bonanova

And Children's Activities had some cool features on the last page - cartoon, riddle or puzzle - as I recall.

Yes, there can be a "bonus" row that contains 4 trees. Here's an adequate proof of the killer:

@CaptainEd - OMG no. Awhile ago I next-to-worshiped Martin Gardner (who wrote the math games column in Sci American for so many years) because he worded his puzzles perfectly, simply and clearly. His, unlike mine, (try tho I may) never needed editing. When I wrap prose around mine to make them perhaps interesting or, sometimes, to camouflage the solution, stuff gets added that has often has to be clarified later. My bad on this one.

Clarification: Dick asserts that he had been out running, and that one of his three brothers has just lied. Inspector just called in and needs a final answer ... Fame awaits the brave.

Sorry guys, "fuller" should have read "at least as full." Examples always help, so here is an example. a b c 7 9 12 <- 14 2 12 -----> 2 2 24 -> 0 4 24 So { 7 9 12 } is a starting point where a plate can be emptied. Can any { a < b < c } lead to an empty plate? A Yes answer needs proof; a No answer just needs a counter example.

On a table are three plates, containing a, b and c jelly beans, in some order, where a < b < c. At any time you may double the number of jelly beans on a plate, by transferring beans to it from a fuller (or equally populated) plate. After one such move, for example, the plates could have 2a, b-a, and c beans. Using a series of these moves, Is it possible to remove all the jelly beans from one of the plates?

I find probability questions interesting, because they often defy intuition. Particularly for me are those that involve waiting times. Other than the basic idea of an event of probability p needing on average 1/p trials to occur. But here's one not that trivial, yet still fairly easy to solve -- with the right approach. On average, how many times do you need to flip a fair coin before you have seen a (continuous) run of an odd number of heads followed by a tail? For example, T T H H H H T H H H T took 11 flips.

There are six dice. They are marked A. B, C, D, E and F in some order. Initially no dice are on the table. Peter's game: You pay $1 and roll die A or die B or die C or die D or die E or die F. You get to choose which die to roll. Paul's game: You pay $1 and roll die A and die B and die C and die D and die E and die F. You must roll all of them. Winning condition: All dice are on the table and collectively they show 1, 2, 3, 4, 5 and 6 dots.

To clarify: (and I'll add it to the OP) Dick: I went out and ran 3 miles in the woods, and I've figured out that one of my 3 (living) brothers is lying.

You're at a carnival and two people offer you money for getting six fair dice collectively to show all six numbers. Each charges you $1 per roll. Peter pays you $20 when you succeed, while Paul will pay you $50. There's another difference. Paul lets you roll all six dice each time, but Peter makes you roll just one die each time. Do you stop and play? If so, with whom?

A discrete event (like rolling a fair die and wanting a 3 to appear) has a probability p of success (1/6 in this case.) The first roll is likely to fail, so let's keep rolling the die until we do get a 3, Then stop and write down the number of rolls that it took. Let's repeat the experiment a large number of times, each time recording the required number of rolls. So we have a bunch of 1s (the number of times 3 appeared on the first roll,) 2s (the number of times a 3 appeared on the second roll,) and so forth. What number will most appear most often?

@Molly Mae Bravo. I should give you a solve (and a gold star) for this answer. But the Inspector has a reputation to uphold -- he needs a conviction -- and he did appeal to us for help. So, let me repair my flawed puzzle by adding this phrase about the brothers (and I'll add it to the OP as well.) None of them lied and told the truth in a single day.

OK, I agree with 22. Nice work. Where my thinking was wrong - I considered left- and right-hand knight moves to be in the same class. (I put all mirror images into the same class.) That's wrong, because mirror image solids (if they lack further symmetry) can in fact be distinct. Three small cubes? Maybe look at it at some point. rodomac's images provide an advanced start point. That said, it still means finding distinct ways to remove C, E, F and (possibly) B small blocks from 22 different images.

Here's my argument for 20 distinct shapes. First, I'm in awe of rocdocmac's images! The meager sketch below indicates only the visible small cubes. It does not show the small cube at the center, which I refer to below as B. The ones that do show are labeled as Corner (C) Edge (E) Face (F) We agree that C, E, F and B are the four distinct classes of small cubes, so that removing just one small cube gives rise to four distinct shapes. C / \ E E / \ C F C | \ / | | E E | | \ / | E C E | | | | F | F | | | | C E C \ | / E | E \ | / C And now we're removing two small cubes and identifying the equivalence classes. If we first remove a C, we can remove another C three distinct ways we can remove a E three distinct ways we can remove a F two distinct ways we can remove a B one distinct way If we first remove a E, having already counted the EC case, we can remove another E four distinct ways we can remove a F three distinct ways we can remove a B one distinct way If we first remove a F, having already counted the FE and FC cases, we can remove another F two distinct ways we can remove a B one distinct way There is no BB case, so we're done. And the total is 20. In summary, CC 3 CE 3 EE 4 CF 2 EF 3 FF 2 CB 1 EB 1 FB 1 BB 0 The cases that seem to disagree both involve Edge-cube cases. Namely, Corner-Edge (CE). Some say 4, I say 3. Edge-Edge (EE). Some say 5, I say 4. Here are my enumerations: CE - having removed a Corner, what classes of Edge faces remain? (I claim three.) Three small cubes E that touch C. Six small cubes E that do not touch C, but do lie on the same big-cube face. (Think of a chess knight move.) Three small cubes E that touch C'. C' is the small cube diagonally opposite C. EE - having removed a E, what classes of other Es remain? (I claim four.) Four small cubes E that touch the first E at one of its corners. Four small cubes E that touch E' at one of its corners. E' is the small cube diagonally opposite the first E. Two small cubes E each of which, along with the first E, surround and touch a common F. The (one) final small cube, E'. Again, E-E' passes through B. My class descriptions exhaust the 12 (CE) and other-11 (EE) Edge small cubes are stated in a way intended to suggest (at least) that the classes are homogeneous. I'm eager to hear other class descriptions for these cases.

In my post I meant to assert the Inspector tried but failed to make the sketch following the method suggested by rocdocmac. That is, the antecedent of "He" was meant to be "the Inspector." Bert, of course, had already succeded.

Two pennies can be placed on a table in such a way that every penny on the table touches (tangos with) exactly one other penny. Three pennies can be placed on a table in such a way that every penny on the table touches exactly two other pennies. What is the smallest number of pennies that can be placed on a table in such a way that every penny on the table touches exactly three other pennies? (All pennies lie flat on the table and tango with each other only at their edges.)

Regarding (2), can you make a sketch for the Inspector? "Absolute truth-teller" includes "without mistakes," and he did say random. Remember also that ...

The Threedie brothers, Al, Bert, Chuck, Dick and Eddie, lived in a cabin 3 miles up the old mountain trail, and it was known they didn't get along all that well. This morning, Eddie was found dead behind the cabin, and his brothers, the only suspects in the case, were being questioned by Inspector Sherlock. It was known that, of the four, at least 3 were absolute truth-tellers, and none of them ever lied and told the truth in a single day. All four, of course, denied murdering their brother. The Inspector started by asking each brother what he had done that morning: Al: I was analyzing random groups of 3 numbers, and I found that if the numbers sum to zero then their product is the average of their cubes. Bert: I was analyzing random polygons with 3 sides, and I found that if I trisected all their angles I could make an equilateral triangle. Chuck: I planted a dozen apple trees out in the orchard, and I found a way to make eighteen rows of 3 trees, each row being dead-on straight. Dick: I went out and ran 3 miles in the woods, and I've figured out that one of my 3 (living) brothers is lying. The Inspector called in these clues to one of his friends at BrainDen, and in 3 shakes of a lamb's tail the case was solved. The sound you hear is your phone ringing. It's your chance to be famous!

@aiemdao You're exactly right. OP should have said they made their statements simultaneously, or that they wrote their statements on slips of paper without reading the other two statements. I'm making that change in the OP now. You get Honorable Mention recognition for your answer. Thanks, and Happy New Year!

@aiemdao Hi, and welcome to the den. This might help your thinking: