# Tag Info

3

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 ...

4

The setup and mate in four immediately reminded me of Since Black is threatening White's first move must be Black now has The problem is really about First: Second: Third: Final: So this is related to what Tim Krabbé calls a

18

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 \... 8 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, 711 = 7 + 5 + 2 - 313 = 7 + 5 + 3 - 217 = 7 \cdot 2 + 319 = 7 \cdot 2 + 523 = 7 \cdot 3 + 229 = 7 \cdot 5 - 3!31 = 7 \cdot 5 - 3! + 237 = 7 \cdot 5 + 241 = 7 \...

10

The letters can be replaced as follows: And the remaining letters spell (Arnie's owner gave me some difficulty, because

4

You can get a bit more excitement from the letters if you assign each number to a range, albeit an irregular one. So 1=A, 2=B, 3=C,D,E, 4=F,G, 5=H, 6=I,J, 7=K, 8=L, 9=M,N,O,P,Q, and 0=R,S,T,U,V,W,X,Y,Z. So for example PUZZLING STACK EXCHANGE becomes 90008694 00137 30351943. To decode, reverse the process. With some of the letters fixed, guessing the ...

5

16 moves is the minimum possible. (I checked by computer, but solved it first without one.) Explanation:

4

A brute force solution: We start with the two sixes with six empty places in between. We must fill those six places with exactly one of each number 1 to 5, as well as 7, while following the rules of spacing. Now, filling in the 3: The 4: When placing the 5, we can easily reject any solution that would leave more than 1 empty place in each suit, since we ...

10

This works, although I'm not sure if it's unique: How I found it We have two 6s (red and black) with six cards between them that must be all the same colour (let's say red). Key fact: Now, to begin with, After that, Also,

5

I'm going to say Grandpa was: because

1

3

I think it's which starts it. Then black can do one of the following First Second and Finally The title refers to

2

2

2

If any corner is white, there aren't enough white cells to avoid 2x2 blues in at least one other corner. Likewise for the cells adjacent to the corners. The last cell in each 2x2 corner block must then be white. So we have BB-BB BwwwB Bw-wB BB-BB This uses up all 5 white cells, so the remaining "." cells must be blue. Solution:

6

My solution:

3

At the risk of flogging a dead horse, I wanted to solve it using Dijkstra's algorithm, since an answer implied that it required a computer to do so. First, we change it to a minimize problem. We know that all solutions use 19 steps, so we simply change the values to $10-x$ and then minimize the result: Initialize the top left to a value of 1. Everywhere ...

6

@Mohit Jain's solution is nice and provably correct. It also shows you the best way to reach any square from the top left corner. I happened to solve the puzzle in a way that's pretty much identical, but in reverse. Both methods are equally valid, but somehow I felt that finding the optimum path from any square to the bottom right might be a more ...

17

It is a 10x10 map and you need to go east 9 times and south 9 times. There are total ${18\choose9}$ = 48620 paths which are impractical to be done with pen and paper in a reasonable amount of time. But if you observe This can be done pretty quickly with just 10x10=100 calculations. The answer is Proof of work Method

1

Local maximum solution:

3

I started by considering the corners.

4

Start by considering the leftmost three cells.

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