bonanova

The OP does not say how many apples there are. It says the proportion that are poisonous. Question: was that the intent?

You've probably seen this puzzle. There are 15 holes in triangular array. (See sketch below.) The game begins with pegs in 14 of the holes. The play is to jump pegs over adjacent pegs, removing the "jumped" pegs afterward, as in checkers. The jump is made in a straight line. To make a jump, you need a contiguous group consisting of { peg1, peg2, hole } in a straight line. Peg1 ends up in the hole, and peg2 is removed. The object is to make 13 legal jumps and end up with a single peg. This happens about 6% of the time. That is, about 94% of the time you get a configuration, with more than one peg remaining, that permits no further legal jumps. In some games the peg must end up in the original empty hole, and that happens only about 3% of the time. So, it's not a trivial puzzle. This puzzle asks for something different, and easier: Lose as badly as possible. That is, select a location for the empty hole, and then find a sequence of moves that leaves the greatest number of pegs on the board where there are no more legal jumps. It's simple enough to play, even without the game, by marking hole locations on a sheet of paper and using pennies. As already stated, there are 15 holes. There are also 36 possible jumps. For convenience in writing sequences of jumps, they can be numbered, as follows: Number the jumps like this: and the holes ---------------------------> o ------------- 1 So Jump #1 means the / \ like this: peg in hole #1 jumps 1 2 ----------> 2 3 over the peg in hole #2 into the empty 4 5 6 hole #4. o o / \ / \ 7 8 9 10 Jump #18 is peg 7 3 4 5 6 over peg 8, into 7 13 11 12 13 14 15 hole 9. / \ o-8 o 14-o Holes 4, 6, 13 / \ / \ / \ begin 4 jumps; 9 10 11 12 15 16 the others 17 19 21 23 begin two. / / \ \ o-18 o-20 22-o 24-o There are 36 jumps. 25 27 29 30 33 35 / / \ / \ \ o-26 o-28 31-o-32 34-o 36-o With symmetries taken into account, the holes have four equivalence classes: Corners (1, 11, 15) Adjacent to corners (2, 3, 7, 10, 12, 14) Edge centers (4, 6, 13) Centers (5, 8, 9) This means that there are just four distinct places for the empty hole to start a game: { 1 2 4 5 }. All other holes are symmetrically equivalent to one of these. Just to be sure the numbering above is understood, here is a winning game of the normal type. Start with pegs in every hole except #1. (The top hole is empty.) Then make these jumps: { 7 14 2 17 23 27 34 26 30 6 35 14 7 }. If done correctly, the original hole #1 contains the final peg. Enjoy.

Yes. .

Doctoring the figure a bit (while I think about solving it.) o - o - o - o - o | / o o o o | / o - o - o Question: You mention the number of nails (12) and also the number of nails in the perimeter (here that's 10.) So for this example, would we be given N=12 or N=10? I'm thinking the latter, but would like to confirm.

Welcome back TSLF. Nice puzzle!

Looks like

With that interpretation,

Assuming the above, we note that

Just to be clear, n = 3 4 5 7 8 9 11 12 13 ..., and CW and CCW alternate untethered (my new favorite word) to parity?

Nice solve. And the "few years back" is actually two years ago, when Feb 28 was Sunday and March 2 was Wednesday.

The professor writes a problem on the whiteboard, thus: 25 - 55 + (85 + 65) = ? He then inexplicably states that, even though you might disagree, the correct answer is actually 5! Explanation?

A few years back while visiting friends, we celebrated the birthdays of their identical twin daughters Joan and Jane, born just 5 minutes apart. Joan had her party on Sunday, and Jane had hers on Wednesday. Explanation?

This was my take.

Nice. One thing I liked about this puzzle is that it's open to clear thinking. Even tho at first it seems too complex.

But actually ...

It certainly should. It clearly fails a units check. Good catch.

Do any of the clues imply that the indicated needles are adjacent? I would take "on the left side of" to imply that "pain" is adjacent to "seek," while "to the left of" simply means "the other pain" is not to the right of "interesting little creation." But I'd like to be certain of that. Also, is "interesting little creation" one of the needles? If so, you have described nine, not eight, needles. Thanks.

You have 10 sets of ten coins. One set of the ten is counterfeit, the others are genuine. The genuine coins weigh exactly 0.10 ounces. The counterfeit coins are exactly a 0.01 ounces off, making the entire set of ten coins 0.10 ounces off. You may use an extremely accurate digital scale only once. How do you determine which set is counterfeit?

Clue.

@ThunderCloud you're homing in on it, but now you're a little high. @BMAD It's certainly true that if the FIRST child was a boy born on a tuesday, then it's just the prob that the second child is a boy. But ... the OP does not tell you that. That is, "one is a boy" does not imply "my oldest child is a boy." So your "second" child simply means the "other" child.

Not that I know of. Tables of powers, or spreadsheet where different abc values can be simply typed in, or inspired guesswork? It's not my fav type of puzzle, but some like this type.

@ThunderCloudThat's close, but a bit low.