# Tag Info

29

The given sum is what we would get if we were to compute the definite But we can also compute this directly: Now by amazing coincidence $9.3$ squared happens to evaluate to which is good enough to pass for $86.5$. One more lucky coincidence: $9.3$ is easily multiplied by $\frac 23$. So we need to do $6.2$ times $86.5$ which yields The $.3$ cancels ...

17

It is not possible to move all 16 bars. Consider the total distance that each person moves, in each direction, carrying each possible number of bars. For example $R_2^+$ is the total distance that Rod moves forward while carrying 2 bars. We can set up a system of equations. Both Rod and Lia must move a net distance of $1$ mile forwards:  R_0^+ - R_0^- + ...

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Yet another one, with all 16 words different: (I admit that this is a computer-assisted answer, in the sense that I wrote a program to extract the list of 4-letter heteropalindromes from YAWL. I constructed the square manually from that list.) For side information: there are only 16 unique such squares containing 16 unique words (not counting rotations and ...

10

You can turn Life into Shit in 6 steps: or or This is the optimal solution. Why 6 steps? My thinking solution:

9

This approach brings 14 bars and leaves them 20 minutes before the train arrives. This lets them bring a 15th bar and catch the train with less than 5 minutes to spare.

8

From OEIS: A000448 For a circle centered at the origin, this gives these lattice points: For a total of

7

I'll take a shot at it Arrived at as follows First Then So But So

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5

Since black has many checks available, we can This limits our options to four possible moves. Nf7+ seems particularly promising, so Black has only one move that doesn't immediately end in a smothered checkmate at Nf6, so we check the checks (heh) after that move to find after which we can somewhat incredibly finish with either or

3

Here is the finished tiling: To get started: Next step: Looking at the left pear: With those chokepoints: For the next step I took a guess: From there, I just looked at the pieces I had left and found something that worked.

3

A little experimentation with Euclids formula, demonstrates that there are only primitive Pythagorean triples with largest element less than or equal to 80. If, from these, we choose the sets and note that 65=5*13, then we can produce the triples Using them, plus their reflections about the lines y=x, y=-x, y=0, and x=0, plus the four extra points sitting ...

2

An alternative method to Kate's to get the same amount of bars.

1

Sure, a 1×4√5 rectangle and a 1×√5 rectangle.

1