid is The steps I found are Also I have found another grid, which may require even more steps to solve, but I am not sure how many exactly: It would be great if someone could confirm/verify my answers. Edit: these grids are not that hard after all. See comments below.


5

Since we are talking about multiplication (products grow really fast), and our data gathering is relatively cheap, it stands to reason that the best strategy is to I had this idea yesterday, but it took me this long to get through the final annoying special case. However, I'm pretty convinced now that This is going to be a longish read because of the ...


3

I believe Note that


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Since Hehe knows his own number by watching the others, one can test the different factors that could lead to it (watched hats, with ascending sort -> possible hats Hehe could have): So we only keep some of these entries: And just consider: Hehe knows his number even before the other ones talk. This means: a-priori he knows, with no ambiguity, the number ...


8

The maximum is Reasoning


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This gives quite a simple solution:


2

I took a slightly different approach to Jaap Scherphuis: The two smaller numbers are factors of the largest number, and all 3 numbers are different. This means that the largest number cannot be prime - ruling out 1,2,3,5,7, and leaving us with 4, 6, 8, and 9. There are 32 possible pairs of factors of the numbers 4, 6, 8, and 9, but only 8 of these ...


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18:


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I thought about something here : There is another answer I thought of first, and I wanted to share. And last but not least, the ultimate move :


6

This grid requires 18 steps: Explanation


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Solution: Finding the Solution: This gets us: We now have: This gives us: Now, we have: Here we have: In conclusion:


3

Intuitively, it seems like


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What combinations of numbers are possible? From this list of possibilities, we need to find one where one person can know what their number is by seeing two of the other numbers. Here are the possible numbers a person could see others having (lowest number first), and the possible numbers that the person could be. Now we have 3 possibilities for two ...


6

It is not possible to do this in 12 steps. In fact, a simple argument shows that at least 20 steps are necessary: Tiles 1 and 8 need to be swapped. They are however a distance of 3 moves apart, so each tile needs to be moved at least 3 times. Similarly, tiles 2 and 7 also each need 3 moves to reach their goal locations. So these 4 tiles together need at ...


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Below I explain all the deductions I have been seeing until I reach the solution: Possibilities: Doubts and deductions: Description: Solution and vision of each:


15

This was trickier than it looked. I have a feeling that there should be a quicker way to find the answer than the list of cases I worked through. Note: I interpret "two of which are factors of the third" to mean that two of the numbers divide the third, and that either of those two factors might be 1.


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Assuming that the walls replace a cell of a grid like in the 9x9 version and are not located between two cells. How I got there:


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Looks like we have mazes we need to solve simultaneously: and These are the difficult cases, so all the other configurations (that are solvable in the first place) should be automatically solved if we can get all of these. So we need a move sequence that has Here's one: Which has Is it optimal? No idea. I'm going to need some more coffee if I'm to ...


13

I have a Here are the details: And the final UUL makes sure every path has touched the T.


5

Taking the basics of Jaap Scherphuis solution, I made some changes. "Once the robot reaches the target, the game will terminate immediately." If I understood this rule correctly, not all robots have to end up on the target square after the same number of moves. Once a robot reaches the target, he has completed the game and we don't have to care about him ...


24

My first attempt was done by hand. It used 28 commands: but this was not optimal. I have now done a computer search to find optimal solutions. It found 180 solutions of length No shorter solutions exist. I'll illustrate the following solution: To see how it works, start with a robot on every available square, and send the commands. The robots should all ...


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Step 1 Step 2 Step 3 And the mutual sum is $86$, as required.


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(And for completeness, the answer)


1

Here are some extra lines that should be enough to calculate the answer with very little math. The area of one of the shaded little triangles is and since there are two of them in the overlapping part, the total overlapping area is


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11

Answer Attempt 1 At first I thought This can be achieved as follows, using the same technique as Daniel Mathias previously: Adjusting this a little, we can get After a lot more fiddling around with this, I realised it seems that Attempt 2 I then tried Attempt 3 Finally, the third option is The mistake was


12

Here's my go (click to enlarge): The pattern of reds is and the white and black patterns are so it's enough to count half the pieces only. :-)


2

Good to have a different way of answering. My logic is little bit easy: 8 $\hspace{2cm}$9$\hspace{2cm}$10$\hspace{2cm}$11$\hspace{2cm}$12 $\hspace{1cm}$0 $\hspace{2cm}$24$\hspace{2cm}$120$\hspace{2cm}$336$\hspace{2cm}$___? $\hspace{0.4cm}$$1^3-1$$\hspace{1cm}$ $3^3-3$ $\hspace{1.2cm}$ $5^3-5$ $\hspace{1cm}$ $7^3-7$ $\hspace{1.2cm}$ $9^3-9$ Next term ...


1

Somebody has already solved it, but here is the detail explanation again: 8 9 10 11 12 0 24 120 336 24 * 0 24 * 1 24 * 5 24 * 14 0 1 5 14 1 4 9 1^2 ...


5

I have to admit that I looked at hint 3, without which I would perhaps not be able to solve it in reasonable time. With hint 3 it's somehow an easy guess that Starting with the 5 euros and 10 euros, I can bound the value of the smallest note so without hesitation I immediately guess the rest are just try and try (reasonably fast, in several minutes). ...


4

The answer is: Reason:


9

Can this be improved?


11

This is a type of puzzle called a "cryptarithm" or an "alphametic". We have many examples of this type of puzzle under the alphametic tag.


3

This should be optimal, but haven’t proved it...


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